text:
Abstract

Locally resonant materials allow for wave propagation control in the subwavelength regime. Even though these materials do not need periodicity, they are usually designed as periodic systems since this allows for the application of the Bloch theorem and analysis of the entire system based on a single unit cell. However, geometries that are invariant to translation result in equations of motion with periodic coefficients only if we assume plane wave propagation. When wave fronts are cylindrical or spherical, a system realized through tessellation of a unit cell does not result in periodic coefficients and the Bloch theorem cannot be applied. Therefore, most studies of periodic locally resonant systems are limited to plane wave propagation. In this article, we address this limitation by introducing a locally resonant effective phononic crystal composed of a radially varying matrix with attached torsional resonators. This material is not geometrically periodic but exhibits effective periodicity, i.e., its equations of motion are invariant to radial translations, allowing the Bloch theorem to be applied to radially propagating torsional waves. We show that this material can be analyzed under the already developed framework for metamaterials. To show the importance of using an effectively periodic system, we compare its behavior to a system that is not effectively periodic but has geometric periodicity. We show considerable differences in transmission as well as in the negative effective properties of these two systems. Locally resonant effective phononic crystals open possibilities for subwavelength elastic wave control in the near field of sources.

1 Introduction

Engineered periodic materials, usually referred to as phononic crystals (PCs), allow for unprecedented ways of wave propagation control [1,2]. PCs can be designed to obtain attenuation in selective frequency ranges [3,4] or even to actively tune these ranges to adjust to external stimuli on demand [57]. PCs can achieve negative refraction [8], break the diffraction limit [9], and even provide robust topological wave guiding [1013].

Since the functionality of PCs relies on the phenomenon of Bragg scattering, the wavelengths at which PCs work are closely related to the length scale of the system [14]. This becomes a limitation when trying to control waves at low frequencies because it results in impractically large systems or the need for extremely soft materials. The introduction of locally resonant materials, typically referred to as acoustic metamaterials, has provided a means of controlling waves in the subwavelength regime [15]. This not only allowed for low-frequency wave propagation control using relatively small features but also introduced other interesting phenomena such as negative effective properties [15] and enhanced damping [16].

Locally resonant materials do not need periodicity and even a single resonator can display negative effective properties [17] or provide vibration suppression [18]. However, involving periodicity in these systems allows for the application of the Bloch theorem and calculation of the band structure [14]. In this way, the behavior of the infinite system can be understood through the analysis of a single unit cell. This becomes particularly useful when dealing with complex geometries, multiple resonators, and two-dimensional (2D) and three-dimensional structures and has motivated the designs of locally resonant metamaterials to be periodic. In fact, many of the novel behaviors of locally resonant elastic metamaterials have been achieved using periodic systems. For example, negative refraction and super resolution have been achieved for square [19] and hexagonal [20,21] periodic resonator arrays, and the phenomenon of metadamping [16] was first shown for one-dimensional (1D) periodic systems.

All these studies use the Bloch theorem for analysis, and this theorem requires periodic coefficients in the equations of motion. This is why the study of periodic locally resonant materials in all the aforementioned references is limited to periodic systems with lattice vectors in a Cartesian coordinate system (or alternatively, in the far field of sources); it is for these lattice vectors that the equations of motion result in periodic coefficients when the system is geometrically periodic (i.e., geometry is invariant to translations along lattice vectors). On the contrary, when the material is invariant to translation along lattice vectors in cylindrical or spherical coordinates (or alternatively, in the near field of sources), the equations of motion that describe their behavior lack periodic coefficients and Bloch analysis cannot be applied [2224]. In fact, most studies that deal with PCs for cylindrical waves rely on analysis of the finite structures [2527] or far-field approximations [28,29].

When it comes to near-field vibrations, where wave fronts cannot be approximated as plane, we must appeal to other solutions to be able to apply Bloch analysis and thus utilize typical well-studied properties of local resonant systems mentioned in references herein. In this area, radial wave crystals have shown that through anisotropic radially varying material properties, the acoustic and electromagnetic equations of motion can be forced to have periodic coefficients [23], resulting in high Q-factor Fabri-Perot-like resonances. In elastic waves, the authors’ recent work proposed a combination of periodic plus isotropic radially varying material properties that cause the equations of motion for radially propagating torsional waves to be equivalent to those that describe wave propagation in a Cartesian phononic material [22]. In this way, we extrapolated the typical Bloch analysis in Cartesian systems to realize band gaps and topological interface modes in cylindrical coordinates [22]. These materials are termed effective phononic crystals (EPCs), since, even though they are not geometrically periodic, their equations of motion effectively describe a PC. However, both radial wave crystals and EPCs are based on the Bragg scattering mechanism (with the limitations this entails) and have not included nor analyzed the effects of local resonances in the near field of sources.

In this article, we extend the concept of EPCs [22] to include local resonances in what we term a locally resonant effective phononic crystal (LREPC). We present a method that enables properties of Cartesian periodic locally resonant systems to be realized with cylindrical wave fronts that are present in the near field of point sources in 2D. We show that by prescribing radially varying material properties in the host medium and torsional local resonances, we can reduce the equations of motion of radial torsional waves to those of axial waves in strings [30] or bars [31] with local resonances, allowing the calculation of band structures based on methods already developed in these references. We focus here on radial torsional waves, which are relevant in rotating machinery such as turbines, compressors, and engines [32]; however, extrapolation to other polarizations can be done using an equivalent analytical treatment [22]. In fact, even though torsional vibrations are problematic in many mechanical systems, the work on locally resonant materials for torsional waves is limited to only a few studies [33,34], both of which deal with plane wave propagation in Cartesian directions (i.e., torsional waves propagating along a Cartesian axis).

To investigate LREPCs, we first construct their equations of motion by describing them as radially dependent heterogeneous material with distributed moments that account for the torsional resonators. We then use the finite element method to solve the 1D equation of motion of torsional waves propagating through a finite LREPC and calculate its transmission. We show that the behavior of the LREPC is independent of unit cell truncation and can be predicted by the analytical band structure analysis. To show the advantages of using LREPCs, we compare transmission of the finite LREPCs with that of systems that are geometrically periodic but not effectively periodic. We refer to the latter as the homogenous system (HS) due to its independence of material properties on radius. We present three different comparisons. For the first two, we compare using the nondimensional parameters for stiffness, moment of inertia, and frequency that are typically used for the parametric analysis of locally resonant systems [30]. We first compare transmission of the LREPC with an HS that has equivalent stiffness and moment of inertia nondimensional parameters. However, in this case, the resonators of the HS have different resonant frequencies. Thus, we construct a second comparison between an LREPC and an HS that has equal stiffness and frequency nondimensional parameters. In both of these comparisons, imposing the respective nondimensional parameters to be equal forces each of the resonators of the HS to have different stiffness and moment of inertia. Thus, we make a third comparison with an HS with resonators that are all equal and the same as those of the LREPC. Since the response of locally resonant material is strongly affected by its overall stiffness, we set the static torsional stiffness of the finite LREPC and HS to be the same in this comparison. Finally, because many of the novel wave phenomena of locally resonant systems arise from their negative effective properties [35], we calculate the ranges of negative effective dynamic moment of inertia for the LREPC and all three cases of HS. By comparing transmission and effective properties of the LREPC and the HS, we aim to show the possible advantages of using an LREPC when dealing with radial torsional waves.

2 LREPC Derivation

To derive and analyze the LREPCs, we first derive the equations of motion for a cylindrical torsional wave propagating through a heterogeneous material with varying thickness and a distributed external moment, the latter of which will ultimately capture the effects of local resonances. Assuming small tangential displacements independent of out-of-plane thickness and axisymmetric wave propagation, the global equilibrium of an infinitesimal cylindrical element (Fig. 1(a)) results in,
$I∂2θ(r,t)∂t2=∂T(r,t)∂rdr+M(r,t)dr$
(1)
where θ is the angular displacement, I is the polar mass moment of inertia of the ring around its center, T is the torque on the inner ring of the infinitesimal element, and M(r, t) is a distributed external moment (moment per unit radius). The torque can be expressed as follows:
$T(r,t)=2πrh(r)rσrθ(r,t)$
(2)
where h(r) is the out-of-plane thickness and σ is the shear stress. Assuming a linear isotropic material, the stresses can be written in terms of the tangential displacements as follows:
$T(r,t)=2πr2h(r)μ(r)(∂uθ(r,t)∂r−uθ(r,t)r)=2πr3h(r)μ(r)∂θ(r,t)∂r$
(3)
where μ(r) is the material shear modulus and $uθ=rθ$ is the tangential displacement. Given that the polar mass moment of inertia of the infinitesimal ring is
$I=ρ(r)(2πrdr)h(r)r2$
(4)
we can combine Eqs. (1), (3), and (4) to obtain the equations of motion for radially propagating torsional waves,
$2πr3ρ(r)h(r)∂2θ(r,t)∂t2=∂∂r(2πr3μ(r)h(r)∂θ(r,t)∂r)+M(r,t)$
(5)
We add an infinite set of resonators of equal torsional stiffness, βr, and equal polar mass moment of inertia, Ir, periodically spaced a distance Δr from each other (Fig. 1(b)) to Eq. (5), by considering them as moments distributed along an infinitesimal ring and adding an extra degree-of-freedom per resonator. Thus, the set of equations that describes this system is
${r3B(r)∂2θ(r,t)∂t2=∂∂r(r3A(r)∂θ∂r)+∑iβr(θ(ri,t)−θi(t))δ(r−ri)Ird2θi(t)dt2=βr(θi(t)−θ(ri,t))$
(6)
where B(r) = 2πρ(r)h(r) is the matrix density function, A(r) = 2πμ(r)h(r) is the matrix stiffness function, δ(r) is the Kronecker-Delta function, θi is the angular displacement of resonator i, ri = r0 + iΔr is the location of resonator i, and r0 is the internal radius (Fig. 1(d)). Note that μ, ρ, and h are the shear modulus, density, and out-of-plane thickness of the material to which the resonators are attached (light gray in Fig. 1), and they are assumed to be a function of the radius. For the remainder of this article, we refer to the material to which resonators are attached as the matrix.
Fig. 1
Fig. 1
Close modal
Inspection of Eq. (6) shows that periodic material properties and/or thicknesses (i.e., periodic A(r) and B(r)) do not yield equations with periodic coefficients. Neither will we have periodic coefficients if the material properties and thickness of the matrix are constant. This is different from the case of plane waves in for example strings [30], bars [31], beams [36], shafts [34], or plates [37] with local resonances. Generally, the Bloch theorem in our case cannot be applied, and the analysis used in these references becomes invalid unless we find a way to force periodic coefficients in the equations of motion. Thus, we define A(r) and B(r) not to be periodic or constant, but to be such that the equations of motion are periodic. We can do this by setting,
${A(r)=A0r3B(r)=B0r3$
(7)
where A0 and B0 are constants that scale the stiffness and density, respectively, of the underlying media in the LREPC. Essentially, to physically realize the LREPC, the products of the matrix modulus and the matrix density with its thickness must be inversely proportional to r3. For example, one way of physical realization is to have a matrix made of a homogenous material and vary its thickness with r−3. The equations of motion for this selection of A(r) and B(r) reduce to
${B0∂2θ(r,t)∂t2=A0∂2θ∂r2+∑iβr(θ(ri,t)−θi(t))δ(r−ri)Ird2θi(t)dt2=βr(θi(t)−θ(ri,t))$
(8)

These equations are now invariant to translations of the form rr + iΔr, and thus, we can apply the Bloch theorem. In fact, these equations are equivalent to those of the systems in Refs. [30,31,34,36], and thus, analysis therein is valid here as well without parametric limitations. Note that the system we have developed does not have geometric periodicity (i.e., we cannot translate a unit cell along its radial lattice vector to recover the full system), but its equations of motion have periodic coefficients, and thus, its behavior will mimic that of a Cartesian geometrically periodic system. This system is effectively periodic and contains local resonances, and thus, we refer to it as a LREPC.

The analysis of LREPCs is straightforward, since their equations of motion are equivalent to the well-known one-dimensional Cartesian system with periodic local resonances. The dispersion relation for this system can be obtained by applying the Bloch theorem using an LREPC unit cell (Fig. 1(c)) [30]:
$cos(kΔr)=cos(κΔr)−P2sin(κΔr)$
(9)
where k is the Bloch wave number, $κ=ωB0A0$, and $P=ω2IrβrA0κ(βr−ω2Ir)$. Here, the definition of the unit cell is strictly mathematical and in terms of the equations of motion. We define the unit cell of the LREPC as the geometry that when tessellated in space by its lattice vectors (in this case, $Δre^r$; see Fig. 1(b)), keeps its equations of motion invariant. This is different from the typical geometrical definition of unit cell, where tessellation of the unit cell results in the entire geometry. Following the normalization given in Ref. [30], we define the following set of nondimensional parameters for the LREPC,
$I¯LREPC=IrB0Δr$
(10)
$β¯LREPC=βrΔrA0$
(11)
$ω¯LREPC=ωrω0$
(12)
where $ω0=πΔrA0B0$ is the Bragg frequency associated with the unit cell and the overbar indicates a nondimensional parameter. Note that the quasi-static torsional stiffness of the matrix material of a unit cell of the LREPC is $βmLREPC=A0Δr$ and its moment of inertia $ImLREPC=B0Δr$ (see Appendices  A and  B for the calculation of stiffness and inertia of the matrix). Thus, we can interpret Eqs. (10) and (11) as inertia and stiffness ratios of resonators to the matrix material, respectively. The dispersion relation can then be written in terms of these nondimensional quantities as follows:
$cos(kΔr)=cos(ωω0π)−Q2sin(ωω0π)$
(13)
where $Q=I¯LREPCπω¯LREPC2(ω/ω0)ω¯LREPC2−(ω/ω0)2$. Note that $β¯LREPC$ does not appear in this expression for the dispersion relation because only two of the three nondimensional parameters in Eqs. (10)(12) are independent. For example, algebraic manipulation results in $ω¯LREPC=1πβ¯LREPCI¯LREPC$, and we could rewrite Eq. (13) in terms of $β¯LREPC$ instead of $I¯LREPC$. We introduce all three of these parameters because they will later become useful when comparing the LREPC with a noneffectively periodic system.

3 Comparisons Between LREPCs and Homogenous Systems

To understand the advantages of using an LREPC, particularly when it comes to near-field vibrations, we compare it to a material whose matrix consists of constants μ, ρ, and h (Eq. (6)):
${AHS(r)=A0HSBHS(r)=B0HS$
(14)
where $A0HS$ and $B0HS$ are stiffness and density constants, respectively. We refer to this system as the HS. The HS has material properties that are independent of radius, and thus, we can define a unit cell in the HS that results in the entire geometry when tessellated along the radial lattice vector, $Δre^r$ (Fig. 1(c) with constant h and matrix material properties). In this way, we can say the HS is a geometrically periodic system. However, this system does not have periodic coefficients in the equations of motion (Eq. (6)) and cannot be analyzed using the Bloch theorem, and therefore, it is not effectively periodic. Thus, it is not possible to obtain the behavior of the HS based on a unit cell analysis and the response of the finite HS will depend on how we decide to truncate the system [22]. To compare the LREPC to the HS, we must do so in terms of the response of their finite versions. Therefore, we compare the LREPC and the HS by calculating the transmission of a four-unit cell system (Fig. 1(d)) for different ratios of internal radius (r0) to periodicity constant (Δr). We choose four-unit cells because (1) typical periodic systems approximate the infinite behavior predicted from the band structure and we expect a four-unit cell LREPC behavior to have enough units to approximate the band structure prediction and (2) when using a limited number of unit cells, most of the system remains in the near field, where effective periodicity becomes most important. We use a 1D finite element code programmed in matlab to solve for the frequency response of the finite systems (Eq. (6)) assuming material properties of Eqs. (7) and (14), for the LREPC and HS, respectively. We calculate transmission as the ratio of input to output angular displacements.

In locally resonant systems, the behavior is strongly influenced by how stiff and dense the matrix material is compared to the stiffness and moment of inertia of the resonators (Eqs. (10)(13)). Thus, a fair comparison between LREPCs and HS is one where the relationship between stiffness and moment of inertia of the matrix materials and those of the resonators of the two systems are somehow comparable. We can do this by using two different approaches: (1) define a set of nondimensional parameters in Eqs. (10)(12) for the HS and compare the LREPC to the HS with equal nondimensional parameters and (2) compare the LREPC and HS with equal torsional static stiffness and equal resonators. For all cases, we calculate the transmission for different internal radii to periodicity constant ratios, r0r (Fig. 1(d)). The purpose of these case studies is to show that the LREPC response is different than the HS in the near field, i.e., at small r0r, and gradually converges to the HS in the far field, i.e., as r0r increases.

3.1 Comparison Between LREPC and Homogenous System Based on Stiffness, Mass, and Frequency Ratios.

To compare the LREPC and HS with equal nondimensional parameters, we first define these parameters for the HS. Equations (10) and (11) can be interpreted in terms of resonator to matrix stiffness and inertia ratios, respectively. Thus, we define the stiffness and inertia nondimensional parameters of the HS based on this interpretation.
$β¯HS=βrβmHS=βr2A0HS(r22r12r22−r12)$
(15)
$I¯HS=IrImHS=IrB0HS4(r24−r14)$
(16)
where $βmHS$ and $ImHS$ are quasi-static torsional stiffness and inertia of the unit cell of the matrix of the HS (see Appendices  A and  B for further details), r1 = r0 + (n − 1)Δr and r2 = r0 + nΔr, where n is the unit cell number. Finally, for $ω¯$, we define an analogous ω0 based on the matrix material properties,
$ω¯HS=ωrω0HS=ωrπΔrA0HSB0HS$
(17)
Note that $ω0HS$ is not the Bragg frequency of the HS, but it is defined in an analogous way as ω0. In fact, a Bragg frequency does not exist for the HS since it is not effectively periodic. Both $β¯HS$ and $I¯HS$ are dependent on the radial location of the unit cell. This is a direct result of the material not being effectively periodic. Furthermore, they each have a different dependence on radius. Thus, forcing these two parameters to be the same in all units changes the resonance frequencies of the resonators, resulting in different $ω¯HS$ for each unit cell. Similarly, if we fix $ω¯HS$ and $β¯HS$(or $I¯HS$) to be the same in all units, then $I¯HS$ (or $β¯HS$) will vary. In summary, only two of these three parameters can be fixed to be equal to those in the LREPC in all units, and the other one must be allowed to vary.

Here, we analyze two cases: one where $β¯HS$ and $I¯HS$ are fixed to be equal to those of the LREPC and $ω¯HS$ is allowed to vary, and one where $β¯HS$ and $ω¯HS$ are fixed to be equal to those of the LREPC and $I¯HS$ is allowed to vary. Fixing $I¯HS$ and $ω¯HS$ has an analogous result to fixing $β¯HS$ and $ω¯HS$, and thus, we exclude it from this article since it does not add any further understanding. For all comparisons, we keep the shear wave speed of the matrix the same in the LREPC and HS $(i.e.,A0B0=A0HSB0HS)$, and thus, $ω0HS=ω0$.

3.1.1 $β¯LREPC=β¯HS$ and $I¯LREPC=I¯HS$.

We first impose $β¯LREPC=β¯HS$ and $I¯LREPC=I¯HS$ in all units. To see the effect these constants have on the LREPC and HS systems, we calculate the band structure of the LREPC, $ω¯$, and transmission for both the LREPC and HS for different values of $β¯LREPC$ and $I¯LREPC$ (Fig. 2).

Fig. 2
Fig. 2
Close modal

Forcing stiffness and inertia parameters to be equal in the LREPC and HS imposes $ω¯HS≠ω¯LREPC$, which is more pronounced at smaller r0r (Fig. 2(b)). Furthermore, $ω¯HS→ω¯LREPC$ as the unit cell number increases (Fig. 2(b)). Thus, the behavior of the last few units of the HS should approach those of the LREPC. This will affect the effective dynamic moment of inertias of each unit cell of the HS (Sec. 4).

As r0r increases, $β¯HSI¯HS→β¯LREPCI¯LREPC$ and the resonator resonances of the HS converge to those of the LREPC (Fig. 2(b)). Thus, the transmission response of the HS approaches that of the LREPC (Figs. 2(c) and 2(d)). This is analogous to the far field of sources where waves are approximately plane.

Transmission reduction of the LREPC (Figs. 2(c) and 2(d)) agrees well with the predicted band gaps from dispersion (Fig. 2(a)) for all $β¯LREPC$ and $I¯LREPC$. As expected, as these nondimensional parameters increase, the band gap width increases (Fig. 2(a)), which is also reflected in the transmission curves (Fig. 2(c)). We do not include the response of the LREPC for different r0r in Figs. 2(c) and 2(d) because the response of the LREPC is independent of this ratio since it is effectively periodic. The varying material properties of the LREPC compensate for diffraction typical of cylindrical waves, and thus, we see no reduction in transmission outside the band gap region. Essentially, radial torsional waves in the LREPC behave as plane waves, and thus, they do not show fundamental features of cylindrical waves such as complex acoustic impedance or spatial wave attenuation from diffraction in the nondissipative system. The response of the LREPC to torsional waves is equivalent to that of a Cartesian system with periodic local resonances. Therefore, any properties of metamaterials that exist through periodic assumptions and band structure analysis apply to the LREPC in the near and far fields.

The behavior of the HS, on the other hand, strongly depends on r0r. In fact, the response of this system is quite different from that of the LRPEC in the near field, i.e., small r0r, for all $β¯$ and $I¯$ (Figs. 2(c) and 2(d)). The detuning of resonators of the different unit cells (Fig. 2(b)) results in a reduction in transmission that is not aligned with the band gap predicted by the dispersion curve of the LREPC. In fact, peaks followed by narrow dips are present in the lower frequency range (Fig. 2(c)). These dips widen as the inertia and stiffness ratios increase (Figs. 2(ci)2(ciii)) due to the larger resonator masses. Interestingly, at large $β¯$ and $I¯$, these peaks and dips widen such that they overlap for the last three resonators (Fig. 2(diii)), resulting in a rainbow-trapping like effect [38].

The reason for the differences found in this comparison is that the HS does not have the same resonance frequency in all its resonators. The lack of effective periodicity of the HS system forces the resonant frequencies of each of its resonators to be different enough to considerably affect the response when keeping $β¯$ and $I¯$ constant across its units.

3.1.2 $β¯LREPC=β¯HS$ and $ω¯LREPC=ω¯HS$.

In this next case, we set $β¯HS=β¯LREPC$ and $ω¯HS=ω¯LREPC$ to understand what happens when the resonance frequencies of resonators are all the same in the HS. Note that since $ω¯HS$ is constant, all resonators of the HS have the same resonance frequency (Eq. (17)). In addition, for $β¯HS$ to be constant, the resonators of the HS must have different stiffness values, since $βmHS$ depends on radius (Eq. (15)). This forces each of the resonators of the HS to also have different moments of inertia to keep their resonance frequencies the same. The LREPC used for this case is the same as for the case of $β¯LREPC=β¯HS$ and $I¯LREPC=I¯HS$ (Sec. 3.1.1).

Keeping $β¯HS$ and $ω¯HS$ constant results in $I¯HS$ that varies with radius (Fig. 3(b)). $I¯HS$ asymptotically approaches $I¯LREPC$ as unit cell number increases (Fig. 3(b)) and thus similar to the previous comparison, we expect the last few units of the HS to behave more similarly to those of LREPC compared to the first few units. As we shall see in Sec. 4, this will affect the dynamic effective moment of inertia of each unit cell of the HS. As r0r increases, the difference in $I¯HS$ and $I¯LREPC$ becomes smaller (Fig. 3(b)). Thus, the HS response approaches that of the LREPC as r0r increases (Fig. 3(c)).

Fig. 3
Fig. 3
Close modal

In the near field, we observe a different response between the LREPC and HS, even close to the resonator resonance frequency (Figs. 3(c) and 3(d)). For all $β¯$ and $ω¯$, we observe resonant peaks of the HS inside the band gap range of the LREPC close to both the upper and lower edges of the band gap (Figs. 3(c) and 3(d)). These resonant peaks seem to penetrate further inside the band gap as stiffness and inertia nondimensional parameters increase (Figs. 3(di)3(diii)). For higher nondimensional parameters, there is a peak in HS transmission in the neighborhood of the resonance frequency of the resonator (Fig. (3diii)), which is not present for the LREPC. This is clear evidence that the behaviors of the systems are quite different even close to the resonance frequency of the resonators, frequency ranges where these materials are typically designed to work.

Both of these comparisons show that in the near field of sources, the HS transmission is quite different from that of the LREPC. Thus, typical properties of periodic locally resonant systems may not be applicable for the HSs. These transmission results are evidence that to replicate well-known phenomena in periodic locally resonant systems in the near field of sources, one must use an LREPC.

3.2 Comparison Between LRPEC and Homogenous System With Equal Static Torsional Stiffness and Resonators.

In both previous comparisons (Sec. 3.1), the resonators of the HS are different in each unit cell. Thus, one might argue that differences in behavior arise from differences in the resonators. In this new case, we aim to compare an LREPC with an HS that has the same resonators, i.e., each resonator has the same stiffness and moment of inertia. To make an accurate comparison, we must somehow relate the stiffnesses and densities of the matrix materials of the LREPC and HS. To do this, we set the static torsional stiffness of both finite systems equal to each other, a possible engineering requirement in the application of these systems.

The objective of this comparison is to study the possible advantages of LREPCs for vibration control applications. When designing a component, there are usually requirements for loading bearing capabilities, for example, a minimum static stiffness. One limitation of locally resonant materials is that the larger the stiffness of the matrix material (which is what carries the load in these systems), the larger the resonator masses and stiffnesses needed for a certain band gap width. This results in more massive systems (in terms of their total mass, i.e., matrix plus resonators) when larger matrix stiffnesses are required, which can be undesirable for performance and cost. So for the final comparison, we compare a four-unit cell LREPC with an HS that has the same quasi-static torsional stiffness and the same resonators.

The static torsional stiffnesses of the finite four-unit cell of the LREPC and the HS are (see Appendix  A for more details):
$βLREPC=A0rout−r0$
(18)
$βHS=2A0HSrout2r02rout2−r02$
(19)
where rout = r0 + 4Δr, is the outer radius of the finite system. Equating Eqs. (18) and (19) results in,
$A0HS=A0(rout+r0)2rout2r02$
(20)
Note that these equations only define values of $A0HS$ but not those of $B0HS$. To define $B0HS$, we consider that in the limit as r → ∞, both systems should have the same response. This is true if $B0HS=B0A0HS/A0$, which is equivalent to setting the shear wave speed of the matrix of the LREPC and HS to be the same.

Interestingly, there is an extra advantage, apart from dynamic effects, of using the LREPC over the HS. For equal quasi-static stiffness and internal radius, the LREPC has both a lower moment of inertia and a lower mass compared to the HS (Fig. (4a)) (refer to Appendices  B and  C for more details). This means that the LREPC uses mass more efficiently. This can be explained by looking at the solution to the quasi-static problem with a fixed internal radius and a prescribed angular displacement on the external radius (Fig. (4b)). In the HS, the angular displacements significantly increase with radius at lower radii, while displacements are approximately constant at larger radii. This means that the deformation is concentrated only at very small radii, and most of the material does not deform: This indicates an inefficient distribution of the stiffness and thus of the mass of the matrix of the HS. However, in the LREPC, the angular displacement increases monotonically, meaning the deformation is evenly distributed over the radius. This indicates an efficient distribution of mass, resulting in a lighter system for equal static torsional stiffness. Lower inertia and mass can be advantageous in rotating systems that need to be accelerated and decelerated since they require lower forces and thus lower power to change angular speeds.

Fig. 4
Fig. 4
Close modal

Transmissions of LREPC and HS based on this comparison are shown in Fig. 5. As expected, the LREPC transmission has no dependence on r0r, and its transmission reduction region agrees well with the one predicted from dispersion (Figs. 5(a) and 5(b)). Even though all resonators of the HS and the LREPC are the same, their response is still considerably different at lower r0r (Figs. 5(b) and 5(c)). At low ratios, there are large differences close to the resonator resonance frequencies, where metamaterials support band gaps. Like the HSs from Sec. 3.1, this HS cannot support properties that stem from Bloch analysis.

Fig. 5
Fig. 5
Close modal

To further understand the system behaviors, we calculate $β¯$, $I¯$, and $ω¯$ for the HS and the LREPC (Figs. 5(d)5(f)). Since resonator frequencies are the same, both systems have the same $ω¯$; however, $β¯$ and $I¯$ are quite different. Contrary to comparisons with equal nondimensional parameters (Sec. 3.1), for a fixed value of r0r, $β¯HS$ and $I¯HS$ do not asymptotically approximate $β¯LREPC$ and $I¯LREPC$, respectively, as the unit cell number increases (Figs. 5(d) and 3(e)). This is because the matrix of the HS becomes stiffer and more massive (in terms of the moment of inertia) as radius increases (Eqs. (15) and (16)). In fact, any HS with equal resonators on all its units will have decreasing $β¯HS$ and $I¯HS$ as unit cell increases. As shown in Sec. 4, this will affect the effective dynamic properties of this HS system. However, as r0r increases, the HS transmission approximates that of the LRPEC (Fig. (5c)). This is because as r0r increases, $β¯HS$, $I¯HS$, and $ω¯HS$ approach $β¯LREPC$, $I¯LREPC$, and $ω¯LREPC$, respectively (Figs. 5(d) and 5(e)).

To understand the effect of the resonators on each system, we use as reference the transmission for both HS and LREPC when resonators are rigidly attached to the matrix (Fig. 5(c)—dashed lines). The reason we do this is so we can normalize for the lower overall transmission due to diffraction in the HS, which is compensated for in the LREPC with the radially dependent matrix material properties. At low r0r, only a narrow range of frequencies show differences in transmission of the HS (Fig. 5(c)). The effect of the resonators in the transmission of the HS is narrowband compared to the LREPC. This is because of two reasons: (1) the HS has more overall mass and inertia (Fig. 4(a)), which is also reflected in $I¯$ values (Fig. 5(e)), and (2) the matrix of the HS is softer for low radii but stiffer at larger radii (Fig. 4(b)) compared to the matrix of the LREPC, thus, while for the first resonator $β¯HS>β¯LREPC$, the other three show $β¯HS<β¯LREPC$ (Fig. 5(d)). Although the masses and stiffnesses of the resonators are the same for LREPC and HS, they produce a smaller effect in the HS and are thus less efficient in affecting response compared to the resonators in the LREPC.

Note that in the locally resonant case, even for r0r as high as 1 (Fig. 5(b)), where our previous study on Bragg scattering-based EPCs [22] have shown that the behavior of the EPCs to be quite close to their homogenous counterparts, the transmission of the LREPC and the HS is quite different. This is because the near-field effects are more significant at lower frequencies and lower radii. The Bragg scattering-based EPC requires large unit cells (i.e., larger radii) to obtain low-frequency band gaps. Even if band gaps frequencies in EPCs were lower, most of the structure would still be operating at large radii where near-field effects are not as significant. Instead, in the LREPC, band gap frequencies depend on the resonant frequencies of the resonators and are independent of unit cell size. Thus, low-frequency band gaps can be obtained for metamaterials with small radii where near-field effects are more significant. Note that the band gaps of the LREPC presented here are about 4% of the Bragg frequency (Fig. 5(a)).

4 Effective Dynamic Moment of Inertia

Capabilities of locally resonant materials such as negative refraction [20,39], mode conversion [20], and low-frequency vibration absorption [15] arise from their negative effective dynamic properties. Here, we compare the effective dynamic moments of inertia (Ieff) of the LREPC and HS as final evidence of how LREPCs can be advantageous for novel wave propagation control in the near field of sources.

Ieff for a unit cell is numerically calculated by analyzing its dynamic response to a unit harmonic displacement at its boundary nodes [40]. Under the long wavelength assumption,
$Ieff(ω)=T¯(ω)−ω2θavg$
(21)
where $T¯$ is the resultant torque on the boundary nodes of the unit cell (calculated using Eq. (2)) and θavg is the average of the angular displacements at the boundary nodes of the unit cell.

Figure 6 shows the frequency ranges of negative Ieff of each unit cell of the LREPC and HS for each of the three comparisons (Sec. 3), taking $r0Δr=0.1$, $β¯LREPC=0.01$, and $I¯LREPC=1$. Due to its effective periodicity, Ieff for all unit cells of the LREPC are the same. Similar to a Cartesian periodic locally resonant system, e.g., Ref. [17], the frequency range of negative Ieff of all units of the LREPC align with the band gap frequency range. Because the LREPC is effectively periodic, the negative Ieff occurs in the entire system and is independent of radius. On the contrary, the HS equations of motion are not invariant to translations, and thus, each of the HS units will have different dynamic effective properties. In fact, the frequency ranges of negative Ieff are different for all HSs studied.

Fig. 6
Fig. 6
Close modal

For the case where $β¯HS=β¯LREPC$ and $I¯HS=I¯LREPC$ (Sec. 3.1.1), the resonators of the HS all have different resonance frequencies (Fig. 2(bi)). The lower bounds of the range of negative Ieff of this HS are equal to the resonance frequency of the resonator of each unit cell (Fig. 6(a)). As the unit cell number increases, the resonance frequency of the resonators of the HS, and thus, its Ieff, approaches that of the LREPC. The range of negative Ieff is different for each unit cell, and there is no overlap of ranges between the first unit and the other three units.

The HS where $β¯HS=β¯LREPC$ and $ω¯HS=ω¯LREPC$(Sec. 3.1.2) also shows differences in the Ieff of its unit cells (Fig. 6(b)). Since all its resonators have the same resonance frequency (and equal to that of the LREPC), the lower bound of the frequency range of negative Ieff for the HS is the same as that of the LREPC. However, these ranges are smaller in the HS compared to the LREPC, particularly in the first few unit cells. This can be explained by looking at the dependence of $I¯HS$ on unit cell (Fig. 3(bi)). In the inner unit cells, $I¯HS$ are small compared to $I¯LREPC$, and thus, the Ieff becomes negative in a narrower frequency region. In the outer unit cells (approaching the far field), $I¯HS$ converges to $I¯LREPC$ and thus so do the Ieff of LREPC and HS.

In the HS with equal static torsional stiffness and resonators (Sec. 3.2), all resonators are the same (Fig. 5(f)), but the stiffness and moment of inertias of the HS matrix increase with each unit cell (Eqs. (15) and (16)). Thus, $β¯HS$ and $I¯HS$ decrease as unit cell increases (Figs. 5(d) and 5(e)), resulting in a frequency range of negative Ieff that decreases as the unit cell increases (Fig. 6(c)). In fact, this range becomes almost negligible in the fourth and final unit cells.

This analysis shows that negative effective properties of all the HSs occur over a limited frequency range and also depend on radius. On the contrary, the LREPC behavior is independent of radius, and all its units show a negative effective moment of inertia inside the frequency ranges predicted by its band structure, just like Cartesian metamaterials.

5 Conclusion

In this article, we extend the concept of EPCs [22] to include local resonances, in what we term LREPCs. We show that by using radially varying matrix properties with attached resonators, we can obtain a locally resonant material with equations of motions that are invariant to translations. This allows the application of the Bloch theorem, not only simplifying the analysis but also enabling properties of periodic locally resonant systems in the near field of sources, where plane wave assumptions are not valid.

We compare the behavior of the LREPC to three different HSs that are not “effectively periodic” but instead have geometrically periodic matrix material properties. Because of this, the equations of motion of the HSs are not invariant to translations. For all three cases, the transmission of the LREPC is quite different than the HS, particularly in the near field. This shows that using the LREPC is the only way to realize the typical properties of locally resonant metamaterials for cylindrical wave fronts. To further support this point, we show that the negative Ieff of the HS strongly depend on the radius and are limited to small frequency ranges, particularly in the near field. However, frequency ranges of negative Ieff of the LREPC are independent of radius and coincide with its band gap frequencies, just like typical Cartesian locally resonant systems.

Interestingly, cylindrical and spherical waves are not the only examples where geometrically periodic systems result in nonperiodic coefficients; for example, this is also the case in periodic pendula. In fact, periodically architected pendula approximate a periodic behavior if a large mass is added at its end, and recent work has shown that a “pseudo” unit cell can be defined from the dynamics of the finite system [41]. Even though there are differences in terms of mathematical approaches and compositions, radial wave crystals [23], EPCs [22] and LREPCs, and periodically architected pendula [41] all have a similar overarching concept: moving away from geometric periodicity and instead defining a mathematically periodic system, thus expanding the boundary of where the Bloch theorem applies. In this way, these studies broaden the concepts of PCs and metamaterials to a new range of mechanical systems.

Our study considers torsional waves and axisymmetric wave propagation, which has applications to rotational machinery or microelectromechanical systems, opening opportunities to embedded wave propagation control mechanisms in waves that propagate outward from a central source or axis. However, similar mathematical treatment could extend these concepts to other wave polarizations and nonaxisymmetric wave propagation. In this way, we can further exploit the platform presented herein to address near-field effects in metamaterials.

Acknowledgment

This material is based upon work supported by the National Science Foundation under Grant No. CMMI-2031110.

Conflict of Interest

There are no conflicts of interest.

Data Availability Statement

The datasets generated and supporting the findings of this article are obtainable from the corresponding author upon reasonable request. The authors attest that all data for this study are included in the paper.

Funding Data

• National Science Foundation (NSF) (Grant No. CMMI-2031110; Funder ID: 10.13039/501100008982).

Nomenclature

• h =

out-of-plane thickness

•
• A =

matrix stiffness function LREPC

•
• B =

matrix density function LREPC

•
• I =

polar mass moment of inertia

•
• M =

external moment

•
• T =

torque

•
• $I¯$ =

moment of inertia nondimensional parameter

•
• r0 =

inner radius

•
• rout =

outer radius

•
• $uθ$ =

tangential displacements

•
• A0 =

matrix stiffness constant of LREPC

•
• A0HS =

matrix stiffness constant of HS

•
• B0 =

matrix density constant of LREPC

•
• B0HS =

matrix density constant of HS

•
• Ieff =

effective dynamic moment of inertia

•
• Ir =

polar mass moment of inertia of the resonators

•
• $I¯HS$ =

moment of inertia nondimensional parameter of the HS

•
• $I¯LREPC$ =

moment of inertia nondimensional parameter of the LREPC

•
• βr =

torsional stiffness of resonators

•
• βHS =

static torsional stiffness of four-unit cell HS

•
• βLREPC =

static torsional stiffness of four-unit cell LREPC

•
• $β¯$ =

stiffness nondimensional parameter

•
• $β¯HS$ =

stiffness nondimensional parameter of the HS

•
• $β¯LREPC$ =

stiffness nondimensional parameter of the LREPC

•
• Δr =

periodicity constant

•
• θ =

angular displacement

•
• μ =

shear modulus

•
• ρ =

density

•
• σ =

shear stress

•
• ω =

angular frequency

Appendix A: Static Stiffness of LREPC and Homogenous System

Static torsional stiffness of LREPC and HS finite systems, as well as their unit cells, are calculated by solving the boundary value problem arising from Eq. (6), neglecting inertial effects and imposing angular displacements in both inner and outer ring surfaces:
${∂∂r(r3A(r)∂θ∂r)=0θ(rin)=θ1θ(rout)=θ2$
(A1)
We substitute material properties for the LREPC (Eq. (7)) and the HS (Eq. (14)) and solve for θ(r). We define the static torsional stiffness as follows:
$β=T(rout)θ2−θ1$
(A2)
where T(r) is the moment as defined in Eq. (3). We replace rout and rin by inner and outer radii of the unit cell, respectively, and by r0 and r0 + NΔr for static stiffness of the unit cells and systems of N unit cells, respectively.

Appendix B: Polar Mass Moment of Inertia of LREPC and Homogenous System

Moment of inertia of LREPC and HS systems as well as their unit cells are calculated as follows:
$∫Ωρ(r)r2dV=∫rinrout∫02π∫0h(r)ρ(r)r3drdθdz=2π∫rinroutρ(r)h(r)r3dr=2π∫rinroutB(r)r3dr$
(B1)

We substitute material properties for the LREPC (Eq. (7)) and the HS (Eq. (14)) and replace rout and rin by inner and outer radii of the unit cell and by r0 and r0 + NΔr for the moment of inertia of the unit cells and systems of N unit cells, respectively.

Appendix C: Mass of LREPC and Homogenous System

Mass of LREPC and HS systems are follows:
$∫Ωρ(r)dV=∫r0r0+NΔr∫02π∫0h(r)ρ(r)rdrdθdz=2π∫r0r0+NΔrρ(r)h(r)rdr=2π∫r0r0+NΔrB(r)rdr$
(C1)

We substitute material properties for the LREPC (Eq. (7)) and the HS (Eq. (14)) and replace rout and rin by inner and outer radii of the unit cell and by r0 and r0 + NΔr for the mass of the unit cells and whole systems, respectively.

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Found in: Page 772

Precalculus Enhanced with Graphing Utilities

Book edition 6th
Author(s) Sullivan
Pages 1200 pages
ISBN 9780321795465

In Problems 5–24, graph each equation of the system. Then solve the system to find the points of intersection.$y=\sqrt{4-{x}^{2}};\phantom{\rule{0ex}{0ex}}y=2x+4$

he graph of the system of equations $y=\sqrt{4-{x}^{2}};y=2x+4$ is:

The points of intersection are $\left(-\frac{6}{5},\frac{8}{5}\right),\left(-2,0\right)$

See the step by step solution

Step 1. Given

The system of non-linear equation:

$y=\sqrt{4-{x}^{2}};\phantom{\rule{0ex}{0ex}}y=2x+4$

To graph the equation and to find the point of intersection.

Step 2. Graph the equations

Graph the equations in the same plane.

Step 3. To find the point of intersection.

Equate both the equations,

$\sqrt{4-{x}^{2}}=2x+4$

Squaring on both sides, we get,

$4-{x}^{2}={\left(2x+4\right)}^{2}\phantom{\rule{0ex}{0ex}}4-{x}^{2}=4{x}^{2}+16x+16\phantom{\rule{0ex}{0ex}}5{x}^{2}+16x+12=0$

Step 4. Solve the equations

$5{x}^{2}+16x+12=0\phantom{\rule{0ex}{0ex}}\left(5x+6\right)\left(x+2\right)=0\phantom{\rule{0ex}{0ex}}x=-\frac{6}{5},-2$

The value of $x$ is $-\frac{6}{5},-2$

Step 5. Find y

When

$x=-\frac{6}{5},\phantom{\rule{0ex}{0ex}}y=2x+4\phantom{\rule{0ex}{0ex}}=2\left(-\frac{6}{5}\right)+4\phantom{\rule{0ex}{0ex}}=-\frac{12}{5}+4\phantom{\rule{0ex}{0ex}}=\frac{8}{5}$

When $x=-2,$

$y=2x+4\phantom{\rule{0ex}{0ex}}=2\left(-2\right)+4\phantom{\rule{0ex}{0ex}}=0$

The point of intersections are $\left(-\frac{6}{5},\frac{8}{5}\right),\left(-2,0\right)$

format: plain


text:
Giáo trình

# Precalculus

Mathematics and Statistics

## Solving Systems with Inverses

Tác giả: OpenStaxCollege

Nancy plans to invest \$10,500 into two different bonds to spread out her risk. The first bond has an annual return of 10%, and the second bond has an annual return of 6%. In order to receive an 8.5% return from the two bonds, how much should Nancy invest in each bond? What is the best method to solve this problem?

There are several ways we can solve this problem. As we have seen in previous sections, systems of equations and matrices are useful in solving real-world problems involving finance. After studying this section, we will have the tools to solve the bond problem using the inverse of a matrix.

# Finding the Inverse of a Matrix

We know that the multiplicative inverse of a real number$\text{\hspace{0.17em}}a\text{\hspace{0.17em}}$is$\text{\hspace{0.17em}}{a}^{-1},\text{\hspace{0.17em}}$and$\text{\hspace{0.17em}}a{a}^{-1}={a}^{-1}a=\left(\frac{1}{a}\right)a=1.\text{\hspace{0.17em}}$For example,$\text{\hspace{0.17em}}{2}^{-1}=\frac{1}{2}\text{\hspace{0.17em}}$ and$\text{\hspace{0.17em}}\left(\frac{1}{2}\right)2=1.\text{\hspace{0.17em}}$The multiplicative inverse of a matrix is similar in concept, except that the product of matrix$A\text{\hspace{0.17em}}$and its inverse$\text{\hspace{0.17em}}{A}^{-1}\text{\hspace{0.17em}}$equals the identity matrix. The identity matrix is a square matrix containing ones down the main diagonal and zeros everywhere else. We identify identity matrices by$\text{\hspace{0.17em}}{I}_{n}\text{\hspace{0.17em}}$where$\text{\hspace{0.17em}}n\text{\hspace{0.17em}}$represents the dimension of the matrix. [link] and [link] are the identity matrices for a$\text{\hspace{0.17em}}2\text{}×\text{}2\text{\hspace{0.17em}}$matrix and a$\text{\hspace{0.17em}}3\text{}×\text{}3\text{\hspace{0.17em}}$matrix, respectively.

$I 2 =[ 1 0 0 1 ]$
$I 3 =[ 1 0 0 0 1 0 0 0 1 ]$

The identity matrix acts as a 1 in matrix algebra. For example,$\text{\hspace{0.17em}}AI=IA=A.$

A matrix that has a multiplicative inverse has the properties

A matrix that has a multiplicative inverse is called an invertible matrix. Only a square matrix may have a multiplicative inverse, as the reversibility,$\text{\hspace{0.17em}}A{A}^{-1}={A}^{-1}A=I,\text{\hspace{0.17em}}$is a requirement. Not all square matrices have an inverse, but if$\text{\hspace{0.17em}}A\text{\hspace{0.17em}}$is invertible, then$\text{\hspace{0.17em}}{A}^{-1}\text{\hspace{0.17em}}$is unique. We will look at two methods for finding the inverse of a$\text{\hspace{0.17em}}2\text{}×\text{}2\text{\hspace{0.17em}}$matrix and a third method that can be used on both$\text{\hspace{0.17em}}2\text{}×\text{}2\text{\hspace{0.17em}}$and$3\text{}×\text{}3\text{\hspace{0.17em}}$matrices.

The Identity Matrix and Multiplicative Inverse

The identity matrix,$\text{\hspace{0.17em}}{I}_{n},\text{\hspace{0.17em}}$is a square matrix containing ones down the main diagonal and zeros everywhere else.

If$\text{\hspace{0.17em}}A\text{\hspace{0.17em}}$is an$\text{\hspace{0.17em}}n\text{\hspace{0.17em}}×\text{\hspace{0.17em}}n\text{\hspace{0.17em}}$ matrix and$\text{\hspace{0.17em}}B\text{\hspace{0.17em}}$is an$\text{\hspace{0.17em}}n\text{\hspace{0.17em}}×\text{\hspace{0.17em}}n\text{\hspace{0.17em}}$ matrix such that$\text{\hspace{0.17em}}AB=BA={I}_{n},\text{\hspace{0.17em}}$then$\text{\hspace{0.17em}}B={A}^{-1},\text{\hspace{0.17em}}$the multiplicative inverse of a matrix$\text{\hspace{0.17em}}A.$

Showing That the Identity Matrix Acts as a 1

Given matrix A, show that$\text{\hspace{0.17em}}AI=IA=A.$

Use matrix multiplication to show that the product of$\text{\hspace{0.17em}}A\text{\hspace{0.17em}}$and the identity is equal to the product of the identity and A.

Given two matrices, show that one is the multiplicative inverse of the other.

1. Given matrix$\text{\hspace{0.17em}}A\text{\hspace{0.17em}}$of order$\text{\hspace{0.17em}}n\text{\hspace{0.17em}}×\text{\hspace{0.17em}}n\text{\hspace{0.17em}}$and matrix$\text{\hspace{0.17em}}B\text{\hspace{0.17em}}$of order$\text{\hspace{0.17em}}n\text{\hspace{0.17em}}×\text{\hspace{0.17em}}n\text{\hspace{0.17em}}$multiply$\text{\hspace{0.17em}}AB.$
2. If$\text{\hspace{0.17em}}AB=I,\text{\hspace{0.17em}}$then find the product$\text{\hspace{0.17em}}BA.\text{\hspace{0.17em}}$If$\text{\hspace{0.17em}}BA=I,\text{\hspace{0.17em}}$then$\text{\hspace{0.17em}}B={A}^{-1}\text{\hspace{0.17em}}$and$\text{\hspace{0.17em}}A={B}^{-1}.$
Showing That Matrix A Is the Multiplicative Inverse of Matrix B

Show that the given matrices are multiplicative inverses of each other.

Multiply$\text{\hspace{0.17em}}AB\text{\hspace{0.17em}}$and$\text{\hspace{0.17em}}BA.\text{\hspace{0.17em}}$If both products equal the identity, then the two matrices are inverses of each other.

$A\text{\hspace{0.17em}}$and$B$are inverses of each other.

Show that the following two matrices are inverses of each other.

## Finding the Multiplicative Inverse Using Matrix Multiplication

We can now determine whether two matrices are inverses, but how would we find the inverse of a given matrix? Since we know that the product of a matrix and its inverse is the identity matrix, we can find the inverse of a matrix by setting up an equation using matrix multiplication.

Finding the Multiplicative Inverse Using Matrix Multiplication

Use matrix multiplication to find the inverse of the given matrix.

For this method, we multiply$\text{\hspace{0.17em}}A\text{\hspace{0.17em}}$by a matrix containing unknown constants and set it equal to the identity.

Find the product of the two matrices on the left side of the equal sign.

Next, set up a system of equations with the entry in row 1, column 1 of the new matrix equal to the first entry of the identity, 1. Set the entry in row 2, column 1 of the new matrix equal to the corresponding entry of the identity, which is 0.

Using row operations, multiply and add as follows:$\text{\hspace{0.17em}}\left(-2\right){R}_{1}+{R}_{2}\to {R}_{2}.\text{\hspace{0.17em}}$Add the equations, and solve for$\text{\hspace{0.17em}}c.$

Back-substitute to solve for$\text{\hspace{0.17em}}a.$

Write another system of equations setting the entry in row 1, column 2 of the new matrix equal to the corresponding entry of the identity, 0. Set the entry in row 2, column 2 equal to the corresponding entry of the identity.

Using row operations, multiply and add as follows:$\text{\hspace{0.17em}}\left(-2\right){R}_{1}+{R}_{2}={R}_{2}.\text{\hspace{0.17em}}$Add the two equations and solve for$\text{\hspace{0.17em}}d.$

Once more, back-substitute and solve for$\text{\hspace{0.17em}}b.$

## Finding the Multiplicative Inverse by Augmenting with the Identity

Another way to find the multiplicative inverse is by augmenting with the identity. When matrix$\text{\hspace{0.17em}}A\text{\hspace{0.17em}}$is transformed into$\text{\hspace{0.17em}}I,\text{\hspace{0.17em}}$the augmented matrix$\text{\hspace{0.17em}}I\text{\hspace{0.17em}}$transforms into$\text{\hspace{0.17em}}{A}^{-1}.$

For example, given

augment$\text{\hspace{0.17em}}A\text{\hspace{0.17em}}$with the identity

Perform row operations with the goal of turning$\text{\hspace{0.17em}}A\text{\hspace{0.17em}}$into the identity.

1. Switch row 1 and row 2.
2. Multiply row 2 by$\text{\hspace{0.17em}}-2\text{\hspace{0.17em}}$and add to row 1.
3. Multiply row 1 by$\text{\hspace{0.17em}}-2\text{\hspace{0.17em}}$and add to row 2.
4. Add row 2 to row 1.
5. Multiply row 2 by$\text{\hspace{0.17em}}-1.$

The matrix we have found is$\text{\hspace{0.17em}}{A}^{-1}.$

## Finding the Multiplicative Inverse of 2×2 Matrices Using a Formula

When we need to find the multiplicative inverse of a$\text{\hspace{0.17em}}2\text{\hspace{0.17em}}×\text{\hspace{0.17em}}2\text{\hspace{0.17em}}$matrix, we can use a special formula instead of using matrix multiplication or augmenting with the identity.

If$\text{\hspace{0.17em}}A\text{\hspace{0.17em}}$is a$\text{\hspace{0.17em}}2×2\text{\hspace{0.17em}}$matrix, such as

the multiplicative inverse of$\text{\hspace{0.17em}}A\text{\hspace{0.17em}}$is given by the formula

$A −1 = 1 ad−bc [ d −b −c a ]$

where$\text{\hspace{0.17em}}ad-bc\ne 0.\text{\hspace{0.17em}}$If$\text{\hspace{0.17em}}ad-bc=0,\text{\hspace{0.17em}}$then$\text{\hspace{0.17em}}A\text{\hspace{0.17em}}$has no inverse.

Using the Formula to Find the Multiplicative Inverse of Matrix A

Use the formula to find the multiplicative inverse of

Using the formula, we have

Analysis

We can check that our formula works by using one of the other methods to calculate the inverse. Let’s augment$A\text{\hspace{0.17em}}$with the identity.

Perform row operations with the goal of turning$\text{\hspace{0.17em}}A\text{\hspace{0.17em}}$into the identity.

1. Multiply row 1 by$\text{\hspace{0.17em}}-2\text{\hspace{0.17em}}$and add to row 2.
2. Multiply row 1 by 2 and add to row 1.

So, we have verified our original solution.

Use the formula to find the inverse of matrix$\text{\hspace{0.17em}}A.\text{\hspace{0.17em}}$Verify your answer by augmenting with the identity matrix.

${A}^{-1}=\left[\begin{array}{cc}\frac{3}{5}& \frac{1}{5}\\ -\frac{2}{5}& \frac{1}{5}\end{array}\right]$

Finding the Inverse of the Matrix, If It Exists

Find the inverse, if it exists, of the given matrix.

We will use the method of augmenting with the identity.

1. Switch row 1 and row 2.
2. Multiply row 1 by −3 and add it to row 2.
3. There is nothing further we can do. The zeros in row 2 indicate that this matrix has no inverse.

## Finding the Multiplicative Inverse of 3×3 Matrices

Unfortunately, we do not have a formula similar to the one for a$\text{\hspace{0.17em}}2\text{}×\text{}2\text{\hspace{0.17em}}$matrix to find the inverse of a$\text{\hspace{0.17em}}3\text{}×\text{}3\text{\hspace{0.17em}}$matrix. Instead, we will augment the original matrix with the identity matrix and use row operations to obtain the inverse.

Given a$\text{\hspace{0.17em}}3\text{}×\text{}3\text{\hspace{0.17em}}$ matrix

augment$\text{\hspace{0.17em}}A\text{\hspace{0.17em}}$with the identity matrix

To begin, we write the augmented matrix with the identity on the right and$\text{\hspace{0.17em}}A\text{\hspace{0.17em}}$on the left. Performing elementary row operations so that the identity matrix appears on the left, we will obtain the inverse matrix on the right. We will find the inverse of this matrix in the next example.

Given a$\text{\hspace{0.17em}}3\text{\hspace{0.17em}}×\text{\hspace{0.17em}}3\text{\hspace{0.17em}}$matrix, find the inverse

1. Write the original matrix augmented with the identity matrix on the right.
2. Use elementary row operations so that the identity appears on the left.
3. What is obtained on the right is the inverse of the original matrix.
4. Use matrix multiplication to show that$\text{\hspace{0.17em}}A{A}^{-1}=I\text{\hspace{0.17em}}$and$\text{\hspace{0.17em}}{A}^{-1}A=I.$
Finding the Inverse of a 3 × 3 Matrix

Given the$\text{\hspace{0.17em}}3\text{\hspace{0.17em}}×\text{\hspace{0.17em}}3\text{\hspace{0.17em}}$matrix$\text{\hspace{0.17em}}A,\text{\hspace{0.17em}}$find the inverse.

Augment$\text{\hspace{0.17em}}A\text{\hspace{0.17em}}$with the identity matrix, and then begin row operations until the identity matrix replaces$\text{\hspace{0.17em}}A.\text{\hspace{0.17em}}$The matrix on the right will be the inverse of$\text{\hspace{0.17em}}A.\text{\hspace{0.17em}}$

Thus,

Analysis

To prove that$\text{\hspace{0.17em}}B={A}^{-1},\text{\hspace{0.17em}}$let’s multiply the two matrices together to see if the product equals the identity, if$A{A}^{-1}=I\text{\hspace{0.17em}}$and$\text{\hspace{0.17em}}{A}^{-1}A=I.$

Find the inverse of the$\text{\hspace{0.17em}}3×3\text{\hspace{0.17em}}$matrix.

${A}^{-1}=\left[\begin{array}{ccc}1& 1& \text{\hspace{0.17em}}\text{\hspace{0.17em}}2\\ 2& 4& -3\\ 3& 6& -5\end{array}\right]$

# Solving a System of Linear Equations Using the Inverse of a Matrix

Solving a system of linear equations using the inverse of a matrix requires the definition of two new matrices:$\text{\hspace{0.17em}}X\text{\hspace{0.17em}}$is the matrix representing the variables of the system, and$\text{\hspace{0.17em}}B\text{\hspace{0.17em}}$is the matrix representing the constants. Using matrix multiplication, we may define a system of equations with the same number of equations as variables as

To solve a system of linear equations using an inverse matrix, let$\text{\hspace{0.17em}}A\text{\hspace{0.17em}}$be the coefficient matrix, let$\text{\hspace{0.17em}}X\text{\hspace{0.17em}}$be the variable matrix, and let$\text{\hspace{0.17em}}B\text{\hspace{0.17em}}$be the constant matrix. Thus, we want to solve a system$\text{\hspace{0.17em}}AX=B.\text{\hspace{0.17em}}$For example, look at the following system of equations.

From this system, the coefficient matrix is

The variable matrix is

And the constant matrix is

Then$\text{\hspace{0.17em}}AX=B\text{\hspace{0.17em}}$looks like

Recall the discussion earlier in this section regarding multiplying a real number by its inverse,$\text{\hspace{0.17em}}\left({2}^{-1}\right)\text{\hspace{0.17em}}2=\left(\frac{1}{2}\right)\text{\hspace{0.17em}}2=1.\text{\hspace{0.17em}}$To solve a single linear equation$\text{\hspace{0.17em}}ax=b\text{\hspace{0.17em}}$for$\text{\hspace{0.17em}}x,\text{\hspace{0.17em}}$we would simply multiply both sides of the equation by the multiplicative inverse (reciprocal) of$\text{\hspace{0.17em}}a.\text{\hspace{0.17em}}$Thus,

The only difference between a solving a linear equation and a system of equations written in matrix form is that finding the inverse of a matrix is more complicated, and matrix multiplication is a longer process. However, the goal is the same—to isolate the variable.

We will investigate this idea in detail, but it is helpful to begin with a$\text{\hspace{0.17em}}2\text{\hspace{0.17em}}×\text{\hspace{0.17em}}2\text{\hspace{0.17em}}$system and then move on to a$\text{\hspace{0.17em}}3\text{\hspace{0.17em}}×\text{\hspace{0.17em}}3\text{\hspace{0.17em}}$system.

Solving a System of Equations Using the Inverse of a Matrix

Given a system of equations, write the coefficient matrix$\text{\hspace{0.17em}}A,\text{\hspace{0.17em}}$the variable matrix$\text{\hspace{0.17em}}X,\text{\hspace{0.17em}}$and the constant matrix$\text{\hspace{0.17em}}B.\text{\hspace{0.17em}}$Then

Multiply both sides by the inverse of$\text{\hspace{0.17em}}A\text{\hspace{0.17em}}$to obtain the solution.

If the coefficient matrix does not have an inverse, does that mean the system has no solution?

No, if the coefficient matrix is not invertible, the system could be inconsistent and have no solution, or be dependent and have infinitely many solutions.

Solving a 2 × 2 System Using the Inverse of a Matrix

Solve the given system of equations using the inverse of a matrix.

Write the system in terms of a coefficient matrix, a variable matrix, and a constant matrix.

Then

First, we need to calculate$\text{\hspace{0.17em}}{A}^{-1}.\text{\hspace{0.17em}}$Using the formula to calculate the inverse of a 2 by 2 matrix, we have:

So,

Now we are ready to solve. Multiply both sides of the equation by$\text{\hspace{0.17em}}{A}^{-1}.$

The solution is$\text{\hspace{0.17em}}\left(-1,1\right).$

Can we solve for$\text{\hspace{0.17em}}X\text{\hspace{0.17em}}$by finding the product$\text{\hspace{0.17em}}B{A}^{-1}?$

No, recall that matrix multiplication is not commutative, so$\text{\hspace{0.17em}}{A}^{-1}B\ne B{A}^{-1}.\text{\hspace{0.17em}}$Consider our steps for solving the matrix equation.

Notice in the first step we multiplied both sides of the equation by$\text{\hspace{0.17em}}{A}^{-1},\text{\hspace{0.17em}}$but the$\text{\hspace{0.17em}}{A}^{-1}\text{\hspace{0.17em}}$was to the left of$\text{\hspace{0.17em}}A\text{\hspace{0.17em}}$on the left side and to the left of$\text{\hspace{0.17em}}B\text{\hspace{0.17em}}$on the right side. Because matrix multiplication is not commutative, order matters.

Solving a 3 × 3 System Using the Inverse of a Matrix

Solve the following system using the inverse of a matrix.

Write the equation$\text{\hspace{0.17em}}AX=B.\text{\hspace{0.17em}}$

First, we will find the inverse of$\text{\hspace{0.17em}}A\text{\hspace{0.17em}}$by augmenting with the identity.

Multiply row 1 by$\text{\hspace{0.17em}}\frac{1}{5}.$

Multiply row 1 by 4 and add to row 2.

Add row 1 to row 3.

Multiply row 2 by −3 and add to row 1.

Multiply row 3 by 5.

Multiply row 3 by$\text{\hspace{0.17em}}\frac{1}{5}\text{\hspace{0.17em}}$and add to row 1.

Multiply row 3 by$\text{\hspace{0.17em}}-\frac{19}{5}\text{\hspace{0.17em}}$and add to row 2.

So,

Multiply both sides of the equation by$\text{\hspace{0.17em}}{A}^{-1}.\text{\hspace{0.17em}}$We want$\text{\hspace{0.17em}}{A}^{-1}AX={A}^{-1}B:$

Thus,

The solution is$\text{\hspace{0.17em}}\left(1,2,0\right).$

Solve the system using the inverse of the coefficient matrix.

$X=[ 4 38 58 ]$

Given a system of equations, solve with matrix inverses using a calculator.

1. Save the coefficient matrix and the constant matrix as matrix variables$\text{\hspace{0.17em}}\left[A\right]\text{\hspace{0.17em}}$and$\text{\hspace{0.17em}}\left[B\right].$
2. Enter the multiplication into the calculator, calling up each matrix variable as needed.
3. If the coefficient matrix is invertible, the calculator will present the solution matrix; if the coefficient matrix is not invertible, the calculator will present an error message.
Using a Calculator to Solve a System of Equations with Matrix Inverses

Solve the system of equations with matrix inverses using a calculator

On the matrix page of the calculator, enter the coefficient matrix as the matrix variable$\text{\hspace{0.17em}}\left[A\right],\text{\hspace{0.17em}}$and enter the constant matrix as the matrix variable$\text{\hspace{0.17em}}\left[B\right].$

On the home screen of the calculator, type in the multiplication to solve for$\text{\hspace{0.17em}}X,\text{\hspace{0.17em}}$calling up each matrix variable as needed.

Evaluate the expression.

Access these online resources for additional instruction and practice with solving systems with inverses.

# Key Equations

 Identity matrix for a$2\text{}×\text{}2$matrix ${I}_{2}=\left[\begin{array}{cc}1& 0\\ 0& 1\end{array}\right]$ Identity matrix for a$\text{3}\text{}×\text{}3$matrix ${I}_{3}=\left[\begin{array}{ccc}1& 0& 0\\ 0& 1& 0\\ 0& 0& 1\end{array}\right]$ Multiplicative inverse of a$2\text{}×\text{}2$matrix

# Key Concepts

• An identity matrix has the property$\text{\hspace{0.17em}}AI=IA=A.\text{\hspace{0.17em}}$See [link].
• An invertible matrix has the property$\text{\hspace{0.17em}}A{A}^{-1}={A}^{-1}A=I.\text{\hspace{0.17em}}$See [link].
• Use matrix multiplication and the identity to find the inverse of a$\text{\hspace{0.17em}}2×2\text{\hspace{0.17em}}$matrix. See [link].
• The multiplicative inverse can be found using a formula. See [link].
• Another method of finding the inverse is by augmenting with the identity. See [link].
• We can augment a$\text{\hspace{0.17em}}3×3\text{\hspace{0.17em}}$matrix with the identity on the right and use row operations to turn the original matrix into the identity, and the matrix on the right becomes the inverse. See [link].
• Write the system of equations as$\text{\hspace{0.17em}}AX=B,\text{\hspace{0.17em}}$and multiply both sides by the inverse of$\text{\hspace{0.17em}}A:{A}^{-1}AX={A}^{-1}B.\text{\hspace{0.17em}}$See [link] and [link].
• We can also use a calculator to solve a system of equations with matrix inverses. See [link].

# Section Exercises

## Verbal

In a previous section, we showed that matrix multiplication is not commutative, that is,$\text{\hspace{0.17em}}AB\ne BA\text{\hspace{0.17em}}$in most cases. Can you explain why matrix multiplication is commutative for matrix inverses, that is,$\text{\hspace{0.17em}}{A}^{-1}A=A{A}^{-1}?$

If$\text{\hspace{0.17em}}{A}^{-1}\text{\hspace{0.17em}}$is the inverse of$\text{\hspace{0.17em}}A,\text{\hspace{0.17em}}$then$\text{\hspace{0.17em}}A{A}^{-1}=I,\text{\hspace{0.17em}}$the identity matrix. Since$\text{\hspace{0.17em}}A\text{\hspace{0.17em}}$is also the inverse of$\text{\hspace{0.17em}}{A}^{-1},{A}^{-1}A=I.\text{\hspace{0.17em}}$You can also check by proving this for a$\text{\hspace{0.17em}}2×2\text{\hspace{0.17em}}$matrix.

Does every$\text{\hspace{0.17em}}2×2\text{\hspace{0.17em}}$matrix have an inverse? Explain why or why not. Explain what condition is necessary for an inverse to exist.

Can you explain whether a$\text{\hspace{0.17em}}2×2\text{\hspace{0.17em}}$matrix with an entire row of zeros can have an inverse?

No, because$\text{\hspace{0.17em}}ad\text{\hspace{0.17em}}$and$\text{\hspace{0.17em}}bc\text{\hspace{0.17em}}$are both 0, so$\text{\hspace{0.17em}}ad-bc=0,\text{\hspace{0.17em}}$which requires us to divide by 0 in the formula.

Can a matrix with an entire column of zeros have an inverse? Explain why or why not.

Can a matrix with zeros on the diagonal have an inverse? If so, find an example. If not, prove why not. For simplicity, assume a$\text{\hspace{0.17em}}2×2\text{\hspace{0.17em}}$matrix.

Yes. Consider the matrix$\text{\hspace{0.17em}}\left[\begin{array}{cc}0& 1\\ 1& 0\end{array}\right].\text{\hspace{0.17em}}$The inverse is found with the following calculation:$\text{\hspace{0.17em}}{A}^{-1}=\frac{1}{0\left(0\right)-1\left(1\right)}\left[\begin{array}{cc}0& -1\\ -1& 0\end{array}\right]=\left[\begin{array}{cc}0& 1\\ 1& 0\end{array}\right].$

## Algebraic

In the following exercises, show that matrix$\text{\hspace{0.17em}}A\text{\hspace{0.17em}}$is the inverse of matrix$\text{\hspace{0.17em}}B.$

$A=\left[\begin{array}{cc}1& 0\\ -1& 1\end{array}\right],\text{\hspace{0.17em}}B=\left[\begin{array}{cc}1& 0\\ 1& 1\end{array}\right]$

$A=\left[\begin{array}{cc}1& 2\\ 3& 4\end{array}\right],\text{\hspace{0.17em}}B=\left[\begin{array}{cc}-2& 1\\ \frac{3}{2}& -\frac{1}{2}\end{array}\right]$

$AB=BA=\left[\begin{array}{cc}1& 0\\ 0& 1\end{array}\right]=I$

$A=\left[\begin{array}{cc}4& 5\\ 7& 0\end{array}\right],\text{\hspace{0.17em}}B=\left[\begin{array}{cc}0& \frac{1}{7}\\ \frac{1}{5}& -\frac{4}{35}\end{array}\right]$

$A=\left[\begin{array}{cc}-2& \frac{1}{2}\\ 3& -1\end{array}\right],\text{\hspace{0.17em}}B=\left[\begin{array}{cc}-2& -1\\ -6& -4\end{array}\right]$

$AB=BA=\left[\begin{array}{cc}1& 0\\ 0& 1\end{array}\right]=I$

$A=\left[\begin{array}{ccc}1& 0& 1\\ 0& 1& -1\\ 0& 1& 1\end{array}\right],\text{\hspace{0.17em}}B=\frac{1}{2}\left[\begin{array}{ccc}2& 1& -1\\ 0& 1& 1\\ 0& -1& 1\end{array}\right]$

$A=\left[\begin{array}{ccc}1& 2& 3\\ 4& 0& 2\\ 1& 6& 9\end{array}\right],\text{\hspace{0.17em}}B=\frac{1}{4}\left[\begin{array}{ccc}6& 0& -2\\ 17& -3& -5\\ -12& 2& 4\end{array}\right]$

$AB=BA=\left[\begin{array}{ccc}1& 0& 0\\ 0& 1& 0\\ 0& 0& 1\end{array}\right]=I$

$A=\left[\begin{array}{ccc}3& 8& 2\\ 1& 1& 1\\ 5& 6& 12\end{array}\right],\text{\hspace{0.17em}}B=\frac{1}{36}\left[\begin{array}{ccc}-6& 84& -6\\ 7& -26& 1\\ -1& -22& 5\end{array}\right]$

For the following exercises, find the multiplicative inverse of each matrix, if it exists.

$\left[\begin{array}{cc}3& -2\\ 1& 9\end{array}\right]$

$\frac{1}{29}\left[\begin{array}{cc}9& 2\\ -1& 3\end{array}\right]$

$\left[\begin{array}{cc}-2& 2\\ 3& 1\end{array}\right]$

$\left[\begin{array}{cc}-3& 7\\ 9& 2\end{array}\right]$

$\frac{1}{69}\left[\begin{array}{cc}-2& 7\\ 9& 3\end{array}\right]$

$\left[\begin{array}{cc}-4& -3\\ -5& 8\end{array}\right]$

$\left[\begin{array}{cc}1& 1\\ 2& 2\end{array}\right]$

There is no inverse

$\left[\begin{array}{cc}0& 1\\ 1& 0\end{array}\right]$

$\left[\begin{array}{cc}0.5& 1.5\\ 1& -0.5\end{array}\right]$

$\frac{4}{7}\left[\begin{array}{cc}0.5& 1.5\\ 1& -0.5\end{array}\right]$

$\left[\begin{array}{ccc}1& 0& 6\\ -2& 1& 7\\ 3& 0& 2\end{array}\right]$

$\left[\begin{array}{ccc}0& 1& -3\\ 4& 1& 0\\ 1& 0& 5\end{array}\right]$

$\frac{1}{17}\left[\begin{array}{ccc}-5& 5& -3\\ 20& -3& 12\\ 1& -1& 4\end{array}\right]$

$\left[\begin{array}{ccc}1& 2& -1\\ -3& 4& 1\\ -2& -4& -5\end{array}\right]$

$\left[\begin{array}{ccc}1& 9& -3\\ 2& 5& 6\\ 4& -2& 7\end{array}\right]$

$\frac{1}{209}\left[\begin{array}{ccc}47& -57& 69\\ 10& 19& -12\\ -24& 38& -13\end{array}\right]$

$\left[\begin{array}{ccc}1& -2& 3\\ -4& 8& -12\\ 1& 4& 2\end{array}\right]$

$\left[\begin{array}{ccc}\frac{1}{2}& \frac{1}{2}& \frac{1}{2}\\ \frac{1}{3}& \frac{1}{4}& \frac{1}{5}\\ \frac{1}{6}& \frac{1}{7}& \frac{1}{8}\end{array}\right]$

$\left[\begin{array}{ccc}18& 60& -168\\ -56& -140& 448\\ 40& 80& -280\end{array}\right]$

$\left[\begin{array}{ccc}1& 2& 3\\ 4& 5& 6\\ 7& 8& 9\end{array}\right]$

For the following exercises, solve the system using the inverse of a$\text{\hspace{0.17em}}2\text{\hspace{0.17em}}×\text{\hspace{0.17em}}2\text{\hspace{0.17em}}$matrix.

$\left(-5,6\right)$

$\begin{array}{l}8x+4y=-100\\ 3x-4y=1\end{array}$

$\begin{array}{l}\text{\hspace{0.17em}}3x-2y=6\hfill \\ -x+5y=-2\hfill \end{array}$

$\left(2,0\right)$

$\begin{array}{l}5x-4y=-5\hfill \\ \text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}4x+y=2.3\hfill \end{array}$

$\begin{array}{l}-3x-4y=9\hfill \\ \text{\hspace{0.17em}}12x+4y=-6\hfill \end{array}$

$\left(\frac{1}{3},-\frac{5}{2}\right)$

$\begin{array}{l}-2x+3y=\frac{3}{10}\hfill \\ \text{\hspace{0.17em}}\text{\hspace{0.17em}}-x+5y=\frac{1}{2}\hfill \end{array}$

$\begin{array}{l}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\frac{8}{5}x-\frac{4}{5}y=\frac{2}{5}\hfill \\ -\frac{8}{5}x+\frac{1}{5}y=\frac{7}{10}\hfill \end{array}$

$\left(-\frac{2}{3},-\frac{11}{6}\right)$

$\begin{array}{l}\frac{1}{2}x+\frac{1}{5}y=-\frac{1}{4}\\ \frac{1}{2}x-\frac{3}{5}y=-\frac{9}{4}\end{array}$

For the following exercises, solve a system using the inverse of a$\text{\hspace{0.17em}}3\text{}×\text{}3\text{\hspace{0.17em}}$ matrix.

$\begin{array}{l}3x-2y+5z=21\hfill \\ \text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}5x+4y=37\hfill \\ \text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}x-2y-5z=5\hfill \end{array}$

$\left(7,\frac{1}{2},\frac{1}{5}\right)$

$\left(5,0,-1\right)$

$\begin{array}{l}6x-5y+2z=-4\hfill \\ \text{\hspace{0.17em}}\text{\hspace{0.17em}}2x+5y-z=12\hfill \\ \text{\hspace{0.17em}}\text{\hspace{0.17em}}2x+5y+z=12\hfill \end{array}$

$\begin{array}{l}4x-2y+3z=-12\hfill \\ 2x+2y-9z=33\hfill \\ \text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}6y-4z=1\hfill \end{array}$

$\frac{1}{34}\left(-35,-97,-154\right)$

$\begin{array}{l}\frac{1}{10}x-\frac{1}{5}y+4z=\frac{-41}{2}\\ \frac{1}{5}x-20y+\frac{2}{5}z=-101\\ \frac{3}{10}x+4y-\frac{3}{10}z=23\end{array}$

$\begin{array}{l}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\frac{1}{2}x-\frac{1}{5}y+\frac{1}{5}z=\frac{31}{100}\hfill \\ -\frac{3}{4}x-\frac{1}{4}y+\frac{1}{2}z=\frac{7}{40}\hfill \\ -\frac{4}{5}x-\frac{1}{2}y+\frac{3}{2}z=\frac{1}{4}\hfill \end{array}$

$\frac{1}{690}\left(65,-1136,-229\right)$

$\begin{array}{l}0.1x+0.2y+0.3z=-1.4\hfill \\ 0.1x-0.2y+0.3z=0.6\hfill \\ \text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}0.4y+0.9z=-2\hfill \end{array}$

## Technology

For the following exercises, use a calculator to solve the system of equations with matrix inverses.

$\begin{array}{l}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}2x-y=-3\hfill \\ -x+2y=2.3\hfill \end{array}$

$\left(-\frac{37}{30},\frac{8}{15}\right)$

$\begin{array}{l}-\frac{1}{2}x-\frac{3}{2}y=-\frac{43}{20}\hfill \\ \text{\hspace{0.17em}}\text{\hspace{0.17em}}\frac{5}{2}x+\frac{11}{5}y=\frac{31}{4}\hfill \end{array}$

$\begin{array}{l}12.3x-2y-2.5z=2\hfill \\ 36.9x+7y-7.5z=-7\hfill \\ \text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}8y-5z=-10\hfill \end{array}$

$\left(\frac{10}{123},-1,\frac{2}{5}\right)$

$\begin{array}{l}0.5x-3y+6z=-0.8\hfill \\ \text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}0.7x-2y=-0.06\hfill \\ 0.5x+4y+5z=0\hfill \end{array}$

## Extensions

For the following exercises, find the inverse of the given matrix.

$\left[\begin{array}{cccc}1& 0& 1& 0\\ 0& 1& 0& 1\\ 0& 1& 1& 0\\ 0& 0& 1& 1\end{array}\right]$

$\frac{1}{2}\left[\begin{array}{rrrr}\hfill 2& \hfill 1& \hfill -1& \hfill -1\\ \hfill 0& \hfill 1& \hfill 1& \hfill -1\\ \hfill 0& \hfill -1& \hfill 1& \hfill 1\\ \hfill 0& \hfill 1& \hfill -1& \hfill 1\end{array}\right]$

$\left[\begin{array}{rrrr}\hfill -1& \hfill 0& \hfill 2& \hfill 5\\ \hfill 0& \hfill 0& \hfill 0& \hfill 2\\ \hfill 0& \hfill 2& \hfill -1& \hfill 0\\ \hfill 1& \hfill -3& \hfill 0& \hfill 1\end{array}\right]$

$\left[\begin{array}{rrrr}\hfill 1& \hfill -2& \hfill 3& \hfill 0\\ \hfill 0& \hfill 1& \hfill 0& \hfill 2\\ \hfill 1& \hfill 4& \hfill -2& \hfill 3\\ \hfill -5& \hfill 0& \hfill 1& \hfill 1\end{array}\right]$

$\frac{1}{39}\left[\begin{array}{rrrr}\hfill 3& \hfill 2& \hfill 1& \hfill -7\\ \hfill 18& \hfill -53& \hfill 32& \hfill 10\\ \hfill 24& \hfill -36& \hfill 21& \hfill 9\\ \hfill -9& \hfill 46& \hfill -16& \hfill -5\end{array}\right]$

$\left[\begin{array}{rrrrr}\hfill 1& \hfill 2& \hfill 0& \hfill 2& \hfill 3\\ \hfill 0& \hfill 2& \hfill 1& \hfill 0& \hfill 0\\ \hfill 0& \hfill 0& \hfill 3& \hfill 0& \hfill 1\\ \hfill 0& \hfill 2& \hfill 0& \hfill 0& \hfill 1\\ \hfill 0& \hfill 0& \hfill 1& \hfill 2& \hfill 0\end{array}\right]$

$\left[\begin{array}{rrrrrr}\hfill 1& \hfill 0& \hfill 0& \hfill 0& \hfill 0& \hfill 0\\ \hfill 0& \hfill 1& \hfill 0& \hfill 0& \hfill 0& \hfill 0\\ \hfill 0& \hfill 0& \hfill 1& \hfill 0& \hfill 0& \hfill 0\\ \hfill 0& \hfill 0& \hfill 0& \hfill 1& \hfill 0& \hfill 0\\ \hfill 0& \hfill 0& \hfill 0& \hfill 0& \hfill 1& \hfill 0\\ \hfill 1& \hfill 1& \hfill 1& \hfill 1& \hfill 1& \hfill 1\end{array}\right]$

$\left[\begin{array}{rrrrrr}\hfill 1& \hfill 0& \hfill 0& \hfill 0& \hfill 0& \hfill 0\\ \hfill 0& \hfill 1& \hfill 0& \hfill 0& \hfill 0& \hfill 0\\ \hfill 0& \hfill 0& \hfill 1& \hfill 0& \hfill 0& \hfill 0\\ \hfill 0& \hfill 0& \hfill 0& \hfill 1& \hfill 0& \hfill 0\\ \hfill 0& \hfill 0& \hfill 0& \hfill 0& \hfill 1& \hfill 0\\ \hfill -1& \hfill -1& \hfill -1& \hfill -1& \hfill -1& \hfill 1\end{array}\right]$

## Real-World Applications

For the following exercises, write a system of equations that represents the situation. Then, solve the system using the inverse of a matrix.

2,400 tickets were sold for a basketball game. If the prices for floor 1 and floor 2 were different, and the total amount of money brought in is \$64,000, how much was the price of each ticket?

In the previous exercise, if you were told there were 400 more tickets sold for floor 2 than floor 1, how much was the price of each ticket?

Infinite solutions.

A food drive collected two different types of canned goods, green beans and kidney beans. The total number of collected cans was 350 and the total weight of all donated food was 348 lb, 12 oz. If the green bean cans weigh 2 oz less than the kidney bean cans, how many of each can was donated?

Students were asked to bring their favorite fruit to class. 95% of the fruits consisted of banana, apple, and oranges. If oranges were twice as popular as bananas, and apples were 5% less popular than bananas, what are the percentages of each individual fruit?

50% oranges, 25% bananas, 20% apples

A sorority held a bake sale to raise money and sold brownies and chocolate chip cookies. They priced the brownies at \$1 and the chocolate chip cookies at \$0.75. They raised \$700 and sold 850 items. How many brownies and how many cookies were sold?

A clothing store needs to order new inventory. It has three different types of hats for sale: straw hats, beanies, and cowboy hats. The straw hat is priced at \$13.99, the beanie at \$7.99, and the cowboy hat at \$14.49. If 100 hats were sold this past quarter, \$1,119 was taken in by sales, and the amount of beanies sold was 10 more than cowboy hats, how many of each should the clothing store order to replace those already sold?

10 straw hats, 50 beanies, 40 cowboy hats

Anna, Ashley, and Andrea weigh a combined 370 lb. If Andrea weighs 20 lb more than Ashley, and Anna weighs 1.5 times as much as Ashley, how much does each girl weigh?

Three roommates shared a package of 12 ice cream bars, but no one remembers who ate how many. If Tom ate twice as many ice cream bars as Joe, and Albert ate three less than Tom, how many ice cream bars did each roommate eat?

Tom ate 6, Joe ate 3, and Albert ate 3.

A farmer constructed a chicken coop out of chicken wire, wood, and plywood. The chicken wire cost \$2 per square foot, the wood \$10 per square foot, and the plywood \$5 per square foot. The farmer spent a total of \$51, and the total amount of materials used wasHe usedmore chicken wire than plywood. How much of each material in did the farmer use?

Jay has lemon, orange, and pomegranate trees in his backyard. An orange weighs 8 oz, a lemon 5 oz, and a pomegranate 11 oz. Jay picked 142 pieces of fruit weighing a total of 70 lb, 10 oz. He picked 15.5 times more oranges than pomegranates. How many of each fruit did Jay pick?

124 oranges, 10 lemons, 8 pomegranates

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# Intro to two-step equations

CCSS.Math:

## Video transcript

Now we mixed up things a little bit more: on the left side of the scale, not only do we have these identical unknown masses with mass X, these three blue things, we also have some of the 1kg masses over here, actually, we have two of them. Now, we are going to figure out what X is. But before we even do that, I want you to think about a mathematical equation that can represent what is going on; that equates what we have on the left hand, with what we have on the right side of the scale. I will give you a few seconds to think about it... So let's think about what we have on the left side: we have 3 masses with mass X, so you can say we have 3x and then we have 2 masses of 1 kilogram, so in total we have 2 kg. So + 2. So one way to think about the total mass on the left-hand side is 3x + 2. Three masses with mass X, plus two kilograms. That is what we have on the left-hand side. Now, let us think about what we have on the right-hand side. We can simply count them: [counts to 14] Fourteen blocks, each has a mass of 1 kg, so the total mass will be 14 kg. And we see that the scale is balanced, not tilting down or upwards. So this mass over here must be equal to this total mass. The scale is balanced, so we can write an 'equal'-sign. (let me do that in a white coulour, I do not like that brown) Now, what I want you to think about, and you can think about it either through the symbols or through the scales, is: how would you go about -- let us think about a few things: how would you first go about at least getting rid of these little 1kg blocks? I will give you a second to think about that... Well, the simplest thing is: you can take these 1kg blocks off of the left-hand side, but remember, if you just took these blocks off of the left-hand side, and it was balanced before, now the left-hand side will be lighter and it will move up. But we want to keep it balanced so we can keep saying 'equal'. That this mass is equal to that mass. So, if we remove 2 block from the left-hand side, we need to remove 2 from the right-hand side. So, we can remove two there, and then we can remove two over there. Mathematically, what we are doing is: we are subtracting 2 kilograms from each side. We are subtracting 2 from this side, So on the left-hand side we now have 3x + 2, minus 2 we are left with just 3x, and on the right-hand side we had 14 and we took away 2 (let me write this:) we took away 2, so we are going to be left with 12 blocks. And you see that there, the ones that I have not crossed out, there are 12 left, and here you have 3 of those X-blocks. Since we removed the same amount from both sides, our scale is still balanced. And our equation: 3x is now equal to 12. Now, this turns into a problem very similar to what we saw in the last video, so now I ask you: what can we do to isolate one x, to only have one 'X' on the left-hand side of the scale, while keeping the scale balanced? The easiest way to think about it is: If I want one X on this left-hand side, that is a third of the total X's here. So what if I were to multiply the left-hand side by one-third -- -- but if I want to keep the scale balanced, I have to multiply the right-hand side by one-third. If we can do that mathematically, Over here I can multiply the left-hand side by 1/3, and if I want to keep my scale balanced I also have to multiply the right-hand side by 1/3. Multiplying it physically literally means: just keeping a third of what we had originally We would get rid of two of these. If we want to keep a third of what we had here originally, -- there are 12 blocks left over after removing those first two -- so, 1/3 of 12: we are only going to have four of these little 1kg boxes left. Let me remove all but four. (so, remove those, and those...) And I have left [counts them] 4 here. And so, what you are left with, the only thing you have left, is this 'X' - I will shade it in to show this is the one we actually have left - and then we have these 1 kilogram boxes. You see it mathematically over here: 1/3 * 3x -- or you could have said 3x divided by 3 -- either way, that gives us -- these threes cancel out, so that would give you an 'X' and on the right-hand side: 12 * 1/3 - which is the same as 12/3, is equal to 4. And, since we did the same thing to both sides, the scale is still balanced. So you see that the mass of this thing must be the same as the mass of these 4 left-over blocks. x must be equal to 4 kilograms.


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Area of any region is the space occupied by the figure on a two dimensional plane surface. We are familiar with the area of a square, rectangle which are obtained by multiplying the length by the width. Circle is a closed figure where a point moves such that it has constant distance from the fixed point. The fixed point is called the center of the circle and the constant distance is called the radius of the circle. We have different terms in circle called, diameter, chord, sector, segment and circumference of a circle. In this section let us discuss with the area of a circle and a semicircle.

## Area of a Circle Formula

Diameter and Radius of a circle: Diameter of a circle is the line segment ( chord ) joining any two points on the circumference of the circle, which passes through the center.
Radius of a circle is the line segment joining the center of a circle to any point on the circumference of the circle.
The following diagram describes the above terms.

Area of a circle is calculated using the formula $\pi$ $r^{2}$

where ' r ' is the radius of the circle and $\pi$ an irrational constant = 3.141592653589793238462643383279.
So approximately we take $\pi$ = 3.14 correct to 2 decimal places.

Surface Areas: Surface areas are calculated for three dimensional solid figures like, cube, cuboid, sphere, cylinder, cone etc. Since the circle is a two dimensional figure which is drawn on a plane surface, the surface area of a circle is the same as the area of the circle.

### Annulus

 Area Circle Equation Area of a Sector of a Circle Area of a Segment of a Circle Area of an Arc of a Circle Area of Chord of Circle Area of Concentric Circles A Circle A Sector of a Circle Center of the Circle Circumferance of Circle Congruent Circle Part of a Circle
 Area Calculator Circle Circle Solver Circumference a Circle


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[web] Asp.net-like master pages in php

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As it turns out, I have a small project coming up where I have to build a simple framework for a personal portfolio. The technology in use is PHP and up to this point I was planning on taking my usual template route. However, with a profane amount of time to let my mind wonder at work yesterday I started thinking back on my few experiences with ASP.net. Without a doubt, for me, one of the coolest features that was immediately accessible was the use of master pages and content pages. I've been pondering how I might be able to emulate such a feature since then. My initial thought was that I'd probably have to write some kind of CGI script to intercept the file request and process it there but since I don't know squat about how to do that I figured I should try to solve my problem with Php first. Currently, the only idea I can come up with is to build a master page as you would in Asp.NET by using <contentplaceholder id> tags everywhere that you would expect things to be inserted. At the top of every content page it would then be a simple matter of calling a utility interface, maybe like so:
<?php
include "masterpageutil.php"
DisplayInMasterPage("some-master-page-file");
?>

The ugly side of this however is in how it would be handled. Currently, my meager knowledge of Php has lead me to this point: Upon calling the above display function, I would open up the current page as a file, parse through it's contents and collect everything contained within <content id> tags in an associative array based on their id. Then I would open the given master page file, parse its contents and look for <contentplacholder id> tags. Everywhere I find a placeholder that has an id matching a previously stored content section, i would just insert the text. All of this might be stuffed into a rather giant string and finally echoed out to the browser. Obviously, this has a lot of pitfalls, not the least of which is that I lose all ability to process actual php scripts in the page. Luckily, the person I'm building this for isn't really keen on anything beyond HTML and CSS and so, won't have much use for php (otherwise I would have just gone with the template scheme in the first place). Also, I imagine this would probably not scale well in bigger site. However, since this is for a personal website with a bit of portfolio information I doubt that will be much of an issue. Are there any suggestions on how I might better handle this? Any other serious issues that can be foreseen both in terms of security and functionality?