Solving Systems with Inverses

Learning Objectives

In this section, you will:

• Find the inverse of a matrix.
• Solve a system of linear equations using an inverse matrix.

Soriya 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 Soriya 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 a𝑎 is a−1,𝑎−1, and aa−1=a−1a=(1a)a=1.𝑎𝑎−1=𝑎−1𝑎=(1𝑎)𝑎=1. For example, 2−1=122−1=12 and (12)2=1.(12)2=1. The multiplicative inverse of a matrix is similar in concept, except that the product of matrix A𝐴 and its inverse A−1𝐴−1 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 In𝐼𝑛 where n𝑛 represents the dimension of the matrix. Observe the following equations.

I2=[1001]𝐼2=[1001]

I3=⎡⎣⎢100010001⎤⎦⎥𝐼3=[100010001]

The identity matrix acts as a 1 in matrix algebra. For example, AI=IA=A.𝐴𝐼=𝐼𝐴=𝐴.

A matrix that has a multiplicative inverse has the properties

AA−1=IA−1A=I𝐴𝐴−1=𝐼𝐴−1𝐴=𝐼

A matrix that has a multiplicative inverse is called an invertible matrix. Only a square matrix may have a multiplicative inverse, as the reversibility, AA−1=A−1A=I,𝐴𝐴−1=𝐴−1𝐴=𝐼, is a requirement. Not all square matrices have an inverse, but if A𝐴 is invertible, then A−1𝐴−1 is unique. We will look at two methods for finding the inverse of a 2×22×2 matrix and a third method that can be used on both 2×22×2 and 3×33×3 matrices.

THE IDENTITY MATRIX AND MULTIPLICATIVE INVERSE

The identity matrix, In,𝐼𝑛, is a square matrix containing ones down the main diagonal and zeros everywhere else.

I2=[1001]I3=⎡⎣⎢100010001⎤⎦⎥       2×2                3×3𝐼2=[1001]𝐼3=[100010001]       2×2                3×3

If A𝐴 is an n×n𝑛×𝑛 matrix and B𝐵 is an n×n𝑛×𝑛 matrix such that AB=BA=In,𝐴𝐵=𝐵𝐴=𝐼𝑛, then B=A−1,𝐵=𝐴−1, the multiplicative inverse of a matrix A.𝐴.

EXAMPLE 1

Showing That the Identity Matrix Acts as a 1

Given matrix A, show that AI=IA=A.𝐴𝐼=𝐼𝐴=𝐴.

A=[3−245]𝐴=[34−25]

Solution

Use matrix multiplication to show that the product of A𝐴 and the identity is equal to the product of the identity and A.

AI=[3−245][1001]=[3⋅1+4⋅0−2⋅1+5⋅03⋅0+4⋅1−2⋅0+5⋅1]=[3−245]𝐴𝐼=[34−25][1001]=[3⋅1+4⋅03⋅0+4⋅1−2⋅1+5⋅0−2⋅0+5⋅1]=[34−25]

IA=[1001][3−245]=[1⋅3+0⋅(−2)0⋅3+1⋅(−2)1⋅4+0⋅50⋅4+1⋅5]=[3−245]𝐼𝐴=[1001][34−25]=[1⋅3+0⋅(−2)1⋅4+0⋅50⋅3+1⋅(−2)0⋅4+1⋅5]=[34−25]

HOW TO

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

1. Given matrix A𝐴 of order n×n𝑛×𝑛 and matrix B𝐵 of order n×n𝑛×𝑛 multiply AB.𝐴𝐵.
2. If AB=I,𝐴𝐵=𝐼, then find the product BA.𝐵𝐴. If BA=I,𝐵𝐴=𝐼, then B=A−1𝐵=𝐴−1 and A=B−1.𝐴=𝐵−1.

EXAMPLE 2

Showing That Matrix A Is the Multiplicative Inverse of Matrix B

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

A=[1−25−9],B=[−92−51]𝐴=[15−2−9],𝐵=[−9−521]

Solution

Multiply AB𝐴𝐵 and BA.𝐵𝐴. If both products equal the identity, then the two matrices are inverses of each other.

AB=[1−25−9]⋅[−92−51]=[1(−9)+5(2)−2(−9)−9(2)1(−5)+5(1)−2(−5)−9(1)]=[1001]𝐴𝐵=[15−2−9]·[−9−521]=[1(−9)+5(2)1(−5)+5(1)−2(−9)−9(2)−2(−5)−9(1)]=[1001]

BA=[−92−51]⋅[1−25−9]=[−9(1)−5(−2)2(1)+1(−2)−9(5)−5(−9)2(5)+1(−9)]=[1001]𝐵𝐴=[−9−521]·[15−2−9]=[−9(1)−5(−2)−9(5)−5(−9)2(1)+1(−2)2(5)+1(−9)]=[1001]

A𝐴 and B𝐵 are inverses of each other.

TRY IT #1

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

A=[1−14−3],B=[−31−41]𝐴=[14−1−3],𝐵=[−3−411]

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.

EXAMPLE 3

Finding the Multiplicative Inverse Using Matrix Multiplication

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

A=[12−2−3]𝐴=[1−22−3]

Solution

For this method, we multiply A𝐴 by a matrix containing unknown constants and set it equal to the identity.

[12−2−3]  [acbd]=[1001][1−22−3]  [𝑎𝑏𝑐𝑑]=[1001]

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

[12−2−3]  [acbd]=[1a−2c2a−3c1b−2d2b−3d][1−22−3]  [𝑎𝑏𝑐𝑑]=[1𝑎−2𝑐1𝑏−2𝑑2𝑎−3𝑐2𝑏−3𝑑]

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.

1a−2c=1   R12a−3c=0   R21𝑎−2𝑐=1   𝑅12𝑎−3𝑐=0   𝑅2

Using row operations, multiply and add as follows: (−2)R1+R2→R2.(−2)𝑅1+𝑅2→𝑅2. Add the equations, and solve for c.𝑐.

1a−2c=10+1c=−2c=−21𝑎−2𝑐=10+1𝑐=−2𝑐=−2

Back-substitute to solve for a.𝑎.

a−2(−2)=1a+4=1a=−3𝑎−2(−2)=1𝑎+4=1𝑎=−3

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.

1b−2d=02b−3d=1R1R21𝑏−2𝑑=0𝑅12𝑏−3𝑑=1𝑅2

Using row operations, multiply and add as follows: (−2)R1+R2=R2.(−2)𝑅1+𝑅2=𝑅2. Add the two equations and solve for d.𝑑.

1b−2d=00+1d=1d=11𝑏−2𝑑=00+1𝑑=1𝑑=1

Once more, back-substitute and solve for b.𝑏.

b−2(1)=0b−2=0b=2𝑏−2(1)=0𝑏−2=0𝑏=2

A−1=[−3−221]𝐴−1=[−32−21]

Finding the Multiplicative Inverse by Augmenting with the Identity

Another way to find the multiplicative inverse is by augmenting with the identity. When matrix A𝐴 is transformed into I,𝐼, the augmented matrix I𝐼 transforms into A−1.𝐴−1.

For example, given

A=[2513]𝐴=[2153]

augment A𝐴 with the identity

[2513∣∣∣1001][2153|1001]

Perform row operations with the goal of turning A𝐴 into the identity.

1. Switch row 1 and row 2.[5231∣∣∣0110][5321|0110]
2. Multiply row 2 by −2−2 and add to row 1.[1211∣∣∣−2110][1121|−2110]
3. Multiply row 1 by −2−2 and add to row 2.[101−1∣∣∣−251−2][110−1|−215−2]
4. Add row 2 to row 1.[100−1∣∣∣35−1−2][100−1|3−15−2]
5. Multiply row 2 by −1.−1.[1001∣∣∣3−5−12][1001|3−1−52]

The matrix we have found is A−1.𝐴−1.

A−1=[3−5−12]𝐴−1=[3−1−52]

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

When we need to find the multiplicative inverse of a 2×22×2 matrix, we can use a special formula instead of using matrix multiplication or augmenting with the identity.

If A𝐴 is a 2×22×2 matrix, such as

A=[acbd]𝐴=[𝑎𝑏𝑐𝑑]

the multiplicative inverse of A𝐴 is given by the formula

A−1=1ad−bc[d−c−ba]𝐴−1=1𝑎𝑑−𝑏𝑐[𝑑−𝑏−𝑐𝑎]

where ad−bc≠0.𝑎𝑑−𝑏𝑐≠0. If ad−bc=0,𝑎𝑑−𝑏𝑐=0, then A𝐴 has no inverse.

EXAMPLE 4

Using the Formula to Find the Multiplicative Inverse of Matrix A

Use the formula to find the multiplicative inverse of

A=[12−2−3]𝐴=[1−22−3]

Solution

Using the formula, we have

A−1=1(1)(−3)−(−2)(2)[−3−221]=1−3+4[−3−221]=[−3−221]𝐴−1=1(1)(−3)−(−2)(2)[−32−21]=1−3+4[−32−21]=[−32−21]

Analysis

We can check that our formula works by using one of the other methods to calculate the inverse. Let’s augment A𝐴 with the identity.

[12−2−3∣∣∣1001][1−22−3|1001]

Perform row operations with the goal of turning A𝐴 into the identity.

1. Multiply row 1 by −2−2 and add to row 2.[10−21∣∣∣1−201][1−201|10−21]
2. Multiply row 2 by 2 and add to row 1.[1001∣∣∣−3−221][1001|−32−21]

So, we have verified our original solution.

A−1=[−3−221]𝐴−1=[−32−21]

TRY IT #2

Use the formula to find the inverse of matrix A.𝐴. Verify your answer by augmenting with the identity matrix.

A=[12−13]𝐴=[1−123]

EXAMPLE 5

Finding the Inverse of the Matrix, If It Exists

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

A=[3162]𝐴=[3612]

Solution

We will use the method of augmenting with the identity.

[3162∣∣∣1001][3612|1001]

1. Switch row 1 and row 2.[1332∣∣∣0110][1332|0110]
2. Multiply row 1 by −3 and add it to row 2.[1020∣∣∣1−301][1200|10−31]
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 2×22×2 matrix to find the inverse of a 3×33×3 matrix. Instead, we will augment the original matrix with the identity matrix and use row operations to obtain the inverse.

Given a 3×33×3 matrix

A=⎡⎣⎢232334111⎤⎦⎥𝐴=[231331241]

augment A𝐴 with the identity matrix

A∣∣∣∣I=⎡⎣⎢232334111 ∣∣∣∣ 100010001⎤⎦⎥𝐴|𝐼=[231331241 | 100010001]

To begin, we write the augmented matrix with the identity on the right and A𝐴 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.

HOW TO

Given a 3×33×3 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 AA−1=I𝐴𝐴−1=𝐼 and A−1A=I.𝐴−1𝐴=𝐼.

EXAMPLE 6

Finding the Inverse of a 3 × 3 Matrix

Given the 3×33×3 matrix A,𝐴, find the inverse.

A=⎡⎣⎢232334111⎤⎦⎥𝐴=[231331241]

Solution

Augment A𝐴 with the identity matrix, and then begin row operations until the identity matrix replaces A.𝐴. The matrix on the right will be the inverse of A.𝐴.

⎡⎣⎢232334111∣∣∣∣100010001⎤⎦⎥→Interchange R2and R1⎡⎣⎢322334111∣∣∣∣010100001⎤⎦⎥[231331241|100010001]→Interchange 𝑅2and 𝑅1[331231241|010100001]

−R2+R1=R1→⎡⎣⎢122034011∣∣∣∣−110100001⎤⎦⎥−𝑅2+𝑅1=𝑅1→[100231241|−110100001]

−R2+R3=R3→⎡⎣⎢120031010∣∣∣∣−11−1100001⎤⎦⎥−𝑅2+𝑅3=𝑅3→[100231010|−110100−101]

R3↔ R2→⎡⎣⎢102013001∣∣∣∣−1−11100010⎤⎦⎥𝑅3↔ 𝑅2→[100010231|−110−101100]

−2R1+R3=R3→⎡⎣⎢100013001∣∣∣∣−1−1310−2010⎤⎦⎥−2𝑅1+𝑅3=𝑅3→[100010031|−110−1013−20]

−3R2+R3=R3→⎡⎣⎢100010001∣∣∣∣−1−1610−201−3⎤⎦⎥−3𝑅2+𝑅3=𝑅3→[100010001|−110−1016−2−3]

Thus,

A−1=B=⎡⎣⎢−1−1610−201−3⎤⎦⎥𝐴−1=𝐵=[−110−1016−2−3]

Analysis

To prove that B=A−1,𝐵=𝐴−1, let’s multiply the two matrices together to see if the product equals the identity, if AA−1=I𝐴𝐴−1=𝐼 and A−1A=I.𝐴−1𝐴=𝐼.

AA−1=⎡⎣⎢232334111⎤⎦⎥  ⎡⎣⎢−1−1610−201−3⎤⎦⎥=⎡⎣⎢2(−1)+3(−1)+1(6)3(−1)+3(−1)+1(6)2(−1)+4(−1)+1(6)2(1)+3(0)+1(−2)3(1)+3(0)+1(−2)2(1)+4(0)+1(−2)2(0)+3(1)+1(−3)3(0)+3(1)+1(−3)2(0)+4(1)+1(−3)⎤⎦⎥=⎡⎣⎢100010001⎤⎦⎥𝐴𝐴−1=[231331241]  [−110−1016−2−3]=[2(−1)+3(−1)+1(6)2(1)+3(0)+1(−2)2(0)+3(1)+1(−3)3(−1)+3(−1)+1(6)3(1)+3(0)+1(−2)3(0)+3(1)+1(−3)2(−1)+4(−1)+1(6)2(1)+4(0)+1(−2)2(0)+4(1)+1(−3)]=[100010001]

A−1A=⎡⎣⎢−1−1610−201−3⎤⎦⎥  ⎡⎣⎢232334111⎤⎦⎥=⎡⎣⎢−1(2)+1(3)+0(2)−1(2)+0(3)+1(2)6(2)+−2(3)+−3(2)−1(3)+1(3)+0(4)−1(3)+0(3)+1(4)6(3)+−2(3)+−3(4)−1(1)+1(1)+0(1)−1(1)+0(1)+1(1)6(1)+−2(1)+−3(1)⎤⎦⎥=⎡⎣⎢100010001⎤⎦⎥𝐴−1𝐴=[−110−1016−2−3]  [231331241]=[−1(2)+1(3)+0(2)−1(3)+1(3)+0(4)−1(1)+1(1)+0(1)−1(2)+0(3)+1(2)−1(3)+0(3)+1(4)−1(1)+0(1)+1(1)6(2)+−2(3)+−3(2)6(3)+−2(3)+−3(4)6(1)+−2(1)+−3(1)]=[100010001]

TRY IT #3

Find the inverse of the 3×33×3 matrix.

A=⎡⎣⎢2−10−1711311−7−2⎤⎦⎥𝐴=[2−1711−111−703−2]

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: X𝑋 is the matrix representing the variables of the system, and B𝐵 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

AX=B𝐴𝑋=𝐵

To solve a system of linear equations using an inverse matrix, let A𝐴 be the coefficient matrix, let X𝑋 be the variable matrix, and let B𝐵 be the constant matrix. Thus, we want to solve a system AX=B.𝐴𝑋=𝐵. For example, look at the following system of equations.

a1x+b1y=c1a2x+b2y=c2𝑎1𝑥+𝑏1𝑦=𝑐1𝑎2𝑥+𝑏2𝑦=𝑐2

From this system, the coefficient matrix is

A=[a1a2b1b2]𝐴=[𝑎1𝑏1𝑎2𝑏2]

The variable matrix is

X=[xy]𝑋=[𝑥𝑦]

And the constant matrix is

B=[c1c2]𝐵=[𝑐1𝑐2]

Then AX=B𝐴𝑋=𝐵 looks like

[a1a2b1b2]  [xy]=[c1c2][𝑎1𝑏1𝑎2𝑏2]  [𝑥𝑦]=[𝑐1𝑐2]

Recall the discussion earlier in this section regarding multiplying a real number by its inverse, (2−1)2=(12)2=1.(2−1)2=(12)2=1. To solve a single linear equation ax=b𝑎𝑥=𝑏 for x,𝑥, we would simply multiply both sides of the equation by the multiplicative inverse (reciprocal) of a.𝑎. Thus,

 ax=b (1a)ax=(1a)b(a−1  )ax=(a−1)b[(a−1)a]x=(a−1)b           1x=(a−1)b             x=(a−1)b 𝑎𝑥=𝑏 (1𝑎)𝑎𝑥=(1𝑎)𝑏(𝑎−1  )𝑎𝑥=(𝑎−1)𝑏[(𝑎−1)𝑎]𝑥=(𝑎−1)𝑏           1𝑥=(𝑎−1)𝑏             𝑥=(𝑎−1)𝑏

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 2×22×2 system and then move on to a 3×33×3 system.

SOLVING A SYSTEM OF EQUATIONS USING THE INVERSE OF A MATRIX

Given a system of equations, write the coefficient matrix A,𝐴, the variable matrix X,𝑋, and the constant matrix B.𝐵. Then

AX=B𝐴𝑋=𝐵

Multiply both sides by the inverse of A𝐴 to obtain the solution.

(A−1)AX=(A−1)B[(A−1)A]X=(A−1)BIX=(A−1)BX=(A−1)B(𝐴−1)𝐴𝑋=(𝐴−1)𝐵[(𝐴−1)𝐴]𝑋=(𝐴−1)𝐵𝐼𝑋=(𝐴−1)𝐵𝑋=(𝐴−1)𝐵

Q&A

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.

EXAMPLE 7

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

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

3x+8y=54x+11y=73𝑥+8𝑦=54𝑥+11𝑦=7

Solution

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

A=[34811],X=[xy],B=[57]𝐴=[38411],𝑋=[𝑥𝑦],𝐵=[57]

Then

[34811]  [xy]=[57][38411]  [𝑥𝑦]=[57]

First, we need to calculate A−1.𝐴−1. Using the formula to calculate the inverse of a 2 by 2 matrix, we have:

A−1=1ad−bc[d−c−ba]     =13(11)−8(4)[11−4−83]     =11[11−4−83]𝐴−1=1𝑎𝑑−𝑏𝑐[𝑑−𝑏−𝑐𝑎]     =13(11)−8(4)[11−8−43]     =11[11−8−43]

So,

A−1=[11−4−8​​3]𝐴−1=[11−8−4​​3]

Now we are ready to solve. Multiply both sides of the equation by A−1.𝐴−1.

(A−1)AX=(A−1)B[11−4−83]  [34811]  [xy]=[11−4−83]  [57][1001]  [xy]=[11(5)+(−8)7−4(5)+3(7)][xy]=[−11](𝐴−1)𝐴𝑋=(𝐴−1)𝐵[11−8−43]  [38411]  [𝑥𝑦]=[11−8−43]  [57][1001]  [𝑥𝑦]=[11(5)+(−8)7−4(5)+3(7)][𝑥𝑦]=[−11]

The solution is (−1,1).(−1,1).

Q&A

Can we solve for X𝑋 by finding the product BA−1?𝐵𝐴−1?

No, recall that matrix multiplication is not commutative, so A−1B≠BA−1.𝐴−1𝐵≠𝐵𝐴−1. Consider our steps for solving the matrix equation.

(A−1)AX=(A−1)B[(A−1)A]X=(A−1)BIX=(A−1)BX=(A−1)B(𝐴−1)𝐴𝑋=(𝐴−1)𝐵[(𝐴−1)𝐴]𝑋=(𝐴−1)𝐵𝐼𝑋=(𝐴−1)𝐵𝑋=(𝐴−1)𝐵

Notice in the first step we multiplied both sides of the equation by A−1,𝐴−1, but the A−1𝐴−1 was to the left of A𝐴 on the left side and to the left of B𝐵 on the right side. Because matrix multiplication is not commutative, order matters.

EXAMPLE 8

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

Solve the following system using the inverse of a matrix.

5x+15y+56z=35−4x−11y−41z=−26−x−3y−11z=−75𝑥+15𝑦+56𝑧=35−4𝑥−11𝑦−41𝑧=−26−𝑥−3𝑦−11𝑧=−7

Solution

Write the equation AX=B.𝐴𝑋=𝐵.

⎡⎣⎢5−4−115−11−356−41−11⎤⎦⎥  ⎡⎣⎢xyz⎤⎦⎥=⎡⎣⎢35−26−7⎤⎦⎥[51556−4−11−41−1−3−11]  [𝑥𝑦𝑧]=[35−26−7]

First, we will find the inverse of A𝐴 by augmenting with the identity.

⎡⎣⎢5−4−115−11−356−41−11∣∣∣∣100010001⎤⎦⎥[51556−4−11−41−1−3−11|100010001]

Multiply row 1 by 15.15.

⎡⎣⎢1−4−13−11−3565−41−11∣∣∣∣∣1500010001⎤⎦⎥[13565−4−11−41−1−3−11|1500010001]

Multiply row 1 by 4 and add to row 2.

⎡⎣⎢⎢10−131−3565195−11∣∣∣∣∣15450010001⎤⎦⎥⎥[1356501195−1−3−11|15004510001]

Add row 1 to row 3.

⎡⎣⎢⎢⎢10031056519515∣∣∣∣∣154515010001⎤⎦⎥⎥⎥[13565011950015|150045101501]

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

⎡⎣⎢⎢100010−1519515∣∣∣∣∣−1154515−310001⎤⎦⎥⎥[10−15011950015|−115−3045101501]

Multiply row 3 by 5.

⎡⎣⎢⎢100010−151951∣∣∣∣∣−115451−310005⎤⎦⎥⎥[10−1501195001|−115−304510105]

Multiply row 3 by 1515 and add to row 1.

⎡⎣⎢10001001951∣∣∣∣∣−2451−310105⎤⎦⎥[10001195001|−2−314510105]

Multiply row 3 by −195−195 and add to row 2.

⎡⎣⎢100010001∣∣∣∣−2−31−3101−195⎤⎦⎥[100010001|−2−31−31−19105]

So,

A−1=⎡⎣⎢−2−31−3101−195⎤⎦⎥𝐴−1=[−2−31−31−19105]

Multiply both sides of the equation by A−1.𝐴−1. We want A−1AX=A−1B:𝐴−1𝐴𝑋=𝐴−1𝐵:

⎡⎣⎢−2−31−3101−195⎤⎦⎥  ⎡⎣⎢5−4−115−11−356−41−11⎤⎦⎥  ⎡⎣⎢xyz⎤⎦⎥=⎡⎣⎢−2−31−3101−195⎤⎦⎥  ⎡⎣⎢35−26−7⎤⎦⎥[−2−31−31−19105]  [51556−4−11−41−1−3−11]  [𝑥𝑦𝑧]=[−2−31−31−19105]  [35−26−7]

Thus,

A−1B=⎡⎣⎢−70+78−7−105−26+13335+0−35⎤⎦⎥=⎡⎣⎢120⎤⎦⎥𝐴−1𝐵=[−70+78−7−105−26+13335+0−35]=[120]

The solution is (1,2,0).(1,2,0).

TRY IT #4

Solve the system using the inverse of the coefficient matrix.

 2x−17y+11z=0 −x+11y−7z=8              3y−2z=−2 2𝑥−17𝑦+11𝑧=0 −𝑥+11𝑦−7𝑧=8              3𝑦−2𝑧=−2

HOW TO

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

1. Save the coefficient matrix and the constant matrix as matrix variables [A][𝐴] and [B].[𝐵].
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.

EXAMPLE 9

Using a Calculator to Solve a System of Equations with Matrix Inverses

Solve the system of equations with matrix inverses using a calculator

2x+3y+z=323x+3y+z=−272x+4y+z=−22𝑥+3𝑦+𝑧=323𝑥+3𝑦+𝑧=−272𝑥+4𝑦+𝑧=−2

Solution

On the matrix page of the calculator, enter the coefficient matrix as the matrix variable [A],[𝐴], and enter the constant matrix as the matrix variable [B].[𝐵].

[A]=⎡⎣⎢232334111⎤⎦⎥, [B]=⎡⎣⎢32−27−2⎤⎦⎥[𝐴]=[231331241], [𝐵]=[32−27−2]

On the home screen of the calculator, type in the multiplication to solve for X,𝑋, calling up each matrix variable as needed.

[A]−1×[B][𝐴]−1×[𝐵]

Evaluate the expression.

⎡⎣⎢−59−34252⎤⎦⎥[−59−34252]