Solve Systems with Inverses of Matrices*

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.

Find the Inverse of a Matrix

We know that the multiplicative inverse of a real number [latex]a[/latex] is [latex]{a}^{-1}[/latex], and [latex]a{a}^{-1}={a}^{-1}a=\left(\frac{1}{a}\right)a=1[/latex]. For example, [latex]{2}^{-1}=\frac{1}{2}[/latex] and [latex]\left(\frac{1}{2}\right)2=1[/latex]. The multiplicative inverse of a matrix is similar in concept, except that the product of matrix [latex]A[/latex] and its inverse [latex]{A}^{-1}[/latex] 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 [latex]{I}_{n}[/latex] where [latex]n[/latex] represents the dimension of the matrix. The equations below are the identity matrices for a [latex]2\text{}\times \text{}2[/latex] matrix and a [latex]3\text{}\times \text{}3[/latex] matrix, respectively.

[latex]{I}_{2}=\left[\begin{array}{rrr}\hfill 1& \hfill & \hfill 0\\ \hfill 0& \hfill & \hfill 1\end{array}\right][/latex]

[latex]{I}_{3}=\left[\begin{array}{rrrrr}\hfill 1& \hfill & \hfill 0& \hfill & \hfill 0\\ \hfill 0& \hfill & \hfill 1& \hfill & \hfill 0\\ \hfill 0& \hfill & \hfill 0& \hfill & \hfill 1\end{array}\right][/latex]

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

A matrix that has a multiplicative inverse has the properties

[latex]\begin{array}{l}A{A}^{-1}=I\\ {A}^{-1}A=I\end{array}[/latex]

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

A General Note: The Identity Matrix and Multiplicative Inverse

The identity matrix, [latex]{I}_{n}[/latex], is a square matrix containing ones down the main diagonal and zeros everywhere else.

[latex]\begin{array}{l}\hfill\begin{array}{l}\begin{array}{l}\hfill \\ {I}_{2}=\left[\begin{array}{rr}\hfill 1& \hfill 0\\ \hfill 0& \hfill 1\end{array}\right]\begin{array}{cccc}& & & \end{array}{I}_{3}=\left[\begin{array}{rrr}\hfill 1& \hfill 0& \hfill 0\\ \hfill 0& \hfill 1& \hfill 0\\ \hfill 0& \hfill 0& \hfill 1\end{array}\right]\hfill \end{array}\hfill \\ \text{ }2\times 2\text{ 3}\times 3\hfill \end{array}\hfill \end{array}[/latex]

If [latex]A[/latex] is an [latex]n\times n[/latex] matrix and [latex]B[/latex] is an [latex]n\times n[/latex] matrix such that [latex]AB=BA={I}_{n}[/latex], then [latex]B={A}^{-1}[/latex], the multiplicative inverse of a matrix [latex]A[/latex].

Example: Showing That the Identity Matrix Acts as a 1

Given matrix A, show that [latex]AI=IA=A[/latex].
[latex]A=\left[\begin{array}{cc}3& 4\\ -2& 5\end{array}\right][/latex]

Solution

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

Given matrix [latex]A[/latex] of order [latex]n\times n[/latex] and matrix [latex]B[/latex] of order [latex]n\times n[/latex] multiply [latex]AB[/latex].
If [latex]AB=I[/latex], then find the product [latex]BA[/latex]. If [latex]BA=I[/latex], then [latex]B={A}^{-1}[/latex] and [latex]A={B}^{-1}[/latex].

Example: Showing That Matrix A Is the Multiplicative Inverse of Matrix B

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

[latex]A=\left[\begin{array}{rrr}\hfill 1& \hfill & \hfill 5\\ \hfill -2& \hfill & \hfill -9\end{array}\right],B=\left[\begin{array}{rrr}\hfill -9& \hfill & \hfill -5\\ \hfill 2& \hfill & \hfill 1\end{array}\right][/latex]

Solution

Try It

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

[latex]A=\left[\begin{array}{rrr}\hfill 1& \hfill & \hfill 4\\ \hfill -1& \hfill & \hfill -3\end{array}\right],B=\left[\begin{array}{rrr}\hfill -3& \hfill & \hfill -4\\ \hfill 1& \hfill & \hfill 1\end{array}\right][/latex]

Solution

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: Finding the Multiplicative Inverse Using Matrix Multiplication

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

[latex]A=\left[\begin{array}{rrr}\hfill 1& \hfill & \hfill -2\\ \hfill 2& \hfill & \hfill -3\end{array}\right][/latex]

Solution

For this method, we multiply [latex]A[/latex] by a matrix containing unknown constants and set it equal to the identity.

[latex]\left[\begin{array}{rr}\hfill 1& \hfill -2\\ \hfill 2& \hfill -3\end{array}\right]\text{ }\left[\begin{array}{rr}\hfill a& \hfill b\\ \hfill c& \hfill d\end{array}\right]=\left[\begin{array}{rr}\hfill 1& \hfill 0\\ \hfill 0& \hfill 1\end{array}\right][/latex]

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

[latex]\left[\begin{array}{rr}\hfill 1& \hfill -2\\ \hfill 2& \hfill -3\end{array}\right]\text{ }\left[\begin{array}{rr}\hfill a& \hfill b\\ \hfill c& \hfill d\end{array}\right]=\left[\begin{array}{rr}\hfill 1a - 2c& \hfill 1b - 2d\\ \hfill 2a - 3c& \hfill 2b - 3d\end{array}\right][/latex]

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.

[latex]\begin{array}{c}1a - 2c=1\text{ }{R}_{1}\\ 2a - 3c=0\text{ }{R}_{2}\end{array}[/latex]

Using row operations, multiply and add as follows: [latex]\left(-2\right){R}_{1}+{R}_{2}\to {R}_{2}[/latex]. Add the equations, and solve for [latex]c[/latex].

[latex]\begin{array}{r}\hfill 1a - 2c=1\\ \hfill 0+1c=-2\\ \hfill c=-2\end{array}[/latex]

Back-substitute to solve for [latex]a[/latex].

[latex]\begin{array}{r}\hfill a - 2\left(-2\right)=1\\ \hfill a+4=1\\ \hfill a=-3\end{array}[/latex]

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.

[latex]\begin{array}{rr}\hfill 1b - 2d=0& \hfill {R}_{1}\\ \hfill 2b - 3d=1& \hfill {R}_{2}\end{array}[/latex]

Using row operations, multiply and add as follows: [latex]\left(-2\right){R}_{1}+{R}_{2}={R}_{2}[/latex]. Add the two equations and solve for [latex]d[/latex].

[latex]\begin{array}{r}\hfill 1b - 2d=0\\ \hfill \frac{0+1d=1}{d=1}\\ \hfill \end{array}[/latex]

Once more, back-substitute and solve for [latex]b[/latex].

[latex]\begin{array}{r}\hfill b - 2\left(1\right)=0\\ \hfill b - 2=0\\ \hfill b=2\end{array}[/latex]

[latex]{A}^{-1}=\left[\begin{array}{rrr}\hfill -3& \hfill & \hfill 2\\ \hfill -2& \hfill & \hfill 1\end{array}\right][/latex]

Finding the Multiplicative Inverse by Augmenting with the Identity

Another way to find the multiplicative inverse is by augmenting with the identity. When matrix [latex]A[/latex] is transformed into [latex]I[/latex], the augmented matrix [latex]I[/latex] transforms into [latex]{A}^{-1}[/latex].

For example, given

[latex]A=\left[\begin{array}{rrr}\hfill 2& \hfill & \hfill 1\\ \hfill 5& \hfill & \hfill 3\end{array}\right][/latex]

augment [latex]A[/latex] with the identity

[latex]\left[\begin{array}{rr}\hfill 2& \hfill 1\\ \hfill 5& \hfill 3\end{array}\text{ }|\text{ }\begin{array}{rr}\hfill 1& \hfill 0\\ \hfill 0& \hfill 1\end{array}\right][/latex]

Perform row operations with the goal of turning [latex]A[/latex] into the identity.

Switch row 1 and row 2.
[latex]\left[\begin{array}{rr}\hfill 5& \hfill 3\\ \hfill 2& \hfill 1\end{array}\text{ }|\text{ }\begin{array}{rr}\hfill 0& \hfill 1\\ \hfill 1& \hfill 0\end{array}\right][/latex]
Multiply row 2 by [latex]-2[/latex] and add to row 1.
[latex]\left[\begin{array}{rr}\hfill 1& \hfill 1\\ \hfill 2& \hfill 1\end{array}\text{ }|\text{ }\begin{array}{rr}\hfill -2& \hfill 1\\ \hfill 1& \hfill 0\end{array}\right][/latex]
Multiply row 1 by [latex]-2[/latex] and add to row 2.
[latex]\left[\begin{array}{rr}\hfill 1& \hfill 1\\ \hfill 0& \hfill -1\end{array}\text{ }|\text{ }\begin{array}{rr}\hfill -2& \hfill 1\\ \hfill 5& \hfill -2\end{array}\right][/latex]
Add row 2 to row 1.
[latex]\left[\begin{array}{rr}\hfill 1& \hfill 0\\ \hfill 0& \hfill -1\end{array}\text{ }|\text{ }\begin{array}{rr}\hfill 3& \hfill -1\\ \hfill 5& \hfill -2\end{array}\right][/latex]
Multiply row 2 by [latex]-1[/latex].
[latex]\left[\begin{array}{rr}\hfill 1& \hfill 0\\ \hfill 0& \hfill 1\end{array}\text{ }|\text{ }\begin{array}{rr}\hfill 3& \hfill -1\\ \hfill -5& \hfill 2\end{array}\right][/latex]

The matrix we have found is [latex]{A}^{-1}[/latex].

[latex]{A}^{-1}=\left[\begin{array}{rrr}\hfill 3& \hfill & \hfill -1\\ \hfill -5& \hfill & \hfill 2\end{array}\right][/latex]

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

When we need to find the multiplicative inverse of a [latex]2\times 2[/latex] matrix, we can use a special formula instead of using matrix multiplication or augmenting with the identity.

If [latex]A[/latex] is a [latex]2\times 2[/latex] matrix, such as

[latex]A=\left[\begin{array}{rrr}\hfill a& \hfill & \hfill b\\ \hfill c& \hfill & \hfill d\end{array}\right][/latex]

the multiplicative inverse of [latex]A[/latex] is given by the formula

[latex]{A}^{-1}=\frac{1}{ad-bc}\left[\begin{array}{rrr}\hfill d& \hfill & \hfill -b\\ \hfill -c& \hfill & \hfill a\end{array}\right][/latex]

where [latex]ad-bc\ne 0[/latex]. If [latex]ad-bc=0[/latex], then [latex]A[/latex] has no inverse.

Example: Using the Formula to Find the Multiplicative Inverse of Matrix A

Use the formula to find the multiplicative inverse of

[latex]A=\left[\begin{array}{cc}1& -2\\ 2& -3\end{array}\right][/latex]

Solution

Try It

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

[latex]A=\left[\begin{array}{cc}1& -1\\ 2& 3\end{array}\right][/latex]

Solution

Example: Finding the Inverse of the Matrix, If It Exists

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

[latex]A=\left[\begin{array}{cc}3& 6\\ 1& 2\end{array}\right][/latex]

Solution

Finding the Multiplicative Inverse of 3×3 Matrices

Unfortunately, we do not have a formula similar to the one for a [latex]2\text{}\times \text{}2[/latex] matrix to find the inverse of a [latex]3\text{}\times \text{}3[/latex] matrix. Instead, we will augment the original matrix with the identity matrix and use row operations to obtain the inverse.

Given a [latex]3\text{}\times \text{}3[/latex] matrix

[latex]A=\left[\begin{array}{ccc}2& 3& 1\\ 3& 3& 1\\ 2& 4& 1\end{array}\right][/latex]
augment [latex]A[/latex] with the identity matrix
[latex]A|I=\left[\begin{array}{ccc}2& 3& 1\\ 3& 3& 1\\ 2& 4& 1\end{array}\text{ }|\text{ }\begin{array}{ccc}1& 0& 0\\ 0& 1& 0\\ 0& 0& 1\end{array}\right][/latex]

To begin, we write the augmented matrix with the identity on the right and [latex]A[/latex] 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 [latex]3\times 3[/latex] matrix, find the inverse

Write the original matrix augmented with the identity matrix on the right.
Use elementary row operations so that the identity appears on the left.
What is obtained on the right is the inverse of the original matrix.
Use matrix multiplication to show that [latex]A{A}^{-1}=I[/latex] and [latex]{A}^{-1}A=I[/latex].

Example: Finding the Inverse of a 3 × 3 Matrix

Given the [latex]3\times 3[/latex] matrix [latex]A[/latex], find the inverse.

[latex]A=\left[\begin{array}{ccc}2& 3& 1\\ 3& 3& 1\\ 2& 4& 1\end{array}\right][/latex]

Solution

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

[latex]\left[\begin{array}{ccc}2& 3& 1\\ 3& 3& 1\\ 2& 4& 1\end{array}|\begin{array}{ccc}1& 0& 0\\ 0& 1& 0\\ 0& 0& 1\end{array}\right]\stackrel{\text{Interchange }{R}_{2}\text{and }{R}_{1}}{\to }\left[\begin{array}{ccc}3& 3& 1\\ 2& 3& 1\\ 2& 4& 1\end{array}|\begin{array}{ccc}0& 1& 0\\ 1& 0& 0\\ 0& 0& 1\end{array}\right][/latex]

[latex]-{R}_{2}+{R}_{1}={R}_{1}\to \left[\begin{array}{ccc}1& 0& 0\\ 2& 3& 1\\ 2& 4& 1\end{array}|\begin{array}{rrr}\hfill -1& \hfill 1& \hfill 0\\ \hfill 1& \hfill 0& \hfill 0\\ \hfill 0& \hfill 0& \hfill 1\end{array}\right][/latex]

[latex]-{R}_{2}+{R}_{3}={R}_{3}\to \left[\begin{array}{ccc}1& 0& 0\\ 2& 3& 1\\ 0& 1& 0\end{array}|\begin{array}{rrr}\hfill -1& \hfill 1& \hfill 0\\ \hfill 1& \hfill 0& \hfill 0\\ \hfill -1& \hfill 0& \hfill 1\end{array}\right][/latex]

[latex]{R}_{3}\leftrightarrow {R}_{2}\to \left[\begin{array}{ccc}1& 0& 0\\ 0& 1& 0\\ 2& 3& 1\end{array}|\begin{array}{rrr}\hfill -1& \hfill 1& \hfill 0\\ \hfill -1& \hfill 0& \hfill 1\\ \hfill 1& \hfill 0& \hfill 0\end{array}\right][/latex]

[latex]-2{R}_{1}+{R}_{3}={R}_{3}\to \left[\begin{array}{ccc}1& 0& 0\\ 0& 1& 0\\ 0& 3& 1\end{array}|\begin{array}{rrr}\hfill -1& \hfill 1& \hfill 0\\ \hfill -1& \hfill 0& \hfill 1\\ \hfill 3& \hfill -2& \hfill 0\end{array}\right][/latex]

[latex]-3{R}_{2}+{R}_{3}={R}_{3}\to \left[\begin{array}{ccc}1& 0& 0\\ 0& 1& 0\\ 0& 0& 1\end{array}|\begin{array}{rrr}\hfill -1& \hfill 1& \hfill 0\\ \hfill -1& \hfill 0& \hfill 1\\ \hfill 6& \hfill -2& \hfill -3\end{array}\right][/latex]

Thus,

[latex]{A}^{-1}=B=\left[\begin{array}{ccc}-1& 1& 0\\ -1& 0& 1\\ 6& -2& -3\end{array}\right][/latex]

Analysis of the Solution

To prove that [latex]B={A}^{-1}[/latex], let’s multiply the two matrices together to see if the product equals the identity, if [latex]A{A}^{-1}=I[/latex] and [latex]{A}^{-1}A=I[/latex].

[latex]\begin{array}{l}\begin{array}{l}\begin{array}{l}\hfill \\ \hfill \end{array}\hfill \\ A{A}^{-1}=\left[\begin{array}{ccc}2& 3& 1\\ 3& 3& 1\\ 2& 4& 1\end{array}\right]\text{ }\left[\begin{array}{rrr}\hfill -1& \hfill 1& \hfill 0\\ \hfill -1& \hfill 0& \hfill 1\\ \hfill 6& \hfill -2& \hfill -3\end{array}\right]\hfill \end{array}\hfill \\ =\left[\begin{array}{ccc}2\left(-1\right)+3\left(-1\right)+1\left(6\right)& 2\left(1\right)+3\left(0\right)+1\left(-2\right)& 2\left(0\right)+3\left(1\right)+1\left(-3\right)\\ 3\left(-1\right)+3\left(-1\right)+1\left(6\right)& 3\left(1\right)+3\left(0\right)+1\left(-2\right)& 3\left(0\right)+3\left(1\right)+1\left(-3\right)\\ 2\left(-1\right)+4\left(-1\right)+1\left(6\right)& 2\left(1\right)+4\left(0\right)+1\left(-2\right)& 2\left(0\right)+4\left(1\right)+1\left(-3\right)\end{array}\right]\hfill \\ =\left[\begin{array}{ccc}1& 0& 0\\ 0& 1& 0\\ 0& 0& 1\end{array}\right]\hfill \end{array}[/latex]
[latex]\begin{array}{l}\begin{array}{l}\begin{array}{l}\hfill \\ \hfill \end{array}\hfill \\ {A}^{-1}A=\left[\begin{array}{rrr}\hfill -1& \hfill 1& \hfill 0\\ \hfill -1& \hfill 0& \hfill 1\\ \hfill 6& \hfill -2& \hfill -3\end{array}\right]\text{ }\left[\begin{array}{ccc}2& 3& 1\\ 3& 3& 1\\ 2& 4& 1\end{array}\right]\hfill \end{array}\hfill \\ =\left[\begin{array}{rrr}\hfill -1\left(2\right)+1\left(3\right)+0\left(2\right)& \hfill -1\left(3\right)+1\left(3\right)+0\left(4\right)& \hfill -1\left(1\right)+1\left(1\right)+0\left(1\right)\\ \hfill -1\left(2\right)+0\left(3\right)+1\left(2\right)& \hfill -1\left(3\right)+0\left(3\right)+1\left(4\right)& \hfill -1\left(1\right)+0\left(1\right)+1\left(1\right)\\ \hfill 6\left(2\right)+-2\left(3\right)+-3\left(2\right)& \hfill 6\left(3\right)+-2\left(3\right)+-3\left(4\right)& \hfill 6\left(1\right)+-2\left(1\right)+-3\left(1\right)\end{array}\right]\hfill \\ =\left[\begin{array}{rrr}\hfill 1& \hfill 0& \hfill 0\\ \hfill 0& \hfill 1& \hfill 0\\ \hfill 0& \hfill 0& \hfill 1\end{array}\right]\hfill \end{array}[/latex]

Try It

Find the inverse of the [latex]3\times 3[/latex] matrix.

[latex]A=\left[\begin{array}{ccc}2& -17& 11\\ -1& 11& -7\\ 0& 3& -2\end{array}\right][/latex]

Solution

Solve a System Using an Inverse

Solving a system of linear equations using the inverse of a matrix requires the definition of two new matrices: [latex]X[/latex] is the matrix representing the variables of the system, and [latex]B[/latex] 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 [latex]AX=B[/latex]

To solve a system of linear equations using an inverse matrix, let [latex]A[/latex] be the coefficient matrix, let [latex]X[/latex] be the variable matrix, and let [latex]B[/latex] be the constant matrix. Thus, we want to solve a system [latex]AX=B[/latex]. For example, look at the following system of equations.

[latex]\begin{array}{c}{a}_{1}x+{b}_{1}y={c}_{1}\\ {a}_{2}x+{b}_{2}y={c}_{2}\end{array}[/latex]

From this system, the coefficient matrix is

[latex]A=\left[\begin{array}{cc}{a}_{1}& {b}_{1}\\ {a}_{2}& {b}_{2}\end{array}\right][/latex]

The variable matrix is

[latex]X=\left[\begin{array}{c}x\\ y\end{array}\right][/latex]

And the constant matrix is

[latex]B=\left[\begin{array}{c}{c}_{1}\\ {c}_{2}\end{array}\right][/latex]

Then [latex]AX=B[/latex] looks like

[latex]\left[\begin{array}{cc}{a}_{1}& {b}_{1}\\ {a}_{2}& {b}_{2}\end{array}\right]\text{ }\left[\begin{array}{c}x\\ y\end{array}\right]=\left[\begin{array}{c}{c}_{1}\\ {c}_{2}\end{array}\right][/latex]

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

[latex]\begin{array}{c}\text{ }ax=b\\ \text{ }\left(\frac{1}{a}\right)ax=\left(\frac{1}{a}\right)b\\ \left({a}^{-1}\text{ }\right)ax=\left({a}^{-1}\right)b\\ \left[\left({a}^{-1}\right)a\right]x=\left({a}^{-1}\right)b\\ \text{ }1x=\left({a}^{-1}\right)b\\ \text{ }x=\left({a}^{-1}\right)b\end{array}[/latex]

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 [latex]2\times 2[/latex] system and then move on to a [latex]3\times 3[/latex] system.

A General Note: Solving a System of Equations Using the Inverse of a Matrix

Given a system of equations, write the coefficient matrix [latex]A[/latex], the variable matrix [latex]X[/latex], and the constant matrix [latex]B[/latex]. Then [latex]AX=B[/latex]. Multiply both sides by the inverse of [latex]A[/latex] to obtain the solution.

[latex]\begin{array}{r}\hfill \left({A}^{-1}\right)AX=\left({A}^{-1}\right)B\\ \hfill \left[\left({A}^{-1}\right)A\right]X=\left({A}^{-1}\right)B\\ \hfill IX=\left({A}^{-1}\right)B\\ \hfill X=\left({A}^{-1}\right)B\end{array}[/latex]

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

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

[latex]\begin{array}{r}\hfill 3x+8y=5\\ \hfill 4x+11y=7\end{array}[/latex]

Solution

Q & A

Can we solve for [latex]X[/latex] by finding the product [latex]B{A}^{-1}?[/latex]

No, recall that matrix multiplication is not commutative, so [latex]{A}^{-1}B\ne B{A}^{-1}[/latex]. Consider our steps for solving the matrix equation.

Notice in the first step we multiplied both sides of the equation by [latex]{A}^{-1}[/latex], but the [latex]{A}^{-1}[/latex] was to the left of [latex]A[/latex] on the left side and to the left of [latex]B[/latex] on the right side. Because matrix multiplication is not commutative, order matters.

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

Solve the following system using the inverse of a matrix.

[latex]\begin{array}{r}\hfill 5x+15y+56z=35\\ \hfill -4x - 11y - 41z=-26\\ \hfill -x - 3y - 11z=-7\end{array}[/latex]

Solution

Write the equation [latex]AX=B[/latex].

[latex]\left[\begin{array}{ccc}5& 15& 56\\ -4& -11& -41\\ -1& -3& -11\end{array}\right]\text{ }\left[\begin{array}{c}x\\ y\\ z\end{array}\right]=\left[\begin{array}{r}\hfill 35\\ \hfill -26\\ \hfill -7\end{array}\right][/latex]

First, we will find the inverse of [latex]A[/latex] by augmenting with the identity.

[latex]\left[\begin{array}{rrr}\hfill 5& \hfill 15& \hfill 56\\ \hfill -4& \hfill -11& \hfill -41\\ \hfill -1& \hfill -3& \hfill -11\end{array}|\begin{array}{ccc}1& 0& 0\\ 0& 1& 0\\ 0& 0& 1\end{array}\right][/latex]

Multiply row 1 by [latex]\frac{1}{5}[/latex].

[latex]\left[\begin{array}{ccc}1& 3& \frac{56}{5}\\ -4& -11& -41\\ -1& -3& -11\end{array}|\begin{array}{ccc}\frac{1}{5}& 0& 0\\ 0& 1& 0\\ 0& 0& 1\end{array}\right][/latex]

Multiply row 1 by 4 and add to row 2.

[latex]\left[\begin{array}{ccc}1& 3& \frac{56}{5}\\ 0& 1& \frac{19}{5}\\ -1& -3& -11\end{array}|\begin{array}{ccc}\frac{1}{5}& 0& 0\\ \frac{4}{5}& 1& 0\\ 0& 0& 1\end{array}\right][/latex]

Add row 1 to row 3.

[latex]\left[\begin{array}{ccc}1& 3& \frac{56}{5}\\ 0& 1& \frac{19}{5}\\ 0& 0& \frac{1}{5}\end{array}|\begin{array}{ccc}\frac{1}{5}& 0& 0\\ \frac{4}{5}& 1& 0\\ \frac{1}{5}& 0& 1\end{array}\right][/latex]

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

[latex]\left[\begin{array}{ccc}1& 0& -\frac{1}{5}\\ 0& 1& \frac{19}{5}\\ 0& 0& \frac{1}{5}\end{array}|\begin{array}{ccc}-\frac{11}{5}& -3& 0\\ \frac{4}{5}& 1& 0\\ \frac{1}{5}& 0& 1\end{array}\right][/latex]

Multiply row 3 by 5.

[latex]\left[\begin{array}{ccc}1& 0& -\frac{1}{5}\\ 0& 1& \frac{19}{5}\\ 0& 0& 1\end{array}|\begin{array}{ccc}-\frac{11}{5}& -3& 0\\ \frac{4}{5}& 1& 0\\ 1& 0& 5\end{array}\right][/latex]

Multiply row 3 by [latex]\frac{1}{5}[/latex] and add to row 1.

[latex]\left[\begin{array}{ccc}1& 0& 0\\ 0& 1& \frac{19}{5}\\ 0& 0& 1\end{array}|\begin{array}{ccc}-2& -3& 1\\ \frac{4}{5}& 1& 0\\ 1& 0& 5\end{array}\right][/latex]

Multiply row 3 by [latex]-\frac{19}{5}[/latex] and add to row 2.

[latex]\left[\begin{array}{ccc}1& 0& 0\\ 0& 1& 0\\ 0& 0& 1\end{array}|\begin{array}{ccc}-2& -3& 1\\ -3& 1& -19\\ 1& 0& 5\end{array}\right][/latex]

So,

[latex]{A}^{-1}=\left[\begin{array}{ccc}-2& -3& 1\\ -3& 1& -19\\ 1& 0& 5\end{array}\right][/latex]

Multiply both sides of the equation by [latex]{A}^{-1}[/latex]. We want

[latex]{A}^{-1}AX={A}^{-1}B:[/latex]

[latex]\left[\begin{array}{rrr}\hfill -2& \hfill -3& \hfill 1\\ \hfill -3& \hfill 1& \hfill -19\\ \hfill 1& \hfill 0& \hfill 5\end{array}\right]\text{ }\left[\begin{array}{rrr}\hfill 5& \hfill 15& \hfill 56\\ \hfill -4& \hfill -11& \hfill -41\\ \hfill -1& \hfill -3& \hfill -11\end{array}\right]\text{ }\left[\begin{array}{c}x\\ y\\ z\end{array}\right]=\left[\begin{array}{rrr}\hfill -2& \hfill -3& \hfill 1\\ \hfill -3& \hfill 1& \hfill -19\\ \hfill 1& \hfill 0& \hfill 5\end{array}\right]\text{ }\left[\begin{array}{r}\hfill 35\\ \hfill -26\\ \hfill -7\end{array}\right][/latex]

Thus,

[latex]{A}^{-1}B=\left[\begin{array}{r}\hfill -70+78 - 7\\ \hfill -105 - 26+133\\ \hfill 35+0 - 35\end{array}\right]=\left[\begin{array}{c}1\\ 2\\ 0\end{array}\right][/latex]

The solution is [latex]\left(1,2,0\right)[/latex].

Try It

Solve the system using the inverse of the coefficient matrix.

[latex]\begin{array}{l}\text{ }2x - 17y+11z=0\hfill \\ \text{ }-x+11y - 7z=8\hfill \\ \text{ }3y - 2z=-2\hfill \end{array}[/latex]

Solution

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

Save the coefficient matrix and the constant matrix as matrix variables [latex]\left[A\right][/latex] and [latex]\left[B\right][/latex].
Enter the multiplication into the calculator, calling up each matrix variable as needed.
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: Using a Calculator to Solve a System of Equations with Matrix Inverses

Solve the system of equations with matrix inverses using a calculator

[latex]\begin{array}{l}2x+3y+z=32\hfill \\ 3x+3y+z=-27\hfill \\ 2x+4y+z=-2\hfill \end{array}[/latex]

Solution

Key Equations

Identity matrix for a [latex]2\text{}\times \text{}2[/latex] matrix	[latex]{I}_{2}=\left[\begin{array}{cc}1& 0\\ 0& 1\end{array}\right][/latex]
Identity matrix for a [latex]\text{3}\text{}\times \text{}3[/latex] matrix	[latex]{I}_{3}=\left[\begin{array}{ccc}1& 0& 0\\ 0& 1& 0\\ 0& 0& 1\end{array}\right][/latex]
Multiplicative inverse of a [latex]2\text{}\times \text{}2[/latex] matrix	[latex]{A}^{-1}=\frac{1}{ad-bc}\left[\begin{array}{cc}d& -b\\ -c& a\end{array}\right],\text{ where }ad-bc\ne 0[/latex]

Key Concepts

An identity matrix has the property [latex]AI=IA=A[/latex].
An invertible matrix has the property [latex]A{A}^{-1}={A}^{-1}A=I[/latex].
Use matrix multiplication and the identity to find the inverse of a [latex]2\times 2[/latex] matrix.
The multiplicative inverse can be found using a formula.
Another method of finding the inverse is by augmenting with the identity.
We can augment a [latex]3\times 3[/latex] 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.
Write the system of equations as [latex]AX=B[/latex], and multiply both sides by the inverse of [latex]A:{A}^{-1}AX={A}^{-1}B[/latex].
We can also use a calculator to solve a system of equations with matrix inverses.

Glossary

identity matrix a square matrix containing ones down the main diagonal and zeros everywhere else; it acts as a 1 in matrix algebra

multiplicative inverse of a matrix a matrix that, when multiplied by the original, equals the identity matrix

Module 7: Solving Systems of Equations

Find the Inverse of a Matrix

A General Note: The Identity Matrix and Multiplicative Inverse

Example: Showing That the Identity Matrix Acts as a 1

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

Example: Showing That Matrix A Is the Multiplicative Inverse of Matrix B

Try It

Finding the Multiplicative Inverse Using Matrix Multiplication

Example: Finding the Multiplicative Inverse Using Matrix Multiplication

Finding the Multiplicative Inverse by Augmenting with the Identity

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

Example: Using the Formula to Find the Multiplicative Inverse of Matrix A

Analysis of the Solution

Try It

Example: Finding the Inverse of the Matrix, If It Exists

Finding the Multiplicative Inverse of 3×3 Matrices

How To: Given a [latex]3\times 3[/latex] matrix, find the inverse

Example: Finding the Inverse of a 3 × 3 Matrix

Analysis of the Solution

Try It

Solve a System Using an Inverse

A General Note: Solving a System of Equations Using the Inverse of a Matrix

Q & A

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

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

Q & A

Can we solve for [latex]X[/latex] by finding the product [latex]B{A}^{-1}?[/latex]

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

Try It

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

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

Key Equations

Key Concepts

Glossary

Candela Citations