Use matrix multiplication to show that the product of [latex]A[/latex] and the identity is equal to the product of the identity and A.

[latex]AI=\left[\begin{array}{rrr}\hfill 3& \hfill & \hfill 4\\ \hfill -2& \hfill & \hfill 5\end{array}\right]\begin{array}{r}\hfill \end{array}\left[\begin{array}{rrr}\hfill 1& \hfill & \hfill 0\\ \hfill 0& \hfill & \hfill 1\end{array}\right]=\left[\begin{array}{rrrr}\hfill 3\cdot 1+4\cdot 0& \hfill & \hfill & \hfill 3\cdot 0+4\cdot 1\\ \hfill -2\cdot 1+5\cdot 0& \hfill & \hfill & \hfill -2\cdot 0+5\cdot 1\end{array}\right]=\left[\begin{array}{rrr}\hfill 3& \hfill & \hfill 4\\ \hfill -2& \hfill & \hfill 5\end{array}\right][/latex]

[latex]AI=\left[\begin{array}{rrr}\hfill 1& \hfill & \hfill 0\\ \hfill 0& \hfill & \hfill 1\end{array}\right]\begin{array}{r}\hfill \end{array}\left[\begin{array}{rrr}\hfill 3& \hfill & \hfill 4\\ \hfill -2& \hfill & \hfill 5\end{array}\right]=\left[\begin{array}{rrrr}\hfill 1\cdot 3+0\cdot \left(-2\right)& \hfill & \hfill & \hfill 1\cdot 4+0\cdot 5\\ \hfill 0\cdot 3+1\cdot \left(-2\right)& \hfill & \hfill & \hfill 0\cdot 4+1\cdot 5\end{array}\right]=\left[\begin{array}{rrr}\hfill 3& \hfill & \hfill 4\\ \hfill -2& \hfill & \hfill 5\end{array}\right][/latex]

Multiply [latex]AB[/latex] and [latex]BA[/latex]. If both products equal the identity, then the two matrices are inverses of each other.

[latex]\begin{array}{l}AB & =\left[\begin{array}{rrr}\hfill 1& \hfill & \hfill 5\\ \hfill -2& \hfill & \hfill -9\end{array}\right]\cdot \left[\begin{array}{rrr}\hfill -9& \hfill & \hfill -5\\ \hfill 2& \hfill & \hfill 1\end{array}\right]\hfill \\ & =\left[\begin{array}{rrr}\hfill 1\left(-9\right)+5\left(2\right)& \hfill & \hfill 1\left(-5\right)+5\left(1\right)\\ \hfill -2\left(-9\right)-9\left(2\right)& \hfill & \hfill -2\left(-5\right)-9\left(1\right)\end{array}\right]\hfill \\ & =\left[\begin{array}{ccc}1& & 0\\ 0& & 1\end{array}\right]\hfill \end{array}[/latex]

[latex]\begin{array}{l}BA & =\left[\begin{array}{rrr}\hfill -9& \hfill & \hfill -5\\ \hfill 2& \hfill & \hfill 1\end{array}\right]\cdot \left[\begin{array}{rrr}\hfill 1& \hfill & \hfill 5\\ \hfill -2& \hfill & \hfill -9\end{array}\right]\hfill \\ & =\left[\begin{array}{rrr}\hfill -9\left(1\right)-5\left(-2\right)& \hfill & \hfill -9\left(5\right)-5\left(-9\right)\\ \hfill 2\left(1\right)+1\left(-2\right)& \hfill & \hfill 2\left(-5\right)+1\left(-9\right)\end{array}\right]\hfill \\ & =\left[\begin{array}{ccc}1& & 0\\ 0& & 1\end{array}\right]\hfill \end{array}[/latex]

[latex]A[/latex] and [latex]B[/latex] are inverses of each other.

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]

Using the formula, we have

[latex]\begin{array}{l}{A}^{-1} & =\frac{1}{\left(1\right)\left(-3\right)-\left(-2\right)\left(2\right)}\left[\begin{array}{cc}-3& 2\\ -2& 1\end{array}\right]\hfill \\ & =\frac{1}{-3+4}\left[\begin{array}{cc}-3& 2\\ -2& 1\end{array}\right]\hfill \\ & =\left[\begin{array}{cc}-3& 2\\ -2& 1\end{array}\right]\hfill \end{array}[/latex]

Analysis of the Solution

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

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

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

1. Multiply row 1 by [latex]-2[/latex] and add to row 2.
   [latex]\left[\begin{array}{cc}1& -2\\ 0& 1\end{array}|\begin{array}{cc}1& 0\\ -2& 1\end{array}\right][/latex]
2. Multiply row 1 by 2 and add to row 1.
   [latex]\left[\begin{array}{cc}1& 0\\ 0& 1\end{array}|\begin{array}{cc}-3& 2\\ -2& 1\end{array}\right][/latex]

So, we have verified our original solution.

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

We will use the method of augmenting with the identity.

[latex]\left[\begin{array}{cc}3& 6\\ 1& 2\end{array}|\begin{array}{cc}1& 0\\ 0& 1\end{array}\right][/latex]

1. Switch row 1 and row 2.
   [latex]\left[\begin{array}{cc}1& 2\\ 3& 6\text{ }\end{array}\text{ }\text{ }\text{ }|\begin{array}{cc}0& 1\\ 1& 0\end{array}\right][/latex]
2. Multiply row 1 by −3 and add it to row 2.
   [latex]\left[\begin{array}{cc}1& 2\\ 0& 0\end{array}|\begin{array}{cc}1& 0\\ 1& -3\end{array}\right][/latex]
3. There is nothing further we can do. The zeros in row 2 indicate that this matrix has no inverse.

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}\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 \\ & =\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}\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 \\ & =\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]

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

[latex]A=\left[\begin{array}{cc}3& 8\\ 4& 11\end{array}\right],X=\left[\begin{array}{c}x\\ y\end{array}\right],B=\left[\begin{array}{c}5\\ 7\end{array}\right][/latex]

Then

[latex]\left[\begin{array}{cc}3& 8\\ 4& 11\end{array}\right]\text{ }\left[\begin{array}{c}x\\ y\end{array}\right]=\left[\begin{array}{c}5\\ 7\end{array}\right][/latex]

First, we need to calculate [latex]{A}^{-1}[/latex]. Using the formula to calculate the inverse of a 2 by 2 matrix, we have:

[latex]\begin{array}{l}{A}^{-1} & =\frac{1}{ad-bc}\left[\begin{array}{cc}d& -b\\ -c& a\end{array}\right]\hfill \\ & =\frac{1}{3\left(11\right)-8\left(4\right)}\left[\begin{array}{cc}11& -8\\ -4& 3\end{array}\right]\hfill \\ & =\frac{1}{1}\left[\begin{array}{cc}11& -8\\ -4& 3\end{array}\right]\hfill \end{array}[/latex]

So,

[latex]{A}^{-1}=\left[\begin{array}{cc}11& -8\\ -4& \text{ }\text{ }3\end{array}\right][/latex]

Now we are ready to solve. Multiply both sides of the equation by [latex]{A}^{-1}[/latex].

[latex]\begin{array}{l}\left({A}^{-1}\right)AX=\left({A}^{-1}\right)B\hfill \\ \left[\begin{array}{rr}\hfill 11& \hfill -8\\ \hfill -4& \hfill 3\end{array}\right]\text{ }\left[\begin{array}{cc}3& 8\\ 4& 11\end{array}\right]\text{ }\left[\begin{array}{c}x\\ y\end{array}\right]=\left[\begin{array}{rr}\hfill 11& \hfill -8\\ \hfill -4& \hfill 3\end{array}\right]\text{ }\left[\begin{array}{c}5\\ 7\end{array}\right]\hfill \\ \left[\begin{array}{cc}1& 0\\ 0& 1\end{array}\right]\text{ }\left[\begin{array}{c}x\\ y\end{array}\right]=\left[\begin{array}{r}\hfill 11\left(5\right)+\left(-8\right)7\\ \hfill -4\left(5\right)+3\left(7\right)\end{array}\right]\hfill \\ \left[\begin{array}{c}x\\ y\end{array}\right]=\left[\begin{array}{r}\hfill -1\\ \hfill 1\end{array}\right]\hfill \end{array}[/latex]

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

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].

On the matrix page of the calculator, enter the coefficient matrix as the matrix variable [latex]\left[A\right][/latex], and enter the constant matrix as the matrix variable [latex]\left[B\right][/latex].

[latex]\left[A\right]=\left[\begin{array}{ccc}2& 3& 1\\ 3& 3& 1\\ 2& 4& 1\end{array}\right],\text{ }\left[B\right]=\left[\begin{array}{c}32\\ -27\\ -2\end{array}\right][/latex]

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

[latex]{\left[A\right]}^{-1}\times \left[B\right][/latex]

Evaluate the expression.

[latex]\left[\begin{array}{c}-59\\ -34\\ 252\end{array}\right][/latex]