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]

[latex]\begin{array}{l} \text{1. Augment } A \text{ with the identity matrix:} \\ \left[\begin{array}{cc|cc} 1 & -2 & 1 & 0 \\ 2 & -3 & 0 & 1 \\ \end{array}\right] \\ \\ \text{2. Perform row operations with the goal of turning } A \text{ into the identity matrix.} \\ \text{First, multiply row 1 by } -2 \text{ and add to row 2:} \\ \left[\begin{array}{cc|cc} 1 & -2 & 1 & 0 \\ 2 & -3 & 0 & 1 \\ \end{array}\right] \rightarrow \left[\begin{array}{cc|cc} 1 & -2 & 1 & 0 \\ 0 & 1 & -2 & 1 \\ \end{array}\right] \\ \\ \text{Next, multiply row 2 by } 2 \text{ and add to row 1:} \\ \left[\begin{array}{cc|cc} 1 & -2 & 1 & 0 \\ 0 & 1 & -2 & 1 \\ \end{array}\right] \rightarrow \left[\begin{array}{cc|cc} 1 & 0 & -3 & 2 \\ 0 & 1 & -2 & 1 \\ \end{array}\right] \\ \\ \text{3. The matrix we have found is } A^{-1} : \\ A^{-1} = \left[\begin{array}{rr} -3 & 2 \\ -2 & 1 \\ \end{array}\right] \\ \end{array}[/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|cc}\hfill 1& \hfill -2& \hfill 1& \hfill 0\\ \hfill 2& \hfill -3& \hfill 0& \hfill 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|cc}\hfill 1& \hfill -2& \hfill 1& \hfill 0\\ \hfill 0& \hfill 1& \hfill -2& \hfill 1\\ \end{array}\right][/latex]
2. Multiply row 1 by 2 and add to row 1.
   [latex]\left[\begin{array}{cc|cc}\hfill 1& \hfill 0& \hfill -3& \hfill 2\\ \hfill 0& \hfill 1& \hfill -2& \hfill 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]