College Algebra, Chapter 7, 7.3, Section 7.3, Problem 14

Determine the inverse of the matrix $\left[ \begin{array}{cc}
\displaystyle \frac{1}{2} & \displaystyle \frac{1}{3} \\
5 & 4
\end{array} \right]$ if it exists.

First, let's add the identity matrix to the right of our matrix

$\left[ \begin{array}{cc|cc}
\displaystyle \frac{1}{2} & \displaystyle \frac{1}{3} & 1 & 0 \\
5 & 4 & 0 & 1
\end{array} \right]$

By using Gauss-Jordan Elimination

$\displaystyle 2 R_1$

$\left[ \begin{array}{cc|cc}
1 & \displaystyle \frac{2}{3} & 2 & 0 \\
5 & 4 & 0 & 1
\end{array} \right]$

$\displaystyle R_2 - 5 R_1 \to R_2$

$\left[ \begin{array}{cc|cc}
1 & \displaystyle \frac{2}{3} & 2 & 0 \\
0 & \displaystyle \frac{2}{3} & -10 & 1
\end{array} \right]$

$\displaystyle \frac{3}{2} R_2$

$\left[ \begin{array}{cc|cc}
1 & \displaystyle \frac{2}{3} & 2 & 0 \\
0 & 1 & -15 & \displaystyle \frac{3}{2}
\end{array} \right]$

$\displaystyle R_1 - \frac{2}{3} R_2 \to R_1$

$\left[ \begin{array}{cc|cc}
1 & 0 & 12 & -1 \\
0 & 1 & -15 & \displaystyle \frac{3}{2}
\end{array} \right]$

The inverse matrix can now be found in the right half of our reduced row-echelon matrix. So the inverse matrix is

$\left[ \begin{array}{cc}
12 & -1 \\
-15 & \displaystyle \frac{3}{2}
\end{array} \right]$