Answer
$\begin{array}( \frac{1}{2a} \\ \end{array}
\begin{bmatrix} 1 &1\\-1 &1 \end{bmatrix} $
Work Step by Step
Step 1. Write the matrix together with an identify as:
$\begin{array}( \\A|I= \\ \\ \end{array}
\begin{bmatrix} a & -a &| &1 &0\\a & a &| &0 &1 \end{bmatrix} \begin{array}( \\R_2-R_1\to R_2 \\\end{array}$
Step 2. Use row operations to transform the left half into a reduced row-echelon:
$\begin{bmatrix} a & -a &| &1 &0\\0 & 2a &| &-1 &1 \end{bmatrix} \begin{array}( 2R_1+R_2\to R_1\\ \\\end{array}$
$\begin{bmatrix} 2a & 0 &| &1 &1\\0 & 2a &| &-1 &1 \end{bmatrix} \begin{array}( R_1/(2a)\to R_1\\R_2/(2a)\to R_2 \\\end{array}$
$\begin{bmatrix} 1 & 0 &| &\frac{1}{2a} &\frac{1}{2a}\\0 & 1 &| &-\frac{1}{2a} &\frac{1}{2a} \end{bmatrix} $
Step 3. The inverse of the original matrix is:
$\begin{array}( \\A^{-1}= \\ \\ \end{array}
\begin{bmatrix} \frac{1}{2a} &\frac{1}{2a}\\-\frac{1}{2a} &\frac{1}{2a} \end{bmatrix}
\begin{array}( =\frac{1}{2a} \\ \end{array}
\begin{bmatrix} 1 &1\\-1 &1 \end{bmatrix} $
Step 4. Double check with the formula on page 725, we confirm the inverse to the original matrix is:
$\begin{array}( \frac{1}{2a} \\ \end{array}
\begin{bmatrix} 1 &1\\-1 &1 \end{bmatrix} $
