Answer
$\begin{bmatrix} \frac{1}{x^2} &-\frac{x-1}{x^2}\\\frac{x-1}{x} &\frac{x-1}{x} \end{bmatrix} $
where $x\ne0, 1$
Work Step by Step
Step 1. Given the matrix:
$\begin{array}( \\A= \\ \\ \end{array}\begin{bmatrix} x &1\\-x &\frac{1}{x-1} \end{bmatrix} $
Step 2. Use the formula in page 725 to get:
$\begin{array}( \\A^{-1}=\frac{1}{\frac{x}{x-1}+x} \\ \\ \end{array}\begin{bmatrix} \frac{1}{x-1} &-1\\x &x \end{bmatrix} \begin{array}( \\=\frac{x-1}{x^2} \\ \\ \end{array}\begin{bmatrix} \frac{1}{x-1} &-1\\x &x \end{bmatrix}\begin{array}( \\= \\ \\ \end{array}\begin{bmatrix} \frac{1}{x^2} &-\frac{x-1}{x^2}\\\frac{x-1}{x} &\frac{x-1}{x} \end{bmatrix} $
Step 3. The inverse of the original matrix is:
$\begin{bmatrix} \frac{1}{x^2} &-\frac{x-1}{x^2}\\\frac{x-1}{x} &\frac{x-1}{x} \end{bmatrix} $
and this inverse is not valid for $x=0, 1$
