Answer
$\begin{bmatrix} 7 & 2 & 3\\10 & 3 & 5\end{bmatrix} $
Work Step by Step
Step 1. Represent the matrix equation in the form of $AX=B$, where
$\begin{array}( \\A= \\ \\ \end{array}\begin{bmatrix} 3 & -2\\-4 & 3 \end{bmatrix}, \begin{array}( \\X= \\ \\ \end{array}\begin{bmatrix} x & y & z\\u & v & w \end{bmatrix}, \begin{array}( \\B= \\ \\ \end{array}\begin{bmatrix} 1 & 0 & -1\\2 & 1 & 3\end{bmatrix}$
Step 2. Use the formula on page 725, we have:
$\begin{array}( \\A^{-1}=\frac{1}{3\times3-2\times4} \\ \\ \end{array}\begin{bmatrix} 3 & 2\\4 & 3 \end{bmatrix} \begin{array}( \\= \\ \\ \end{array}\begin{bmatrix} 3 & 2\\4 & 3 \end{bmatrix}$
Step 3. Multiply the results from step 2 to the original equation, we have $A^{-1}AX=A^{-1}B$ and $X=A^{-1}B$, thus
$\begin{array}( \\X= \\ \\ \end{array}\begin{bmatrix} 3 & 2\\4 & 3 \end{bmatrix}\begin{bmatrix} 1 & 0 & -1\\2 & 1 & 3\end{bmatrix} \begin{array}( \\= \\ \\ \end{array}\begin{bmatrix} 7 & 2 & 3\\10 & 3 & 5\end{bmatrix} $
Step 4. The solutions to the original matrix equation are:
$\begin{bmatrix} 7 & 2 & 3\\10 & 3 & 5\end{bmatrix} $
