Answer
$\begin{bmatrix} -39/2 & -39\\15 & 30\\39/2 & 39 \end{bmatrix}$
Work Step by Step
Step 1. Represent the matrix equation in the form of $AX=B$, where
$\begin{array}( \\A= \\ \\ \end{array}\begin{bmatrix} 0 & -2 &2\\3 & 1 & 3\\1&-2&3 \end{bmatrix}, \begin{array}( \\X= \\ \\ \end{array}\begin{bmatrix} x & u \\y & v\\z & w \end{bmatrix}, \begin{array}( \\B= \\ \\ \end{array}\begin{bmatrix} 3 & 6\\6 & 12\\0 & 0 \end{bmatrix}$
Step 2. Note that matrix A is the same as that of Exercise 23. Recall the answer to that question, we have:
$\begin{array}( \\A^{-1}= \\ \\ \end{array}\begin{bmatrix} -9/2 & -1 &4\\3 & 1 & -3\\7/2&1&-3 \end{bmatrix}$
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
$\begin{array}( \\X= \\ \\ \end{array}\begin{bmatrix} -9/2 & -1 &4\\3 & 1 & -3\\7/2&1&-3 \end{bmatrix}\begin{bmatrix} 3 & 6\\6 & 12\\0 & 0 \end{bmatrix}\begin{array}( \\= \\ \\ \end{array}\begin{bmatrix} -39/2 & -39\\15 & 30\\39/2 & 39 \end{bmatrix}$
Step 4. The solutions to the original matrix equation are:
$\begin{bmatrix} -39/2 & -39\\15 & 30\\39/2 & 39 \end{bmatrix}$
