Solve System Using Matrix Inverse

[brinlin]

 

[skeeter]

$
\begin{bmatrix}
0 &-2 &2 \\
3 & 1 &3 \\
1 &-2 &3
\end{bmatrix} \cdot\begin{bmatrix}
x\\
y\\
z
\end{bmatrix}=\begin{bmatrix}
12\\
-2\\
8
\end{bmatrix}$

follow the directions for (b) and (c)

 

[HOI]

There are many ways to find an inverse matrix. One I like
write the matrix and identity matrix next to each other:
\(\displaystyle \begin{bmatrix}0 & -2 & 2 \\ 3 & 1 & 3 \\ 1 & -3 & 1 \end{bmatrix}\begin{bmatrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1\end{bmatrix}\).

Now use "row operations" to reduce the matrix to the identity matrix while applying the same row operation to the identity matrix.

Normally the first thing you would do is divide every number in the first row by the leftmost number in that row. But here, that is 0 so instead swap the first and third rows:
\(\displaystyle \begin{bmatrix}1 & -3 & 1 \\ 3 & 1 & 3 \\ 0 & -2 & 2 \end{bmatrix}\begin{bmatrix}0 & 0 & 1 \\ 0 & 1 & 0 \\ 1 & 0 & 0\end{bmatrix}\)

Now there is already a 1 in the upper left so all we need to do to get the right first column is subtract 3 times the first row from the second row:
\(\displaystyle \begin{bmatrix}1 & -3 & 1 \\ 0 & 10 & 0 \\ 0 & -2 & 2 \end{bmatrix}\begin{bmatrix}0 & 0 & 1 \\ 0 & 1 & -1 \\ 1 & 0 & 0 \end{bmatrix}\)

Divide the second row by 10:
\(\displaystyle \begin{bmatrix}1 & -3 & 1 \\ 0 & 1 & 0 \\ 0 & -2 & 2 \end{bmatrix}\begin{bmatrix}0 & 0 & 1 \\ 0 & \frac{1}{10} & -\frac{1}{10} \\ 1 & 0 & 0 \end{bmatrix}\).

Add 3 times the second row to the first row and add 2 times the second row to the third row:
\(\displaystyle \begin{bmatrix}1 & 0 & 1 \\ 0 & 1 & 0 \\ 0 & 0 & 2 \end{bmatrix}\begin{bmatrix}0 & \frac{3}{10} & \frac{7}{10} \\ 0 & \frac{1}{10} & -\frac{1}{10} \\ 1 & \frac{2}{10} & \frac{8}{10} \end{bmatrix}\)

Divide the third row by 2:
\(\displaystyle \begin{bmatrix}1 & 0 & 1 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{bmatrix}\begin{bmatrix}0 & \frac{3}{10} & \frac{7}{10} \\ 0 & \frac{1}{10} & -\frac{1}{10} \\ \frac{1}{2} & \frac{1}{10} & \frac{4}{10} \end{bmatrix}\).

Finally, subtract the third row from the first row:
\(\displaystyle \begin{bmatrix}1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{bmatrix}\begin{bmatrix}0 & \frac{3}{10} & \frac{3}{10} \\ 0 & \frac{1}{10} & -\frac{1}{10} \\ \frac{1}{2} & \frac{1}{10} & \frac{4}{10} \end{bmatrix}\)

IF I have done everything correctly, \(\displaystyle \begin{bmatrix}0 & \frac{3}{10} & \frac{3}{10} \\ 0 & \frac{1}{10} & -\frac{1}{10} \\ \frac{1}{2} & \frac{1}{10} & \frac{4}{10} \end{bmatrix}\) is the inverse matrix to \(\displaystyle \begin{bmatrix}0 & -2 & 2 \\ 3 & 1 & 3 \\ 1 & -3 & 1 \end{bmatrix}\)