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Solve System Using Matrix Inverse

brinlin

New member
Aug 13, 2021
13
Pre9.PNG
 

skeeter

Well-known member
MHB Math Helper
Mar 1, 2012
1,017
$
\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)
 

Country Boy

Well-known member
MHB Math Helper
Jan 30, 2018
881
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 \(\displaystyle \begin{bmatrix}0 & -2 & 2 \\ 3 & 1 & 3 \\ 1 & -3 & 1 \end{bmatrix}\)