# How do you solve w+4x+3y-11z=42, 6w+9x+8y-9z=31, -5w+6x+3y+13z=2, and 8w+3x-7y+6z=31 using matrices?

Oct 1, 2017

The answer is $\left(\begin{matrix}w \\ x \\ y \\ z\end{matrix}\right) = \left(\begin{matrix}- \frac{12054}{4889} \\ \frac{38342}{4889} \\ - \frac{31301}{4889} \\ - \frac{14357}{4889}\end{matrix}\right)$

#### Explanation:

Perform the Gauss Jordan elimination on the augmented matrix

$A = \left(\begin{matrix}1 & 4 & 3 & - 11 & | & 42 \\ 6 & 9 & 8 & - 9 & | & 31 \\ - 5 & 6 & 3 & 13 & | & 2 \\ 8 & 3 & - 7 & 6 & | & 31\end{matrix}\right)$

Perform the row operations

$R 2 \leftarrow R 2 - 6 R 1$, $R 3 \leftarrow R 3 + 5 R 1$, $R 4 - 8 R 1$

$A = \left(\begin{matrix}1 & 4 & 3 & - 11 & | & 42 \\ 0 & - 15 & - 10 & 57 & | & - 221 \\ 0 & 26 & 18 & - 42 & | & 212 \\ 0 & - 29 & - 31 & 94 & | & - 305\end{matrix}\right)$

$R 2 \leftarrow \frac{R 2}{- 15}$

$A = \left(\begin{matrix}1 & 4 & 3 & - 11 & | & 42 \\ 0 & 1 & \frac{2}{3} & - \frac{19}{5} & | & \frac{221}{15} \\ 0 & 26 & 18 & - 42 & | & 212 \\ 0 & - 29 & - 31 & 94 & | & - 305\end{matrix}\right)$

$R 3 \leftarrow R 3 - 26 R 2$, $R 4 \leftarrow R 4 + 29 R 2$,

$A = \left(\begin{matrix}1 & 4 & 3 & - 11 & | & 42 \\ 0 & 1 & \frac{2}{3} & - \frac{19}{5} & | & \frac{221}{15} \\ 0 & 0 & \frac{2}{3} & \frac{284}{5} & | & - \frac{2566}{15} \\ 0 & 0 & - \frac{35}{3} & - \frac{81}{5} & | & \frac{1834}{15}\end{matrix}\right)$

$R 3 \leftarrow \frac{R 3}{\frac{2}{3}}$

$A = \left(\begin{matrix}1 & 4 & 3 & - 11 & | & 42 \\ 0 & 1 & \frac{2}{3} & - \frac{19}{5} & | & \frac{221}{15} \\ 0 & 0 & 1 & \frac{426}{5} & | & - \frac{1283}{5} \\ 0 & 0 & - \frac{35}{3} & - \frac{81}{5} & | & \frac{1834}{15}\end{matrix}\right)$

$R 4 \leftarrow \left(R 4\right) \cdot \left(- \frac{3}{35}\right) + R 3$

$A = \left(\begin{matrix}1 & 4 & 3 & - 11 & | & 42 \\ 0 & 1 & \frac{2}{3} & - \frac{19}{5} & | & \frac{221}{15} \\ 0 & 0 & 1 & \frac{426}{5} & | & - \frac{1283}{5} \\ 0 & 0 & 0 & \frac{4889}{5} & | & - \frac{14357}{5}\end{matrix}\right)$

$R 4 \leftarrow \left(R 4\right) \cdot \left(\frac{5}{4889}\right)$

$A = \left(\begin{matrix}1 & 4 & 3 & - 11 & | & 42 \\ 0 & 1 & \frac{2}{3} & - \frac{19}{5} & | & \frac{221}{15} \\ 0 & 0 & 1 & \frac{426}{5} & | & - \frac{1283}{5} \\ 0 & 0 & 0 & 1 & | & - \frac{14357}{4889}\end{matrix}\right)$

$R 3 \leftarrow \left(R 3\right) - \left(\frac{426}{5}\right) R 4$

$A = \left(\begin{matrix}1 & 4 & 3 & - 11 & | & 42 \\ 0 & 1 & \frac{2}{3} & - \frac{19}{5} & | & \frac{221}{15} \\ 0 & 0 & 1 & 0 & | & - \frac{31301}{4889} \\ 0 & 0 & 0 & 1 & | & - \frac{14357}{4889}\end{matrix}\right)$

$R 2 \leftarrow \left(R 2\right) - \left(\frac{2}{3}\right) R 3$, $R 2 \leftarrow R 2 + \frac{19}{5} R 4$

$A = \left(\begin{matrix}1 & 4 & 3 & - 11 & | & 42 \\ 0 & 1 & 0 & 0 & | & \frac{38342}{4889} \\ 0 & 0 & 1 & 0 & | & - \frac{31301}{4889} \\ 0 & 0 & 0 & 1 & | & - \frac{14357}{4889}\end{matrix}\right)$

$R 1 \leftarrow \left(R 1\right) - \left(4\right) R 2$, $R 1 \leftarrow R 1 - 3 R 3$, $R 1 \leftarrow R 1 + 11 R 4$

$A = \left(\begin{matrix}1 & 0 & 0 & 0 & | & - \frac{12054}{4889} \\ 0 & 1 & 0 & 0 & | & \frac{38342}{4889} \\ 0 & 0 & 1 & 0 & | & - \frac{31301}{4889} \\ 0 & 0 & 0 & 1 & | & - \frac{14357}{4889}\end{matrix}\right)$