# How do you solve 5x - 5y + 10z = -11, 10x + 5y - 5z = 1 and 15x - 15y -10z = -1 using matrices?

Aug 2, 2018

The solution is $\left(\begin{matrix}x \\ y \\ z\end{matrix}\right) = \left(\begin{matrix}- \frac{2}{5} \\ \frac{1}{5} \\ - \frac{4}{5}\end{matrix}\right)$

#### Explanation:

Perform the Gauss-Jordan Elimination on the augmented matrix

$A = \left(\begin{matrix}5 & - 5 & 10 & | & - 11 \\ 10 & 5 & - 5 & | & 1 \\ 15 & - 15 & - 10 & | & - 1\end{matrix}\right)$

Make the pivot in the first column and first row

$R 1 \leftarrow \left(\frac{R 1}{5}\right)$

$\left(\begin{matrix}1 & - 1 & 2 & | & - 2.2 \\ 10 & 5 & - 5 & | & 1 \\ 15 & - 15 & - 10 & | & - 1\end{matrix}\right)$

Eliminate the first column

$R 2 \leftarrow \left(R 2 - 10 R 1\right)$ and $R 3 \leftarrow \left(R 3 - 15 R 1\right)$

$\left(\begin{matrix}1 & - 1 & 2 & | & - 2.2 \\ 0 & 15 & - 25 & | & 23 \\ 0 & 0 & - 40 & | & 32\end{matrix}\right)$

Make the pivot in the second column and second row

$R 2 \leftarrow \left(\frac{R 2}{15}\right)$

$\left(\begin{matrix}1 & - 1 & 2 & | & - 2.2 \\ 0 & 1 & - \frac{5}{3} & | & \frac{23}{15} \\ 0 & 0 & - 40 & | & 32\end{matrix}\right)$

Eliminate the second column

$R 1 \leftarrow \left(R 1 + R 2\right)$

$\left(\begin{matrix}1 & 0 & \frac{1}{3} & | & - \frac{2}{3} \\ 0 & 1 & - \frac{5}{3} & | & \frac{23}{15} \\ 0 & 0 & - 40 & | & 32\end{matrix}\right)$

Make the pivot in the third column and third row

$R 3 \leftarrow \left(\frac{R 3}{-} 40\right)$

$\left(\begin{matrix}1 & 0 & \frac{1}{3} & | & - \frac{2}{3} \\ 0 & 1 & - \frac{5}{3} & | & \frac{23}{15} \\ 0 & 0 & 1 & | & - \frac{4}{5}\end{matrix}\right)$

Eliminate the third column

$R 1 \leftarrow \left(R 1 - \frac{1}{3} R 3\right)$, and $R 2 \leftarrow \left(R 2 + \frac{5}{3} R 3\right)$,

$\left(\begin{matrix}1 & 0 & 0 & | & - \frac{2}{5} \\ 0 & 1 & 0 & | & \frac{1}{5} \\ 0 & 0 & 1 & | & - \frac{4}{5}\end{matrix}\right)$

The solution is

$\left(\begin{matrix}x \\ y \\ z\end{matrix}\right) = \left(\begin{matrix}- \frac{2}{5} \\ \frac{1}{5} \\ - \frac{4}{5}\end{matrix}\right)$