# How do you solve 24x + 24y + 16z = 106, 8x + 24y + 40z = 135 and 24x + 16y - 8z = 42 using matrices?

Dec 16, 2017

$x = \frac{11}{8}$, $y = \frac{13}{8}$ and $z = \frac{17}{8}$

#### Explanation:

Perform the Gauss Jordan elimination on the augmented matrix

$A = \left(\begin{matrix}24 & 24 & 16 & | & 106 \\ 8 & 24 & 40 & | & 135 \\ 24 & 16 & - 8 & | & 42\end{matrix}\right)$

I have written the equations not in the sequence as in the question in order to get $1$ as pivot.

Perform the folowing operations on the rows of the matrix

$R 1 \leftarrow R 1 - 3 R 2$ ; $R 3 \leftarrow R 3 - 3 R 2$

$A = \left(\begin{matrix}0 & - 48 & - 104 & | & - 299 \\ 8 & 24 & 40 & | & 135 \\ 0 & - 56 & - 128 & | & - 363\end{matrix}\right)$

$R 1 \leftarrow \frac{R 1}{- 48}$

$A = \left(\begin{matrix}0 & 1 & \frac{13}{6} & | & \frac{299}{48} \\ 8 & 24 & 40 & | & 135 \\ 0 & - 56 & - 128 & | & - 363\end{matrix}\right)$

$R 2 \leftarrow R 1 - 24 R 1$ ; $R 3 \leftarrow R 3 + 56 R 1$

$A = \left(\begin{matrix}0 & 1 & \frac{13}{6} & | & \frac{299}{48} \\ 8 & 0 & - 12 & | & - \frac{29}{2} \\ 0 & 0 & - \frac{20}{3} & | & - \frac{85}{6}\end{matrix}\right)$

$R 3 \leftarrow \left(R 3\right) \cdot \left(- \frac{3}{20}\right)$

$A = \left(\begin{matrix}0 & 1 & \frac{13}{6} & | & \frac{299}{48} \\ 8 & 0 & - 12 & | & - \frac{29}{2} \\ 0 & 0 & 1 & | & \frac{17}{8}\end{matrix}\right)$

$R 1 \leftarrow R 1 - \frac{13}{6} R 3$ ; $R 2 \leftarrow R 2 + 12 R 3$

$A = \left(\begin{matrix}0 & 1 & 0 & | & \frac{13}{8} \\ 8 & 0 & 0 & | & 11 \\ 0 & 0 & 1 & | & \frac{17}{8}\end{matrix}\right)$

$R 2 \leftarrow \frac{R 2}{8}$

$A = \left(\begin{matrix}0 & 1 & 0 & | & \frac{13}{8} \\ 1 & 0 & 0 & | & \frac{11}{8} \\ 0 & 0 & 1 & | & \frac{17}{8}\end{matrix}\right)$

Thus, solution of equation system is $x = \frac{11}{8}$, $y = \frac{13}{8}$ and $z = \frac{17}{8}$