# How do you solve the system 4r-4x+4t=-4, 4r+x-2t=5, and -3r-3x-4t=-16?

Feb 23, 2018

$r = 1$, $x = 3$ and $t = 1$

#### Explanation:

Perform the Gauss Jordan elimination on the augmented matrix

$A = \left(\begin{matrix}4 & - 4 & 4 & | & - 4 \\ 4 & 1 & - 2 & | & 5 \\ - 3 & - 3 & - 4 & | & - 16\end{matrix}\right)$

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

$R 1 \leftarrow \frac{R 1}{4}$

$A = \left(\begin{matrix}1 & - 1 & 1 & | & - 1 \\ 4 & 1 & - 2 & | & 5 \\ - 3 & - 3 & - 4 & | & - 16\end{matrix}\right)$

$R 2 \leftarrow R 1 - 4 R 2$; $R 3 \leftarrow R 3 + 3 R 1$

$A = \left(\begin{matrix}1 & - 1 & 1 & | & - 1 \\ 0 & 5 & - 6 & | & 9 \\ 0 & - 6 & - 1 & | & - 19\end{matrix}\right)$

$R 2 \leftarrow R 2 \cdot 6$; $R 3 \leftarrow R 3 \cdot 5$

$A = \left(\begin{matrix}1 & - 1 & 1 & | & - 1 \\ 0 & 30 & - 36 & | & 54 \\ 0 & - 30 & - 5 & | & - 95\end{matrix}\right)$

$R 3 \leftarrow R 3 + R 2$

$A = \left(\begin{matrix}1 & - 1 & 1 & | & - 1 \\ 0 & 30 & - 36 & | & 54 \\ 0 & 0 & - 41 & | & - 41\end{matrix}\right)$

$R 3 \leftarrow \frac{R 3}{- 41}$

$A = \left(\begin{matrix}1 & - 1 & 1 & | & - 1 \\ 0 & 30 & - 36 & | & 54 \\ 0 & 0 & 1 & | & 1\end{matrix}\right)$

R1larrR1-R2; R2larrR2+36R3

$A = \left(\begin{matrix}1 & - 1 & 0 & | & - 2 \\ 0 & 30 & 0 & | & 90 \\ 0 & 0 & 1 & | & 1\end{matrix}\right)$

$R 2 \leftarrow \frac{R 2}{30}$

$A = \left(\begin{matrix}1 & - 1 & 0 & | & - 2 \\ 0 & 1 & 0 & | & 3 \\ 0 & 0 & 1 & | & 1\end{matrix}\right)$

$R 1 \leftarrow R 1 + R 2$

$A = \left(\begin{matrix}1 & 0 & 0 & | & 1 \\ 0 & 1 & 0 & | & 3 \\ 0 & 0 & 1 & | & 1\end{matrix}\right)$

Thus, $r = 1$, $x = 3$ and $t = 1$