# How do you solve [(-x,+2y,-4z,-5u,=,27),(-2x,-3y,+3z,-3u,=,9),(5x,-4y,+2z,-3u,=,5), (-2x,+0y,+4z,-5u,=,29)]?

Mar 5, 2018

$x = 0$, $y = 3$, $z = 1$ and $u = - 5$

#### Explanation:

Perform the Gauss Jordan elimination on the augmented matrix

$A = \left(\begin{matrix}- 1 & 2 & - 4 & - 5 & | & 27 \\ - 2 & - 3 & 3 & - 3 & | & 9 \\ 5 & - 4 & 2 & - 3 & | & 5 \\ - 2 & 0 & 4 & - 5 & | & 29\end{matrix}\right)$

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

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

$A = \left(\begin{matrix}- 1 & 2 & - 4 & - 5 & | & 27 \\ 0 & - 7 & 11 & 7 & | & - 45 \\ 0 & 6 & - 18 & - 28 & | & 140 \\ 0 & - 4 & 12 & 5 & | & - 25\end{matrix}\right)$

$R 2 \leftarrow R 2 - 2 R 4$;

$A = \left(\begin{matrix}- 1 & 2 & - 4 & - 5 & | & 27 \\ 0 & 1 & - 13 & - 3 & | & 5 \\ 0 & 6 & - 18 & - 28 & | & 140 \\ 0 & - 4 & 12 & 5 & | & - 25\end{matrix}\right)$

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

$A = \left(\begin{matrix}- 1 & 0 & 22 & 1 & | & 17 \\ 0 & 1 & - 13 & - 3 & | & 5 \\ 0 & 0 & 60 & - 10 & | & 110 \\ 0 & 0 & - 40 & - 7 & | & - 5\end{matrix}\right)$

$R 1 \leftarrow R 1$, $R 3 \leftarrow R 3 \cdot 2$, $R 4 \leftarrow R 4 \cdot 3$;

$A = \left(\begin{matrix}1 & 0 & - 22 & - 1 & | & - 17 \\ 0 & 1 & - 13 & - 3 & | & 5 \\ 0 & 0 & 120 & - 20 & | & 220 \\ 0 & 0 & - 120 & - 21 & | & - 15\end{matrix}\right)$

$R 4 \leftarrow R 4 + R 3$;

$A = \left(\begin{matrix}1 & 0 & - 22 & - 1 & | & - 17 \\ 0 & 1 & - 13 & - 3 & | & 5 \\ 0 & 0 & 120 & - 20 & | & 220 \\ 0 & 0 & 0 & - 41 & | & 205\end{matrix}\right)$

$R 4 \leftarrow \frac{R 4}{- 41}$;

$A = \left(\begin{matrix}1 & 0 & - 22 & - 1 & | & - 17 \\ 0 & 1 & - 13 & - 3 & | & 5 \\ 0 & 0 & 120 & - 20 & | & 220 \\ 0 & 0 & 0 & 1 & | & - 5\end{matrix}\right)$

$R 1 \leftarrow R 1 + R 4$, $R 2 \leftarrow R 2 + 3 R 4$, $R 3 \leftarrow R 3 + 20 R 4$;

$A = \left(\begin{matrix}1 & 0 & - 22 & 0 & | & - 22 \\ 0 & 1 & - 13 & 0 & | & - 10 \\ 0 & 0 & 120 & 0 & | & 120 \\ 0 & 0 & 0 & 1 & | & - 5\end{matrix}\right)$

$R 3 \leftarrow \frac{R 3}{120}$;

$A = \left(\begin{matrix}1 & 0 & - 22 & 0 & | & - 22 \\ 0 & 1 & - 13 & 0 & | & - 10 \\ 0 & 0 & 1 & 0 & | & 1 \\ 0 & 0 & 0 & 1 & | & - 5\end{matrix}\right)$

$R 1 \leftarrow R 1 + 22 R 3$, $R 2 \leftarrow R 2 + 13 R 3$;

$A = \left(\begin{matrix}1 & 0 & 0 & 0 & | & 0 \\ 0 & 1 & 0 & 0 & | & 3 \\ 0 & 0 & 1 & 0 & | & 1 \\ 0 & 0 & 0 & 1 & | & - 5\end{matrix}\right)$

Thus $x = 0$, $y = 3$, $z = 1$ and $u = - 5$