# How do you solve using gaussian elimination or gauss-jordan elimination, 2x - y + 5z - t = 7, x + 2y - 3t = 6, 3x - 4y + 10z + t = 8?

Aug 12, 2017

The solutions are $\left(\begin{matrix}x \\ y \\ z \\ t\end{matrix}\right) = \left(\begin{matrix}4 - 2 z + t \\ 1 + z + t \\ z \\ t\end{matrix}\right)$

#### Explanation:

We perform the Gauss Jordan elimination with the augmented matrix

$\left(\begin{matrix}1 & 2 & 0 & - 3 & : & 6 \\ 2 & - 1 & 5 & - 1 & : & 7 \\ 3 & - 4 & 10 & 1 & : & 8\end{matrix}\right)$

$R 3 \leftarrow R 3 - 3 R 1$, $\implies$, $\left(\begin{matrix}1 & 2 & 0 & - 3 & : & 6 \\ 2 & - 1 & 5 & - 1 & : & 7 \\ 0 & - 10 & 10 & 10 & : & - 10\end{matrix}\right)$

$R 2 \leftarrow R 2 - 2 R 1$, $\implies$, $\left(\begin{matrix}1 & 2 & 0 & - 3 & : & 6 \\ 0 & - 5 & 5 & 5 & : & - 5 \\ 0 & - 10 & 10 & 10 & : & - 10\end{matrix}\right)$

$R 3 \leftarrow \frac{R 3}{- 10}$, $\implies$, $\left(\begin{matrix}1 & 2 & 0 & - 3 & : & 6 \\ 0 & - 5 & 5 & 5 & : & - 5 \\ 0 & 1 & - 1 & - 1 & : & 1\end{matrix}\right)$

$R 2 \leftarrow \frac{R 2}{- 5}$, $\implies$, $\left(\begin{matrix}1 & 2 & 0 & - 3 & : & 6 \\ 0 & 1 & - 1 & - 1 & : & 1 \\ 0 & 1 & - 1 & - 1 & : & 1\end{matrix}\right)$

$R 3 \leftarrow R 3 - R 2$, $\implies$, $\left(\begin{matrix}1 & 2 & 0 & - 3 & : & 6 \\ 0 & 1 & - 1 & - 1 & : & 1 \\ 0 & 0 & 0 & 0 & : & 0\end{matrix}\right)$

$R 1 \leftarrow R 1 - 2 R 2$, $\implies$, $\left(\begin{matrix}1 & 0 & 2 & - 1 & : & 4 \\ 0 & 1 & - 1 & - 1 & : & 1 \\ 0 & 0 & 0 & 0 & : & 0\end{matrix}\right)$

The solutions are

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

$z$ and $t$ are free