# How do you solve the system 3a-3b+4c=-23, a+2b-3c=25, and 4a-b+c=25?

##### 1 Answer
Aug 6, 2017

The solution is $S = \emptyset$

#### Explanation:

We perform the Gauss Jordan elimination with the augmented matrix

$\left(\begin{matrix}1 & 2 & - 3 & : & 25 \\ 3 & - 3 & 4 & : & - 23 \\ 4 & - 1 & 1 & : & 25\end{matrix}\right)$

$R 3 \leftarrow R 3 - 4 R 1$, $\implies$, $\left(\begin{matrix}1 & 2 & - 3 & : & 25 \\ 3 & - 3 & 4 & : & - 23 \\ 0 & - 9 & 13 & : & - 75\end{matrix}\right)$

$R 2 \leftarrow R 2 - 3 R 1$, $\implies$, $\left(\begin{matrix}1 & 2 & - 3 & : & 25 \\ 0 & - 9 & 13 & : & - 98 \\ 0 & - 9 & 13 & : & - 75\end{matrix}\right)$

$R 3 \leftarrow R 3 - R 2$, $\implies$, $\left(\begin{matrix}1 & 2 & - 3 & : & 25 \\ 0 & - 9 & 13 & : & - 98 \\ 0 & 0 & 0 & : & 23\end{matrix}\right)$

$R 2 \leftarrow \frac{R 2}{- 9}$, $\implies$, $\left(\begin{matrix}1 & 2 & - 3 & : & 25 \\ 0 & 1 & - \frac{13}{9} & : & \frac{98}{9} \\ 0 & 0 & 0 & : & 23\end{matrix}\right)$

$R 1 \leftarrow R 1 - 2 R 2$, $\implies$, $\left(\begin{matrix}1 & 0 & - \frac{1}{9} & : & \frac{29}{9} \\ 0 & 1 & - \frac{13}{9} & : & \frac{98}{9} \\ 0 & 0 & 0 & : & 23\end{matrix}\right)$

The system of equations is inconsistent.