# How do you solve 3a + 6b + c + 4d = 2, a - 2b + 3c - d = -3, 2a - 4c + d= -15 and a + 8b + 2c + d = 6 using matrices?

Dec 15, 2017

The solution is $\left(\begin{matrix}a \\ b \\ c \\ d\end{matrix}\right) = \left(\begin{matrix}- 5 \\ \frac{1}{2} \\ 2 \\ 3\end{matrix}\right)$

#### Explanation:

Perform the Gauss Jordan elimination on the augmented matrix

$A = \left(\begin{matrix}1 & 8 & 2 & 1 & | & 6 \\ 1 & - 2 & 3 & - 1 & | & - 3 \\ 3 & 6 & 1 & 4 & | & 2 \\ 2 & 0 & - 4 & 1 & | & - 15\end{matrix}\right)$

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

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

$A = \left(\begin{matrix}1 & 8 & 2 & 1 & | & 6 \\ 0 & - 10 & 1 & - 2 & | & - 9 \\ 0 & - 18 & - 5 & 1 & | & - 16 \\ 0 & - 16 & - 8 & - 1 & | & - 27\end{matrix}\right)$

$R 3 \leftarrow \frac{R 3}{- 10}$

$A = \left(\begin{matrix}1 & 8 & 2 & 1 & | & 6 \\ 0 & 1 & - \frac{1}{10} & \frac{1}{5} & | & \frac{9}{10} \\ 0 & - 18 & - 5 & 1 & | & - 16 \\ 0 & - 16 & - 8 & - 1 & | & - 27\end{matrix}\right)$

$R 3 \leftarrow \left(R 3 + 18 R 2\right)$ , $R 4 \leftarrow \left(R 4 + 16 R 2\right)$

$A = \left(\begin{matrix}1 & 8 & 2 & 1 & | & 6 \\ 0 & 1 & - \frac{1}{10} & \frac{1}{5} & | & \frac{9}{10} \\ 0 & 0 & - \frac{34}{5} & \frac{23}{5} & | & \frac{1}{5} \\ 0 & 0 & - \frac{48}{5} & \frac{11}{5} & | & - \frac{63}{5}\end{matrix}\right)$

$R 3 \leftarrow \left(R 3 \times \left(- \frac{5}{34}\right)\right)$

$A = \left(\begin{matrix}1 & 8 & 2 & 1 & | & 6 \\ 0 & 1 & - \frac{1}{10} & \frac{1}{5} & | & \frac{9}{10} \\ 0 & 0 & 1 & - \frac{23}{34} & | & - \frac{1}{34} \\ 0 & 0 & - \frac{48}{5} & \frac{11}{5} & | & - \frac{63}{5}\end{matrix}\right)$

$R 4 \leftarrow \left(R 4 + \left(\frac{48}{5}\right) \cdot R 3\right)$

$A = \left(\begin{matrix}1 & 8 & 2 & 1 & | & 6 \\ 0 & 1 & - \frac{1}{10} & \frac{1}{5} & | & \frac{9}{10} \\ 0 & 0 & 1 & - \frac{23}{34} & | & - \frac{1}{34} \\ 0 & 0 & 0 & - \frac{73}{17} & | & - \frac{219}{17}\end{matrix}\right)$

$R 4 \leftarrow \left(R 4 \cdot \left(- \frac{17}{73}\right)\right)$

$A = \left(\begin{matrix}1 & 8 & 2 & 1 & | & 6 \\ 0 & 1 & - \frac{1}{10} & \frac{1}{5} & | & \frac{9}{10} \\ 0 & 0 & 1 & - \frac{23}{34} & | & - \frac{1}{34} \\ 0 & 0 & 0 & 1 & | & 3\end{matrix}\right)$

$R 3 \leftarrow \left(R 3 + \frac{23}{34} R 4\right)$

$A = \left(\begin{matrix}1 & 8 & 2 & 1 & | & 6 \\ 0 & 1 & - \frac{1}{10} & \frac{1}{5} & | & \frac{9}{10} \\ 0 & 0 & 1 & 0 & | & 2 \\ 0 & 0 & 0 & 1 & | & 3\end{matrix}\right)$

$R 2 \leftarrow \left(R 2 + \frac{1}{10} R 3\right)$ , $R 2 \leftarrow R 2 - \frac{1}{5} R 3$

$A = \left(\begin{matrix}1 & 8 & 2 & 1 & | & 6 \\ 0 & 1 & 0 & 0 & | & \frac{1}{2} \\ 0 & 0 & 1 & 0 & | & 2 \\ 0 & 0 & 0 & 1 & | & 3\end{matrix}\right)$

$R 1 \leftarrow \left(R 1 - 8 R 2\right)$ , $R 1 \leftarrow R 1 - 2 R 3$ ,; $R 1 \leftarrow R 1 - R 4$

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