# How do you solve a+3b+2c=3 , 2a-b-3c= -8, 5a+2b+c=9 using matrices?

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Jan 19, 2018

#### Answer:

$a = 2$, $b = - 3$ and $c = 5$

#### Explanation:

Perform the Gauss Jordan elimination on the augmented matrix

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

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

$R 2 \leftarrow R 2 - 2 R 1$; $R 3 \leftarrow R 3 - 5 R 1$

$A = \left(\begin{matrix}1 & 3 & 2 & | & 3 \\ 0 & - 7 & - 7 & | & - 14 \\ 0 & - 13 & - 9 & | & - 6\end{matrix}\right)$

$R 2 \leftarrow \frac{R 2}{- 7}$

$A = \left(\begin{matrix}1 & 3 & 2 & | & 3 \\ 0 & 1 & 1 & | & 2 \\ 0 & - 13 & - 9 & | & - 6\end{matrix}\right)$

$R 3 \leftarrow R 3 + 13 R 2$

$A = \left(\begin{matrix}1 & 3 & 2 & | & 3 \\ 0 & 1 & 1 & | & 2 \\ 0 & 0 & 4 & | & 20\end{matrix}\right)$

$R 3 \leftarrow \frac{R 3}{4}$

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

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

$A = \left(\begin{matrix}1 & 3 & 0 & | & - 7 \\ 0 & 1 & 0 & | & - 3 \\ 0 & 0 & 1 & | & 5\end{matrix}\right)$

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

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

Thus $a = 2$, $b = - 3$ and $c = 5$

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