# Solve the following systems of simultaneous equations using the inverse matrix of Gauss Jordan X1+2x2+3x3=3? 2x1+4x2+5x3=4? 3x1+5x2+6x3=8?

Aug 14, 2018

The solution is $\left(\begin{matrix}{x}_{1} \\ {x}_{2} \\ {x}_{3}\end{matrix}\right) = \left(\begin{matrix}7 \\ - 5 \\ 2\end{matrix}\right)$

#### Explanation:

The augmented matrix is

$A = \left(\begin{matrix}1 & 2 & 3 & | & 3 \\ 2 & 4 & 5 & | & 4 \\ 3 & 5 & 6 & | & 8\end{matrix}\right)$

The main matrix is

${A}_{1} = \left(\begin{matrix}1 & 2 & 3 \\ 2 & 4 & 5 \\ 3 & 5 & 6\end{matrix}\right)$

The inverse is calculated as follows

Write side by side $A$ and ${I}_{3}$ on the right

$\left(\begin{matrix}1 & 2 & 3 \\ 2 & 4 & 5 \\ 3 & 5 & 6\end{matrix}\right) \left(\begin{matrix}1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1\end{matrix}\right)$

Perform the row operations

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

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

$R 3 \leftrightarrow R 2$

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

$R 2 \leftarrow \frac{R 2}{- 1}$ and $R 3 \leftarrow \frac{R 3}{- 1}$

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

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

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

$R 1 \leftarrow R 1 - 2 \times R 2$

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

Therefore,

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

Then,

((x_1),(x_2),(x_3))=((1,-3,2),(-3,3,-1),(2,-1,0))*((3),(4),(8)

$= \left(\begin{matrix}7 \\ - 5 \\ 2\end{matrix}\right)$