Question
Solve system of linear equations, using matrix method.
x - y + 2z=7
3x + 4y - 5z = -5
2x - y + 3z = 12

Answer

Matrix form of given equations is AX = B $\Rightarrow\ \begin{bmatrix}1&-1&2\\3&4&-5\\2&-1&3\end{bmatrix}\begin{bmatrix}x\\y\\z\end{bmatrix}=\begin{bmatrix}7\\-5\\12\end{bmatrix}$
$\text{Here}\ \text{A}=\begin{bmatrix}1&-1&2\\3&4&-5\\2&-1&3\end{bmatrix},\ \text{X}=\begin{bmatrix}x\\y\\z\end{bmatrix}\text{and B}=\begin{bmatrix}7\\-5\\12\end{bmatrix}$
$\therefore\ \text{|A|}=\begin{vmatrix}1&-1&2\\3&4&-5\\2&-1&3\end{vmatrix}=1(12-5)-(-1)(9+10)+2(-3-8)=7+9-22=4\neq0$
Therefore, solution is unique and $\text{X=A}^{-1}\text{B}=\frac{1}{\text{|A|}}\text{(adj. A)B}$
$\Rightarrow\ \begin{bmatrix}x\\y\\z\end{bmatrix}=\frac{1}{4}\begin{bmatrix}7&1&-3\\-19&-1&11\\-11&-1&7\end{bmatrix}\begin{bmatrix}7\\-5\\12\end{bmatrix}$
$=\frac{1}{4}\begin{bmatrix}49-5-36\\-133+5+132\\-77+5+84\end{bmatrix}=\frac{1}{4}\begin{bmatrix}8\\4\\12\end{bmatrix}=\begin{bmatrix}2\\1\\3\end{bmatrix}$
Therefore, x = 2, y = 1 and z = 3

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