Solve system of linear equations, using matrix method. 2x + 3y + … - Vidyadip

Question

Solve system of linear equations, using matrix method.

2x + 3y + 3z = 5

x - 2y + z = -4

3x - y - 2z = 3

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Answer

Matrix form of given equations is AX = B $\Rightarrow\ \begin{bmatrix}2&3&3\\1&-2&1\\3&-1&-2\end{bmatrix}\begin{bmatrix}x\\y\\z\end{bmatrix}=\begin{bmatrix}5\\-4\\3\end{bmatrix}$
$\text{Here}\ \text{A}= \begin{bmatrix}2&3&3\\1&-2&1\\3&-1&-2\end{bmatrix},\text{X}=\begin{bmatrix}x\\y\\z\end{bmatrix}\text{and B}=\begin{bmatrix}5\\-4\\3\end{bmatrix}$
$\therefore\ \text{|A|}=\begin{vmatrix}2&3&3\\1&-2&1\\3&-1&-2\end{vmatrix}=2(4+1)-3(-2-3)+3(-1+6)=10+15+15=40\neq0$
Therefore, solution is unique and $\text{X=A}^{-1}\text{B}=\frac{1}{\text{|A|}}\text{(adj. A)B}$
$\Rightarrow\ \begin{bmatrix}x\\y\\z\end{bmatrix}=\frac{1}{40}\begin{bmatrix}5&3&9\\5&-13&1\\5&11&-7\end{bmatrix}\begin{bmatrix}5\\-4\\3\end{bmatrix}$
$=\frac{1}{40}\begin{bmatrix}25-12+27\\25+52+3\\25-44-21\end{bmatrix}=\frac{1}{40}\begin{bmatrix}40\\80\\-40\end{bmatrix}=\begin{bmatrix}1\\2\\-1\end{bmatrix}$
Therefore, x = 1, y = 2 and z = -1

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