Question

Solve by matrix method 2x – y + z = 4, x + y + z = 1, x – 3y – 2z = 2 ​

Answer 1

The given system can be written as

$\left[\begin{array}{ccc}2&-1&1\\1&1&1\\1&-3&-2\end{array}\right]\left[\begin{array}{c}x\\y\\z\end{array}\right]=\left[\begin{array}{c}4\\1\\2\end{array}\right]$

This is of the form AX=B

Here

$A=\left[\begin{array}{ccc}2&-1&1\\1&1&1\\1&-3&-2\end{array}\right]$

$|A|=2(-2+3)+1(-2-1)+1(-3-1)$

$|A|=2-3-4$

$|A|=-5$

$\text{Cofactor matrix of A is]$

$\left[\begin{array}{ccc}1&3&-3\\-5&-5&5\\-2&-1&3\end{array}\right]$

$\textbf{Adjoint of matrix A = Transpose of cofactor matrix of A}$

$\text{Then }$

$adjA=\left[\begin{array}{ccc}1&-5&-2\\3&-5&-1\\-4&5&3\end{array}\right]$

$A^{-1}=\frac{1}{|A|}(adjA)$

$A^{-1}=\frac{1}{-5}\left[\begin{array}{ccc}1&-5&-2\\3&-5&-1\\-4&5&3\end{array}\right]$

$\text{Now,}$

$X=A^{-1}B$

$X=\frac{1}{-5}\left[\begin{array}{ccc}1&-5&-2\\3&-5&-1\\-4&5&3\end{array}\right]\left[\begin{array}{c}4\\1\\2\end{array}\right]$

$X=\frac{1}{-5}\left[\begin{array}{ccc}4-5-4\\12-5-2\\-16+5+6\end{array}\right]$

$X=\frac{1}{-5}\left[\begin{array}{ccc}-5\\5\\-5\end{array}\right]$

$X=\left[\begin{array}{ccc}1\\-1\\1\end{array}\right]$

$\text{The solution is}$

$\textbf{x= 1, y= -1, z= 1}$