Solve the following system of equations by matrix method:

$x-y+z = 4$; $x+y+z=2$ and $2x+y-3z=0$

$A$ = $\left[ \begin{matrix}


1 & \phantom{-}-1 & \phantom{-}1 \\


1 & \phantom{-}1 & \phantom{-}1 \\


2 & \phantom{-}1 & \phantom{-}-3 \\


\end{matrix} \right]$, $B$ = $\left[ \begin{matrix}


4 \\


2 \\


0 \\


\end{matrix} \right]$, X = $\left[ \begin{matrix}


x \\


y \\


z \\


\end{matrix} \right]$


$|A|$ = $1(-1-3)+1(-3-2)+1(1-2)$ = $-10 \ne 0$

Hence inverse of A exists.


$A_{11} = -4$, $A_{12} = 5$, $A_{13} = -1$

$A_{21} = -2$, $A_{22} = -5$, $A_{23} = -3$

$A_{31} = -2$, $A_{32} = 0$, $A_{33} = 2$


∴ adj $A$ = $\left[ \begin{matrix}


-4 & \phantom{-}5 & \phantom{-}-1 \\


-2 & \phantom{-}-5 & \phantom{-}-3 \\


-2 & \phantom{-}0 & \phantom{-}2 \\


\end{matrix} \right]^T$ = $\left[ \begin{matrix}


-4 & \phantom{-}-2 & \phantom{-}-2 \\


5 & \phantom{-}-5 & \phantom{-}0 \\


-1 & \phantom{-}-3 & \phantom{-}2 \\


\end{matrix} \right] $


$A^{-1}$ = $\dfrac{1}{|A|}$ adj $A$ = $\dfrac{1}{-10}$$\left[ \begin{matrix}


-4 & \phantom{-}-2 & \phantom{-}-2 \\


5 & \phantom{-}-5 & \phantom{-}0 \\


-1 & \phantom{-}-3 & \phantom{-}2 \\


\end{matrix} \right] $


$X = A^{-1}B$ = $\dfrac{1}{-10}$$\left[ \begin{matrix}


-4 & \phantom{-}-2 & \phantom{-}-2 \\


5 & \phantom{-}-5 & \phantom{-}0 \\


-1 & \phantom{-}-3 & \phantom{-}2 \\


\end{matrix} \right] $ $\left[ \begin{matrix}


4 \\


2 \\


0 \\


\end{matrix} \right]$ = $\dfrac{1}{10}$$\left[ \begin{matrix}


20 \\


-10 \\


10 \\


\end{matrix} \right]$


∴ $\left[ \begin{matrix}


x \\


y \\


z \\


\end{matrix} \right]$ = $\left[ \begin{matrix}


2 \\


-1 \\


1 \\


\end{matrix} \right]$


∴ $x=2$, $y=-1$ and $z=1$