Question

Solve the following system of linear equations, using matrix method

$x-y+2z=7,3x+4y-5z=-5,2x-y+3z=12$

Answer 1

The given system can be written as Ax =B, where

$A=\left[\begin{array}{ccc}1& -1& 2\\ 3& 4& -5\\ 2& -1& 3\end{array}\right],X=\left[\begin{array}{c}x\\ y\\ z\end{array}\right]andB=\left[\begin{array}{c}7\\ -5\\ 12\end{array}\right]$
Here, $|A|=\left[\begin{array}{ccc}1& -1& 2\\ 3& 4& -5\\ 2& -1& 3\end{array}\right]=1(12-5-(-1\left)\right)(9+10)+2(-3-8)$
$=7+19-22=4\ne 0$
Thus, A is non-singular, Therefore, its inverse exists.
Therefore, the given system is consistent and has a unique solution given by $X=A^{-1}B$.
Cofactors of A are
$A_{11}=12-5=7,A_{12}=-(9+10)=-19,A_{13}=-3-8=-11A_{21}=-(-3+2)=1,A_{22}=3-4=-1,A_{23}=-(-1+2)=-1A_{31}=5-8=-3,A_{32}=-(-5-6)=11,A_{33}=4+3=7$
$adj\left(A\right)={\left[\begin{array}{ccc}7& -19& -11\\ 1& -1& -1\\ -3& 11& 7\end{array}\right]}^T=\left[\begin{array}{ccc}7& 1& -3\\ -19& -1& 11\\ -11& -1& 7\end{array}\right]$

$\therefore A^{-1}=\frac{1}{|A|}\left(adjA\right)=\frac{1}{4}\left[\begin{array}{ccc}7& 1& -3\\ -19& -1& 11\\ -11& -1& 7\end{array}\right]$
Now, $X=A^{-1}B\Rightarrow \left[\begin{array}{c}x\\ y\\ z\end{array}\right]=\frac{1}{4}\left[\begin{array}{ccc}7& 1& -3\\ -19& -1& 11\\ -11& -1& 7\end{array}\right]\left[\begin{array}{c}7\\ -5\\ 12\end{array}\right]=\frac{1}{4}\left[\begin{array}{c}49-5-36\\ -133+5+132\\ -77+5+84\end{array}\right]$
$\Rightarrow \left[\begin{array}{c}x\\ y\\ z\end{array}\right]=\frac{1}{4}\left[\begin{array}{c}8\\ 4\\ 12\end{array}\right]=\left[\begin{array}{c}2\\ 1\\ 3\end{array}\right]$
Hence, x=2, y=1 and z=3