Question

Solve the following system of linear equations, using matrix method

$x-y+z=4,2x+y-2z=0,x+y+z=2$

Answer 1

The given system can be written as AX=B, where
$A=\left[\begin{array}{ccc}1& -1& 1\\ 2& 1& -3\\ 1& 1& 1\end{array}\right],X=\left[\begin{array}{c}x\\ y\\ z\end{array}\right]andB=\left[\begin{array}{c}4\\ 0\\ 2\end{array}\right]$
Here, $|A|=\left[\begin{array}{ccc}1& -1& 1\\ 2& 1& -3\\ 1& 1& 1\end{array}\right]=1(1+3)-(-1)(2+3)+1(2-1)=4+5+1=10\ne 0$
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
$A_{11}=1+3=4,A_{12}=-(2+3)=-5,A_{13}=2-1=1,A_{21}=-(-1-1)=2,A_{22}=1-1=0,A_{23}=-(1+1)=-2A_{31}=3-1=2,A_{32}=-(-3-2)=5,A_{33}=1+2=3$
$\therefore adj\left(A\right)=\left[\begin{array}{ccc}4& -5& 1\\ 2& 0& -2\\ 2& 5& 3\end{array}\right]=\left[\begin{array}{ccc}4& 2& 2\\ -5& 0& 5\\ 1& -2& 3\end{array}\right]$
$\therefore A^{-1}=\frac{1}{|A|}\left(adjA\right)=\frac{1}{10}\left[\begin{array}{ccc}4& 2& 2\\ -5& 0& 5\\ 1& -2& 3\end{array}\right]$
Now, $X=A^{-1}B\to \left[\begin{array}{c}x\\ y\\ z\end{array}\right]=\frac{1}{10}\left[\begin{array}{ccc}4& 2& 2\\ -5& 0& 5\\ 1& -2& 3\end{array}\right]\left[\begin{array}{c}4\\ 0\\ 2\end{array}\right]$
$\Rightarrow \left[\begin{array}{c}x\\ y\\ z\end{array}\right]=\frac{1}{10}\left[\begin{array}{c}16+0+4\\ -20+0+10\\ 4+0+6\end{array}\right]\Rightarrow \left[\begin{array}{c}x\\ y\\ z\end{array}\right]=\frac{1}{10}\left[\begin{array}{c}20\\ -10\\ 10\end{array}\right]=\left[\begin{array}{c}2\\ -1\\ 1\end{array}\right]$
Hence, $x=2,y=-1$ and $z=1$