Question

Solve the following system of linear equations, using matrix method

$2x+y+z=1,x-2y-z=\frac{3}{2},3y-5x=9$

Answer 1

The given system can be written as AX=B, where
$A=\left[\begin{array}{ccc}2& 1& 1\\ 2& -4& -2\\ 0& 3& -5\end{array}\right],X=\left[\begin{array}{c}x\\ y\\ z\end{array}\right]andB=\left[\begin{array}{c}1\\ 3\\ 9\end{array}\right]$
Here, $|A|=\left[\begin{array}{ccc}2& 1& 1\\ 2& -4& -2\\ 0& 3& -5\end{array}\right]=2(20+6)-1(-10-0)+1(6-0)=52+10+6=68\ne 0$
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
$A_{11}=20+6=26,A_{12}=-(-10+0)=10,A_{13}=6+0=6A_{21}=-(-5-3)=8,A_{22}=-10-0=-10,A_{23}=-(6-0)=-6A_{31}=(-2+4)=2,A_{32}=-(-4-2)=6,A_{33}=-8-2=-10$

$adj\left(A\right)={\left[\begin{array}{ccc}26& 10& 6\\ 8& -10& -6\\ 2& 6& -10\end{array}\right]}^T=\left[\begin{array}{ccc}26& 8& 2\\ 10& -10& 6\\ 6& -6& -10\end{array}\right]$
$\therefore A^{-1}=\frac{1}{|a|}\left(adjA\right)=\frac{1}{68}\left[\begin{array}{ccc}26& 8& 2\\ 10& -10& 6\\ 6& -6& -10\end{array}\right]$
Now, $X=A^{-1}B\Rightarrow \left[\begin{array}{c}x\\ y\\ z\end{array}\right]=\frac{1}{68}\left[\begin{array}{ccc}26& 8& 2\\ 10& -10& 6\\ 6& -6& -10\end{array}\right]\left[\begin{array}{c}1\\ 3\\ 9\end{array}\right]$
$\Rightarrow \left[\begin{array}{c}x\\ y\\ z\end{array}\right]=\frac{1}{68}\left[\begin{array}{c}26+24+18\\ 10-30+54\\ 6-18-90\end{array}\right]=\frac{1}{68}\left[\begin{array}{c}68\\ 34\\ -102\end{array}\right]\Rightarrow \left[\begin{array}{c}x\\ y\\ z\end{array}\right]=\left[\begin{array}{c}1\\ \frac{1}{2}\\ \frac{-3}{2}\end{array}\right]$
Hence, $x=1,y=\frac{1}{2}$ and $z=\frac{-3}{2}$.