Question

Solve the following system of linear equations, using matrix method

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

Answer 1

The given system can be written as AX=B, where
$A=\left[\begin{array}{ccc}2& 3& 3\\ 1& -2& 1\\ 3& -1& -2\end{array}\right],X=\left[\begin{array}{c}x\\ y\\ z\end{array}\right]andB=\left[\begin{array}{c}5\\ -4\\ 3\end{array}\right]$
Here, $|A|=\left[\begin{array}{ccc}2& 3& 3\\ 1& -2& 1\\ 3& -1& -2\end{array}\right]=2(4+1)-3(-2-3))+3(-1+6)$
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
$A_{11}=4+1=5,A_{12}=-(-2-3)=-5,A_{13}=(-1+6)=5A_{21}=-(-6+3)=3,A_{22}(-4-9)=-13,A_{23}=-(-2-9)=11A_{31}=3+6=9,A_{32}=-(2-3)=1,A_{33}=-4-3=-7$
$adj\left(A\right)={\left[\begin{array}{ccc}5& 5& 5\\ 3& -13& 11\\ 9& 1& -7\end{array}\right]}^T=\left[\begin{array}{ccc}5& 3& 9\\ 5& -13& 1\\ 5& 11& -7\end{array}\right]$
$\therefore A^{-1}=\frac{1}{|A|}\left(adjA\right)=\frac{1}{40}\left[\begin{array}{ccc}5& 3& 9\\ 5& -13& 1\\ 5& 11& -7\end{array}\right]\therefore A^{-1}=\frac{1}{|A|}\left(adjA\right)=\frac{1}{40}\left[\begin{array}{ccc}5& 3& 9\\ 5& -13& 1\\ 5& 11& -7\end{array}\right]Now,X=A^{-1}B$
$\Rightarrow \left[\begin{array}{c}x\\ y\\ z\end{array}\right]=\frac{1}{40}\left[\begin{array}{ccc}5& 3& 9\\ 5& -13& 1\\ 5& 11& -7\end{array}\right]\left[\begin{array}{c}5\\ -4\\ 3\end{array}\right]=\frac{1}{40}\left[\begin{array}{c}25-12+27\\ 25+52+3\\ 25-44-21\end{array}\right]=\frac{1}{40}\left[\begin{array}{c}40\\ 80\\ -40\end{array}\right]=\left[\begin{array}{c}1\\ 2\\ -1\end{array}\right]$
Hence, x=1, y=2 and z=-1.