Question

Solve the following system of equations by matrix method.
$3x-2y+3z=8$
$2x+y-z=1$
$4x-3y+2z=4$

Answer 1

Simplification of given data
The system of equation is
$3x-2y+3z=8$
$2x+y-z=1$
$4x-3y+2z=4$
Writing equation as $AX=B$
$\left[\begin{array}{ccc}3& -2& 3\\ 2& 1& -1\\ 4& -3& 2\end{array}\right]\left[\begin{array}{c}x\\ y\\ z\end{array}\right]=\left[\begin{array}{c}8\\ 1\\ 4\end{array}\right]$
$\left(A\right)\left(X\right)\left(B\right)$
Hence,
$A=\left[\begin{array}{ccc}3& -2& 3\\ 2& 1& -1\\ 4& -3& 2\end{array}\right],X=\left[\begin{array}{c}x\\ y\\ z\end{array}\right]\&B=\left[\begin{array}{c}8\\ 1\\ 4\end{array}\right]$
Calculate $A^{-1}$
$A=\left[\begin{array}{ccc}3& -2& 3\\ 2& 1& -1\\ 4& -3& 2\end{array}\right]$
$\left|\begin{array}{c}A\end{array}\right|=\left|\begin{array}{ccc}3& -2& 3\\ 2& 1& -1\\ 4& -3& 2\end{array}\right|$
$=3\left|\begin{array}{cc}1& -1\\ -3& 2\end{array}\right|-(-2)\left|\begin{array}{cc}2& -1\\ 4& 2\end{array}\right|+3\left|\begin{array}{cc}2& 1\\ 4& -3\end{array}\right|$
$=3(2-3)+2(4+4)+3(-6-4)$
$=-3+16-30=-17$
Calculate adjoint $A$
$M_{11}=\left|\begin{array}{cc}1& -1\\ -3& 2\end{array}\right|=2-3=-1$
$M_{12}=\left|\begin{array}{cc}2& -1\\ 4& 2\end{array}\right|=4+4=8$
$M_{13}=\left|\begin{array}{cc}2& 1\\ 4& -3\end{array}\right|=-6-4=-10$
$M_{21}=\left|\begin{array}{cc}-2& 3\\ -3& 2\end{array}\right|=-4+9=5$
$M_{22}=\left|\begin{array}{cc}3& 3\\ 4& 2\end{array}\right|=6-12=-6$
$M_{23}=\left|\begin{array}{cc}3& -2\\ 4& -3\end{array}\right|=-9+8=-1$
$M_{31}=\left|\begin{array}{cc}-2& 3\\ 1& -1\end{array}\right|=2-3=-1$
$M_{32}=\left|\begin{array}{cc}3& 3\\ 2& -1\end{array}\right|=-3-6=-9$
$M_{33}=\left|\begin{array}{cc}3& -2\\ 2& 1\end{array}\right|=3+4=7$
Thus $adj\left(A\right)={\left[\begin{array}{ccc}{A}_{11}& A_{12}& A_{13}\\ A_{21}& A_{22}& A_{23}\\ A_{13}& A_{32}& A_{33}\end{array}\right]}^T=\left[\begin{array}{ccc}{A}_{11}& A_{21}& A_{31}\\ A_{12}& A_{22}& A_{32}\\ A_{13}& A_{23}& A_{33}\end{array}\right]$
$=\left[\begin{array}{ccc}{M}_{11}& -M_{21}& M_{31}\\ -M_{12}& M_{22}& -M_{23}\\ M_{13}& -M_{23}& M_{33}\end{array}\right]$
$=\left[\begin{array}{ccc}-1& -5& -1\\ -8& -6& 9\\ -10& 1& 7\end{array}\right]$
Calculating $A^{-1}$
$A^{-1}=\frac{1}{\left|\begin{array}{c}A\end{array}\right|}adj\left(A\right)$
$=\frac{1}{-17}\left[\begin{array}{ccc}-1& -5& -1\\ -8& -6& 9\\ -10& 1& 7\end{array}\right]$
Solve for the value of $x,y$ and $z$
$X=A^{-1}B$
$\Rightarrow \left[\begin{array}{c}x\\ y\\ z\end{array}\right]=\frac{1}{-17}\left[\begin{array}{ccc}-1& -5& -1\\ -8& -6& 9\\ -10& 1& 7\end{array}\right]\left[\begin{array}{c}8\\ 1\\ 4\end{array}\right]$
$\Rightarrow \left[\begin{array}{c}x\\ y\\ z\end{array}\right]=\frac{1}{-17}\left[\begin{array}{c}-1\left(8\right)+(-5)\left(1\right)+(-1)\left(4\right)\\ -8\left(8\right)+(-6)\left(1\right)+9\left(4\right)\\ -10\left(8\right)+1\left(1\right)+7\left(4\right)\end{array}\right]$
$\Rightarrow \left[\begin{array}{c}x\\ y\\ z\end{array}\right]=\frac{-1}{17}\left[\begin{array}{c}-8-5-4\\ -64-6+36\\ -80+1+28\end{array}\right]$
$\Rightarrow \left[\begin{array}{c}x\\ y\\ z\end{array}\right]=\frac{-1}{17}\left[\begin{array}{c}-17\\ -34\\ -51\end{array}\right]$
$\Rightarrow \left[\begin{array}{c}x\\ y\\ z\end{array}\right]=\left[\begin{array}{c}1\\ 2\\ 3\end{array}\right]$
Therefore, $x=1,y=2$ and $z=3$