Question

Solve system of linear equations, using matrix method.
$x-y+2z=7$
$3x+4y-5z=-5$
$2x-y+3z=12$

Answer 1

Simplification of given data
Given: The system of equations is

$x-y+2z=7$
$3x+4y-5z=-5$
$2x-y+3z=12$
Writing above equation as $AX=B$

$\left[\begin{array}{ccc}1& -1& 2\\ 3& 4& -5\\ 2& -1& 3\end{array}\right]\left[\begin{array}{c}x\\ y\\ z\end{array}\right]=\left[\begin{array}{c}7\\ -5\\ 12\end{array}\right]$

Hence, $A=\left[\begin{array}{ccc}1& -1& 2\\ 3& 4& -5\\ 2& -1& 3\end{array}\right],X=\left[\begin{array}{c}x\\ y\\ z\end{array}\right]\&B=\left[\begin{array}{c}7\\ -5\\ 12\end{array}\right]$
Calculate $A^{-1}$
Calculating $|A|$

$|A|=\left|\begin{array}{ccc}1& -1& 2\\ 3& 4& -5\\ 2& -1& 3\end{array}\right|$

$=1\left|\begin{array}{cc}4& -5\\ -1& 3\end{array}\right|-(-1)\left|\begin{array}{cc}3& -5\\ 2& 3\end{array}\right|+2\left|\begin{array}{cc}3& 4\\ 2& -1\end{array}\right|$

$=1(12-5)+1(9+10)+2(-3-8)$
$=1\left(7\right)+1\left(19\right)+2(-11)$
$=7+19-22=4$

Since $|A|\ne 0$

$\therefore$ The system of equations is consistent & has a unique solution.

$A=\left[\begin{array}{ccc}1& -1& 2\\ 3& 4& -5\\ 2& -1& 3\end{array}\right]$

Calculate $adj\left(A\right)$

$M_{11}=\left[\begin{array}{cc}4& -5\\ -1& 3\end{array}\right]=12-5=7$

$M_{12}=\left[\begin{array}{cc}3& -5\\ 2& 3\end{array}\right]=9+10=19$

$M_{13}=\left[\begin{array}{cc}3& 4\\ 2& -1\end{array}\right]=-3-8=-11$

$M_{21}=\left[\begin{array}{cc}-1& 2\\ -1& 3\end{array}\right]=-3+2=-1$

$M_{22}=\left[\begin{array}{cc}1& 2\\ 2& 3\end{array}\right]=3-4=-1$

$M_{23}=\left[\begin{array}{cc}1& -1\\ 2& -1\end{array}\right]=-1+2=1$

$M_{31}=\left[\begin{array}{cc}-1& 2\\ 4& -5\end{array}\right]=5-8=-3$

$M_{32}=\left[\begin{array}{cc}1& 2\\ 3& -5\end{array}\right]=-5-6=-11$

$M_{33}=\left[\begin{array}{cc}1& -1\\ 3& 4\end{array}\right]=4+3=7$

Thus,
$adj\left(A\right)={\left[\begin{array}{ccc}{A}_{11}& A_{12}& A_{13}\\ A_{21}& A_{22}& A_{23}\\ A_{31}& A_{32}& A_{33}\end{array}\right]}^{\prime }$

$={\left[\begin{array}{ccc}{A}_{11}& A_{21}& A_{31}\\ A_{12}& A_{22}& A_{32}\\ A_{13}& A_{23}& A_{33}\end{array}\right]}^{\prime }$

$=\left[\begin{array}{ccc}{M}_{11}& -M_{21}& M_{31}\\ -M_{12}& M_{22}& -M_{32}\\ M_{13}& -M_{23}& M_{33}\end{array}\right]=\left[\begin{array}{ccc}7& 1& -3\\ -19& -1& 11\\ -11& -1& 7\end{array}\right]$

Now,
$A^{-1}=\frac{1}{|A|}adjA$

$adjA=\left[\begin{array}{ccc}7& 1& -3\\ -19& -1& 11\\ -11& -1& 7\end{array}\right]$

Solve for the values of $X,Y\&Z$
$AX=B$
$\Rightarrow X=A^{-1}B$

$\Rightarrow \left[\begin{array}{c}x\\ y\\ z\end{array}\right]=\frac{1}{4}\left[\begin{array}{ccc}7& 1& -3\\ -19& -1& 11\\ -11& -1& 7\end{array}\right]\left[\begin{array}{c}7\\ -5\\ 12\end{array}\right]$

$\Rightarrow \left[\begin{array}{c}x\\ y\\ z\end{array}\right]=\frac{1}{4}\left[\begin{array}{c}49-5-36\\ -133+5+132\\ -77+51+84\end{array}\right]$

$\Rightarrow \left[\begin{array}{c}x\\ y\\ z\end{array}\right]=\frac{1}{4}\left[\begin{array}{c}8\\ 4\\ 12\end{array}\right]$

$\Rightarrow \left[\begin{array}{c}x\\ y\\ z\end{array}\right]=\left[\begin{array}{c}2\\ 1\\ 3\end{array}\right]$

​​​​​​​$\therefore x=2\&y=1\&z=3$