Question

Solve the system of equations, using matrix method

$x-y+2z=7,3x+4y-5z=-5,2x-y+3z=12$

Answer 1

Given system of equations

$x-y+2z=7$
$3x+4y-5z=-5$

$2x-y+3z=12$

This can be written as

$AX=B$

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

Here, $|A|=1(12-5)+1(9+10)+2(-3-8)$

$\Rightarrow |A|=7+19-22=4$

Since, $|A|\ne 0$

Hence, the system of equations is consistent and has a unique solution given by $X==A^{-1}B$

$A^{-1}=\frac{adjA}{|A|}$ and $adjA=C^T$

$C_{11}=(-1{)}^{1+1}\left|\begin{array}{cc}4& -5\\ -1& 3\end{array}\right|$

$\Rightarrow C_{11}=12-5=7$

$C_{12}=(-1{)}^{1+2}\left|\begin{array}{cc}3& -5\\ 2& 3\end{array}\right|$

$\Rightarrow C_{12}=-(9+10)=-19$

$C_{13}=(-1{)}^{1+3}\left|\begin{array}{cc}3& 4\\ 2& -1\end{array}\right|$

$\Rightarrow C_{13}=-3-8=-11$

$C_{21}=(-1{)}^{2+1}\left|\begin{array}{cc}-1& 2\\ -1& 3\end{array}\right|$

$\Rightarrow C_{21}=-(-3+2)=1$

$C_{22}=(-1{)}^{2+2}\left|\begin{array}{cc}1& 2\\ 2& 3\end{array}\right|$

$\Rightarrow C_{22}=3-4=-1$

$C_{23}=(-1{)}^{2+3}\left|\begin{array}{cc}1& -1\\ 2& -1\end{array}\right|$

$\Rightarrow C_{23}=-(-1+2)=-1$

$C_{31}=(-1{)}^{3+1}\left|\begin{array}{cc}-1& 2\\ 4& -5\end{array}\right|$

$\Rightarrow C_{31}=5-8=-3$

$C_{32}=(-1{)}^{3+2}\left|\begin{array}{cc}1& 2\\ 3& -5\end{array}\right|$

$\Rightarrow C_{32}=-(-5-6)=11$

$C_{33}=(-1{)}^{3+3}\left|\begin{array}{cc}1& -1\\ 3& 4\end{array}\right|$

$\Rightarrow C_{33}=4+3=7$

Hence, the co-factor matrix is $C=\left[\begin{array}{ccc}7& -19& -11\\ 1& -1& -1\\ -3& 11& 7\end{array}\right]$

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

$\Rightarrow A^{-1}=\frac{adjA}{|A|}=\frac{1}{4}\left[\begin{array}{ccc}7& 1& -3\\ -19& -1& 11\\ -11& -1& 7\end{array}\right]$

Solution is given by

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

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

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

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

Hence, $x=2,y=1,z=3$