Question

Solve the system of equations, using matrix method

$x-y+z=4,2x+y-3z=0,x+y+z=2$

Answer 1

Given system of equations

$x-y+z=4$

$2x+y-3z=0$

$x+y+z=2$

This can be written as

$AX=B$

where $A=\left[\begin{array}{ccc}1& -1& 1\\ 2& 1& -3\\ 1& 1& 1\end{array}\right],X=\left[\begin{array}{c}x\\ y\\ z\end{array}\right],B=\left[\begin{array}{c}4\\ 0\\ 2\end{array}\right]$

Here,

$|A|=1(1+3)+1(2+3)+1(2-1)$

$\Rightarrow |A|=4+5+1=10$

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 $adjA=C^T$

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

$\Rightarrow C_{11}=1+3=4$

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

$\Rightarrow C_{12}=-(2+3)=-5$

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

$\Rightarrow C_{13}=2-1=1$

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

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

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

$\Rightarrow C_{22}=1-1=0$

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

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

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

$\Rightarrow C_{31}=3-1=2$

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

$\Rightarrow C_{32}=-(-3-2)=5$

$C_{33}=(-1{)}^{3+3}\left|\begin{array}{cc}1& -1\\ 2& 1\end{array}\right|$

$\Rightarrow C_{33}=1+2=3$

Hence, the co-factor matrix is $C=\left[\begin{array}{ccc}4& -5& 1\\ 2& 0& -2\\ 2& 5& 3\end{array}\right]$

$\Rightarrow adjA=C^T=\left[\begin{array}{ccc}4& 2& 2\\ -5& 0& 5\\ 1& -2& 3\end{array}\right]$

$\Rightarrow A^{-1}=\frac{adjA}{|A|}=\frac{1}{10}\left[\begin{array}{ccc}4& 2& 2\\ -5& 0& 5\\ 1& -2& 3\end{array}\right]$

Solution is given by

$X=A^{-1}B$

$\left[\begin{array}{c}x\\ y\\ z\end{array}\right]=\frac{1}{10}\left[\begin{array}{ccc}4& 2& 2\\ -5& 0& 5\\ 1& -2& 3\end{array}\right]\left[\begin{array}{c}4\\ 0\\ 2\end{array}\right]$

$\left[\begin{array}{c}x\\ y\\ z\end{array}\right]=\frac{1}{10}\left[\begin{array}{c}16+4\\ -20+10\\ 4+6\end{array}\right]$

$\left[\begin{array}{c}x\\ y\\ z\end{array}\right]=\frac{1}{10}\left[\begin{array}{c}20\\ -10\\ 10\end{array}\right]$

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

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