Question

Using matrices, solve the following system of equations:
$x+y-z=3;2x+3y+z=10;3x-y-7z=1.$

Answer 1

The given system of equation can be expressed can be represented in matrix form as AX = B, where

$A=\left|\begin{array}{ccc}1& 1& -1\\ 2& 3& 1\\ 3& -1& -7\end{array}\right|,X=\left|\begin{array}{c}x\\ y\\ z\end{array}\right|,B=\left|\begin{array}{c}3\\ 10\\ 1\end{array}\right|$

Now $\left|A\right|=\left|\begin{array}{ccc}1& 1& -1\\ 2& 3& 1\\ 3& -1& -7\end{array}\right|=1(-21+1)-1(-14-3)-1(-2-9)\Rightarrow -20+17+11=8\ne 0$

$C_{11}=(-1{)}^{1+1}\left|\begin{array}{cc}3& 1\\ -1& -7\end{array}\right|=-21+1=-20$
$C_{12}=(-1{)}^{1+2}\left|\begin{array}{cc}2& 1\\ 3& -7\end{array}\right|=-(-14-3)=17$
$C_{13}=(-1{)}^{1+3}\left|\begin{array}{cc}2& 3\\ 3& -1\end{array}\right|=-2-9=-11$
$C_{21}=(-1{)}^{2+1}\left|\begin{array}{cc}1& -1\\ -1& -7\end{array}\right|=-(-7-1)=8$
$C_{22}=(-1{)}^{2+2}\left|\begin{array}{cc}1& -1\\ 3& -7\end{array}\right|=-7+3=-4$
$C_{23}=(-1{)}^{2+3}\left|\begin{array}{cc}1& 1\\ 3& -1\end{array}\right|=-(-1-3)=4$
$C_{31}=(-1{)}^{3+1}\left|\begin{array}{cc}1& -1\\ 3& 1\end{array}\right|=1+3=4$
$C_{32}=(-1{)}^{3+2}\left|\begin{array}{cc}1& -1\\ 2& 1\end{array}\right|=-(1+2)=-3$
$C_{33}=(-1{)}^{3+3}\left|\begin{array}{cc}1& 1\\ 2& 3\end{array}\right|=3-2=1$
$AdjA={\left|\begin{array}{ccc}-20& 17& -11\\ 8& -4& 4\\ 4& 3& 1\end{array}\right|}^T=\left|\begin{array}{ccc}-20& 8& 4\\ 17& -4& -3\\ -11& 4& 1\end{array}\right|$
$A^{-1}=\frac{1}{\left|A\right|}adjA=\frac{1}{8}\left|\begin{array}{ccc}-20& 8& 4\\ 17& -4& -3\\ -11& 4& 1\end{array}\right|$
$AX=B\Rightarrow X=A^{-1}B$
Therefore, $\left[\begin{array}{c}x\\ y\\ z\end{array}\right]=\frac{1}{8}\left|\begin{array}{ccc}-20& 8& 4\\ 17& -4& -3\\ -11& 4& 1\end{array}\right|\left|\begin{array}{c}3\\ 10\\ 1\end{array}\right|$
$=\frac{1}{8}\left[\begin{array}{c}-60+80+4\\ 51-40-3\\ -33+40+1\end{array}\right]$
$=\frac{1}{8}\left[\begin{array}{c}24\\ 8\\ 8\end{array}\right]$
$\left|\begin{array}{c}x\\ y\\ z\end{array}\right|=\left|\begin{array}{c}3\\ 1\\ 1\end{array}\right|$
Equating the corresponding elements we get
$x=3,y=1,z=1.$