Question

Using matrices, solve the following system of equations:
$2x+3y+3z=5,x-2y+z=-4,3x-y-2z=3.$

Answer 1

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

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

Now $\left|A\right|=\left|\begin{array}{ccc}2& 3& 3\\ 1& -2& 1\\ 3& -1& -2\end{array}\right|=2(4+1)-3(-2-3)+3(-1+6)\Rightarrow 10+15+15=40\ne 0$

$C_{11}=(-1{)}^{1+1}\left|\begin{array}{cc}-2& 1\\ -1& -2\end{array}\right|=4+1=5$

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

$C_{13}=(-1{)}^{1+3}\left|\begin{array}{cc}1& -2\\ 3& -1\end{array}\right|=-1+6=5$

$C_{21}=(-1{)}^{2+1}\left|\begin{array}{cc}3& 3\\ -1& -2\end{array}\right|=-(-6+3)=3$

$C_{22}=(-1{)}^{2+2}\left|\begin{array}{cc}2& 3\\ 3& -2\end{array}\right|=(-4-9)=-13$

$C_{23}=(-1{)}^{2+3}\left|\begin{array}{cc}2& 3\\ 3& -1\end{array}\right|=-(-2-9)=11$

$C_{31}=(-1{)}^{3+1}\left|\begin{array}{cc}3& 3\\ -2& 1\end{array}\right|=3+6=9$

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

$C_{33}=(-1{)}^{3+3}\left|\begin{array}{cc}2& 3\\ 1& -2\end{array}\right|=-4-3=-7$

$AdjA={\left|\begin{array}{ccc}5& 5& 5\\ 3& -13& 11\\ 9& 1& -7\end{array}\right|}^T=\left|\begin{array}{ccc}5& 3& 9\\ 5& -13& 1\\ 5& 11& -7\end{array}\right|$

$A^{-1}=\frac{1}{\left|A\right|}adjA=\frac{1}{40}\left|\begin{array}{ccc}5& 3& 9\\ 5& -13& 1\\ 5& 11& -7\end{array}\right|$

$AX=B\Rightarrow X=A^{-1}B$

Therefore, $\left[\begin{array}{c}x\\ y\\ z\end{array}\right]=\frac{1}{40}\left|\begin{array}{ccc}5& 3& 9\\ 5& -13& 1\\ 5& 11& -7\end{array}\right|\left|\begin{array}{c}5\\ -4\\ 3\end{array}\right|$

$=\frac{1}{40}\begin{bmatrix}25-12+27\\25+52+3 \\25-44-21

\end{bmatrix}$

$=\frac{1}{40}\begin{bmatrix}40\\80 \\-40

\end{bmatrix}$

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

Equating the corresponding elements we get

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