Question

Using matrices, solve the following system of equations:
$3x+4y+7z=4,2x-y+3z=-3;x+2y-3z=8$

Answer 1

The given system of equation can be expressed can be represented in matrix form as AX = B, where
$A=\left|\begin{array}{ccc}3& 4& 7\\ 2& -1& 3\\ 1& 2& -3\end{array}\right|,X=\left|\begin{array}{c}x\\ y\\ z\end{array}\right|,B=\left|\begin{array}{c}4\\ -3\\ 8\end{array}\right|$
Now $\left|A\right|=\left|\begin{array}{ccc}3& 4& 7\\ 2& -1& 3\\ 1& 2& -3\end{array}\right|=3(3-6)-4(-6-3)+7(4+1)\Rightarrow -9+36+35=62\ne 0$
$C_{11}=(-1{)}^{1+1}\left|\begin{array}{cc}-1& 3\\ 2& -3\end{array}\right|=3-6=-3$
$C_{12}=(-1{)}^{1+2}\left|\begin{array}{cc}2& 3\\ 1& -3\end{array}\right|=-(-6-3)=9$
$C_{13}=(-1{)}^{1+3}\left|\begin{array}{cc}2& -1\\ 1& 2\end{array}\right|=4+1=5$
$C_{21}=(-1{)}^{2+1}\left|\begin{array}{cc}4& 7\\ 2& -3\end{array}\right|=-(-12-14)=26$
$C_{22}=(-1{)}^{2+2}\left|\begin{array}{cc}3& 7\\ 1& -3\end{array}\right|=-9-7=-16$
$C_{23}=(-1{)}^{2+3}\left|\begin{array}{cc}3& 4\\ 1& 2\end{array}\right|=-(6-4)=-2$
$C_{31}=(-1{)}^{3+1}\left|\begin{array}{cc}4& 7\\ -1& 3\end{array}\right|=12+7=19$
$C_{32}=(-1{)}^{3+2}\left|\begin{array}{cc}3& 7\\ 2& 3\end{array}\right|=-(9-14)=5$
$C_{33}=(-1{)}^{3+3}\left|\begin{array}{cc}3& 4\\ 2& -1\end{array}\right|=-3-8=-11$
$AdjA={\left|\begin{array}{ccc}-3& 9& 5\\ 26& -16& -2\\ 19& 5& -11\end{array}\right|}^T=\left|\begin{array}{ccc}-3& 26& 19\\ 9& -16& 5\\ 5& -2& -11\end{array}\right|$
$A^{-1}=\frac{1}{\left|A\right|}adjA=\frac{1}{62}\left|\begin{array}{ccc}-3& 26& 19\\ 9& -16& 5\\ 5& -2& -11\end{array}\right|$
$AX=B\Rightarrow X=A^{-1}B$
Therefore, $\left[\begin{array}{c}x\\ y\\ z\end{array}\right]=\frac{1}{62}\left|\begin{array}{ccc}-3& 26& 19\\ 9& -16& 5\\ 5& -2& -11\end{array}\right|\left|\begin{array}{c}4\\ -3\\ 8\end{array}\right|$
$=\frac{1}{62}\left[\begin{array}{c}-12-78+152\\ 36+48+40\\ 20+6-88\end{array}\right]$
$=\frac{1}{62}\left[\begin{array}{c}62\\ 124\\ -62\end{array}\right]$
$\left|\begin{array}{c}x\\ y\\ z\end{array}\right|=\left|\begin{array}{c}1\\ 2\\ -1\end{array}\right|$
Equating the corresponding elements we get
$x=1,y=2,z=-1.$