Question

Solve the following system of equations, using matrix method; $x+2y+z=7,x+3z=11,2x-3y=1$.Find $x+y+z$

Answer 1

The given system of equations is
$x+2y+z=7$
$x+0y+3z=11$
$2x-3y+0z=1$
or
$\left[\begin{array}{ccc}1& 2& 1\\ 1& 0& 3\\ 2& -3& 0\end{array}\right]\left[\begin{array}{c}x\\ y\\ z\end{array}\right]=\left[\begin{array}{c}7\\ 11\\ 1\end{array}\right]$
or $AX=B$, where
$A\left[\begin{array}{ccc}1& 2& 1\\ 1& 0& 3\\ 2& -3& 0\end{array}\right],X=\left[\begin{array}{c}x\\ y\\ z\end{array}\right]$ and $B=\left[\begin{array}{c}7\\ 17\\ 1\end{array}\right]$
Now,
$|A|=\left[\begin{array}{ccc}1& 2& 1\\ 1& 0& 3\\ 2& -3& 0\end{array}\right]=18$
So, the given system of equations has a unique solution given by $X=A^{-1}B$. Therefore,
$adjA=\left[\begin{array}{ccc}9& 6& -3\\ -3& -2& 7\\ 6& -2& -2\end{array}\right]=\left[\begin{array}{ccc}9& -3& 6\\ 6& -2& -2\\ -3& 7& -2\end{array}\right]$
$\Rightarrow A^{-1}=\frac{1}{|A|}adjA=\frac{1}{18}\left[\begin{array}{ccc}9& -3& 6\\ 6& -2& -2\\ -3& 7& -2\end{array}\right]$
Now,
$X=A^{-1}B$
$\Rightarrow X=\frac{1}{18}\left[\begin{array}{ccc}9& -3& 6\\ 6& -2& -2\\ -3& 7& -2\end{array}\right]\left[\begin{array}{c}7\\ 11\\ 1\end{array}\right]=\frac{1}{18}\left[\begin{array}{c}63-33+6\\ 42-22-2\\ -21+77-2\end{array}\right]$

$\Rightarrow \left[\begin{array}{c}x\\ y\\ z\end{array}\right]=\frac{1}{18}\left[\begin{array}{c}36\\ 18\\ 54\end{array}\right]=\left[\begin{array}{c}2\\ 1\\ 3\end{array}\right]\Rightarrow x=2,y=1,z=3$

Then $x+y+z=6$