Question

Solution of the sysytem of equations, $x+2y+z=7,x+3z=11,2x-3y=1$, is $(x,y,z)$ then $x+y-z$ is equal to:

• A. $0$
• B. $-1$
• C. $1$
• D. $6$

Answer 1

The correct option is A $0$

The given system of equation is
$x+2y+z=7x+0y+3z=112x-3y+0z=1$

$\Rightarrow \left[\begin{array}{ccc}1& 2& 1\\ 1& 0& 3\\ 2& -3& 0\end{array}\right]\left[\begin{array}{c}x\\ y\\ z\end{array}\right]=\left[\begin{array}{c}7\\ 11\\ 1\end{array}\right]$
$AX=B$, where
$A=\left[\begin{array}{ccc}1& 2& 1\\ 1& 0& 3\\ 2& -3& 0\end{array}\right],X=\left[\begin{array}{c}x\\ y\\ z\end{array}\right]\text{and}B=\left[\begin{array}{c}7\\ 11\\ 1\end{array}\right]$

Now,
$\left|A\right|=\left[\begin{array}{ccc}1& 2& 1\\ 1& 0& 3\\ 2& -3& 0\end{array}\right]=18$
So, the given system of equations has a unique solution given by $X=A^{-1}B$.

$\therefore adjA={\left[\begin{array}{ccc}9& 6& -3\\ -3& -2& 7\\ 6& -2& -2\end{array}\right]}^T=\left[\begin{array}{ccc}9& -3& 6\\ 6& -2& -2\\ -3& 7& -2\end{array}\right]$

$\Rightarrow A^{-1}=\frac{1}{\left|A\right|}adjA=\frac{1}{18}\left[\begin{array}{ccc}9& -3& 6\\ 6& -2& -2\\ -3& 7& -2\end{array}\right]$

Now,
$X=A^{-1}B$
$\Rightarrow X=\frac{1}{18}\left[\begin{array}{ccc}9& -3& 6\\ 6& -2& -2\\ -3& 7& -2\end{array}\right]\left[\begin{array}{c}7\\ 11\\ 1\end{array}\right]=\frac{1}{18}\left[\begin{array}{c}63-33+6\\ 42-22-2\\ -21+77-2\end{array}\right]$
$\Rightarrow \left[\begin{array}{c}x\\ y\\ z\end{array}\right]=\frac{1}{18}\left[\begin{array}{c}36\\ 18\\ 54\end{array}\right]=\left[\begin{array}{c}2\\ 1\\ 3\end{array}\right]\Rightarrow x=2,y=1,z=3$