Question

Solve the following system of equations $x-y+z=4$, $x-2y+2z=9$ and $2x+y+3z=1$.

• A. $x=-4$, $y=-3$, $z=2$
• B. $x=-1$, $y=-3$, $z-2$
• C. $x=2$, $y=4$, $z=6$
• D. $x=3$, $y=6$ , $z=9$

Answer 1

Answer: (B)

Given system of equations is $x-y+z=4$, $x-2y+2z=9$ and $2x+y+3z=1$, which is written in the matrix form as
$\left[\begin{array}{ccc}1& -1& 1\\ 1& -2& 2\\ 2& 1& 3\end{array}\right]\left[\begin{array}{c}x\\ y\\ z\end{array}\right]=\left[\begin{array}{c}4\\ 9\\ 1\end{array}\right]$
or $AX=B$
where, $A=\left[\begin{array}{ccc}1& -1& 1\\ 1& -2& 2\\ 2& 1& 3\end{array}\right],B=\left[\begin{array}{c}4\\ 9\\ 1\end{array}\right]andX=\left[\begin{array}{c}x\\ y\\ z\end{array}\right]$
Now, $|A|=\left|\begin{array}{ccc}1& -1& 1\\ 1& -2& 2\\ 2& 1& 3\end{array}\right|=1\left(-6-2\right)+1\left(3-4\right)+1\left(1+4\right)=-8-1+5=-4\ne 0$
$\therefore$ A is invertible
So, there exists a unique solution $X=A^{-1}B$.
Now, cofactor matrix of $A=\left[\begin{array}{ccc}-8& 1& 5\\ 4& 1& -3\\ 0& -1& -1\end{array}\right]$
$\therefore adjA={\left[\begin{array}{ccc}-8& 1& 5\\ 4& 1& -3\\ 0& -1& -1\end{array}\right]}^T=\left[\begin{array}{ccc}-8& 4& 0\\ 1& 1& -1\\ 5& -3& -1\end{array}\right]$
So, $A^{-1}=\frac{1}{|A|}\left(adjA\right)=\frac{-1}{4}\left[\begin{array}{ccc}-8& 4& 0\\ 1& 1& -1\\ 5& -3& -1\end{array}\right]$
Now, $X=A^{-1}B=\frac{-1}{4}\left[\begin{array}{ccc}-8& 4& 0\\ 1& 1& -1\\ 5& -3& -1\end{array}\right]\left[\begin{array}{c}4\\ 9\\ 1\end{array}\right]$
$\Rightarrow \left[\begin{array}{c}x\\ y\\ z\end{array}\right]=\frac{-1}{4}\left[\begin{array}{c}-32+36+0\\ 4+9-1\\ 20-27-1\end{array}\right]=\frac{-1}{4}\left[\begin{array}{c}4\\ 12\\ -8\end{array}\right]=\left[\begin{array}{c}-1\\ -3\\ 2\end{array}\right]$
On comparing, we get $x=-1$, $y=-3$ and $z=2$