Question

Solve the system of equations $x+2y+z=4$, $-x+y+z=0$ and $x-3y+z=4$.

• A. $x=2$ , $y=0$, $z=2$
• B. $x=2$,$y=0$, $z=-2$
• C. $x=-2$, $y=2$, $z=0$
• D. $x=-2$, $y-0$, $z=2$

Answer 1

Answer: (A)

Given system of equations is $x+2y+z=4$, $-x+y+z=0$ and $x-3y+z=4$
In matrix form, it can be written as
$\left[\begin{array}{ccc}1& 2& 1\\ -1& 1& 1\\ 1& -3& 1\end{array}\right]\left[\begin{array}{c}x\\ y\\ z\end{array}\right]=\left[\begin{array}{c}4\\ 0\\ 4\end{array}\right]$
or $AX=B$
where, $A=\left[\begin{array}{ccc}1& 2& 1\\ -1& 1& 1\\ 1& -3& 1\end{array}\right],B=\left[\begin{array}{c}4\\ 0\\ 4\end{array}\right]andX=\left[\begin{array}{c}x\\ y\\ z\end{array}\right]$
Now, $|A|=1\left(1+3\right)-2\left(-1-1\right)+1\left(3-1\right)=4+4+2=10$
$\because \left|A\right|\ne 0$, hence unique solution exists.
Cofactor matrix of $A=\left[\begin{array}{ccc}4& 2& 2\\ -5& 0& 5\\ 1& -2& 3\end{array}\right]$
$\therefore adj\left(A\right)={\left[\begin{array}{ccc}4& 2& 2\\ -5& 0& 5\\ 1& -2& 3\end{array}\right]}^T=\left[\begin{array}{ccc}4& -5& 1\\ 2& 0& -2\\ 2& 5& 3\end{array}\right]$
$A^{-1}=\frac{1}{|A|}adj\left(A\right)=\frac{1}{10}\left[\begin{array}{ccc}4& -5& 1\\ 2& 0& -2\\ 2& 5& 3\end{array}\right]<br/>Now,X=A^{-1}B=\frac{1}{10}\left[\begin{array}{ccc}4& -5& 1\\ 2& 0& -2\\ 2& 5& 3\end{array}\right]\left[\begin{array}{c}4\\ 0\\ 4\end{array}\right]$
$\left[\begin{array}{c}x\\ y\\ z\end{array}\right]=\frac{1}{10}\left[\begin{array}{c}16+0+4\\ 8+0-8\\ 8+0+12\end{array}\right]=\frac{1}{10}\left[\begin{array}{c}20\\ 0\\ 20\end{array}\right]=\left[\begin{array}{c}2\\ 0\\ 2\end{array}\right]$
On comparing the corresponding elements, we get $x=2$, $y=0$ and $z=2$