Question

If $\left[\begin{array}{ccc}1& 1& 1\\ 1& -2& -2\\ 1& 3& 1\end{array}\right]$ $\left[\begin{array}{c}x\\ y\\ z\end{array}\right]=\left[\begin{array}{c}0\\ 3\\ 4\end{array}\right]$, then $\left[\begin{array}{c}x\\ y\\ z\end{array}\right]$ is equal to.

• A. $\left[\begin{array}{c}0\\ 1\\ 1\end{array}\right]$
• B. $\left[\begin{array}{c}1\\ 2\\ -3\end{array}\right]$
• C. $\left[\begin{array}{c}5\\ -2\\ 1\end{array}\right]$
• D. $\left[\begin{array}{c}1\\ -2\\ 3\end{array}\right]$

Answer 1

The correct option is B $\left[\begin{array}{c}1\\ 2\\ -3\end{array}\right]$

Given, $\left[\begin{array}{ccc}1& 1& 1\\ 1& -2& -2\\ 1& 2& 1\end{array}\right]\left[\begin{array}{c}x\\ y\\ z\end{array}\right]=\left[\begin{array}{c}0\\ 3\\ 4\end{array}\right]$
$\therefore x+y+z=0$,
$x-2y-2z=3$,
$x+3y=z=4$
On solving these equations, we get
$x=1,y=2$ and $z=-3$
$\therefore \left[\begin{array}{c}x\\ y\\ z\end{array}\right]=\left[\begin{array}{c}1\\ 2\\ -3\end{array}\right]$