Question

If $\left[\begin{array}{cc}x+3& 2y+x\\ z-1& 4a-z\end{array}\right]=\left[\begin{array}{cc}0& -7\\ 3& 2a\end{array}\right],$ then $(x+y+z+a)$ is:

• A. $-1$
• B. $0$
• C. $1$
• D. $8$

Answer 1

The correct option is C $1$

If $A=B$, then all the elements of $A$ and $B$ should be same.

$\therefore x+3=0\Rightarrow x=-3$

$2y+x=-7$..................(i)

$z-1=3\Rightarrow z=4$

$z-4a=-2a\Rightarrow z=2a$

Putting value of $x$ in $eq.\left(i\right)$

$2y=3-7\Rightarrow y=\frac{-4}{2}=-2$

$\therefore y=-2$, $z=4$ & $a=2$

So, $x+y+z+a=-3-2+4+2$ $=1$