Question

If $A=\left[\begin{array}{cc}1& 2\\ 3& 4\end{array}\right],$ $B=\left[\begin{array}{cc}-1& -1\\ -1& -1\end{array}\right],$ $C=\left[\begin{array}{cc}x& y\\ z& r\end{array}\right]$ and $A+3B=C,$ then $x+y+z+r$ is

• A. $1$
• B. $-2$
• C. $-1$
• D. $2$

Answer 1

The correct option is B $-2$

Given, $A=\left[\begin{array}{cc}1& 2\\ 3& 4\end{array}\right],B=\left[\begin{array}{cc}-1& -1\\ -1& -1\end{array}\right],C=\left[\begin{array}{cc}x& y\\ z& r\end{array}\right],$ $A+3B=C$

Therefore, $3B=3\left[\begin{array}{cc}-3& -3\\ -3& -3\end{array}\right]$

Now, $A+3B=C$

$\Rightarrow \left[\begin{array}{cc}1& 2\\ 3& 4\end{array}\right]+\left[\begin{array}{cc}-3& -3\\ -3& -3\end{array}\right]=\left[\begin{array}{cc}x& y\\ z& r\end{array}\right]$

$\Rightarrow \left[\begin{array}{cc}-2& -1\\ 0& 1\end{array}\right]=\left[\begin{array}{cc}x& y\\ z& r\end{array}\right]$

Also, we know that two matrices are said to be equal if all elements (corresponding) are equal.

$\therefore x=-2,y=-1,z=0,r=1$

$x+y+z+r=2-1+0+1=-2$

Option B is correct.