Question

If $A=\left[\begin{array}{cc}9& 1\\ 7& 8\end{array}\right],B=\left[\begin{array}{cc}1& 5\\ 7& 12\end{array}\right],$ then $C$ such that $5A+3B+2C$ is a null matrix, is:

• A. $\left[\begin{array}{cc}-28& -10\\ -4& -24\end{array}\right]$
• B. $\left[\begin{array}{cc}-18& -5\\ -28& -38\end{array}\right]$
• C. $\left[\begin{array}{cc}-12& -10\\ -10& -38\end{array}\right]$
• D. $\left[\begin{array}{cc}-24& -10\\ -28& -38\end{array}\right]$

Answer 1

The correct option is D $\left[\begin{array}{cc}-24& -10\\ -28& -38\end{array}\right]$

$5A+3B+2C=O$

$=5\left[\begin{array}{cc}9& 1\\ 7& 8\end{array}\right]+3\left[\begin{array}{cc}1& 5\\ 7& 12\end{array}\right]+2\left[\begin{array}{cc}x& y\\ z& w\end{array}\right]=\left[\begin{array}{cc}0& 0\\ 0& 0\end{array}\right]$

$=\left[\begin{array}{cc}45& 5\\ 35& 40\end{array}\right]+\left[\begin{array}{cc}3& 15\\ 21& 36\end{array}\right]+\left[\begin{array}{cc}2x& 2y\\ 2z& 2w\end{array}\right]=\left[\begin{array}{cc}0& 0\\ 0& 0\end{array}\right]$

$=\left[\begin{array}{cc}45+3+2x& 5+15+2y\\ 35+21+2z& 40+36+2w\end{array}\right]=\left[\begin{array}{cc}0& 0\\ 0& 0\end{array}\right]$
$2x+3+45=0\cdots \left(1\right)$
$2y+5+15=0\cdots \left(2\right)$
$2z+35+21=0\cdots \left(3\right)$
$2w+40+36=0\cdots \left(4\right)$
From equations $\left(1\right),\left(2\right),\left(3\right),\left(4\right),$ we get
$2x=-48\Rightarrow x=-24;2y=-20\Rightarrow y=-10$
$2z=-56\Rightarrow z=-28,2w=-76\Rightarrow w=-38$

Hence, $C=\left[\begin{array}{cc}-24& -10\\ -28& -38\end{array}\right]$