Question

If A = $\left[\begin{array}{cc}9& 1\\ 4& 3\end{array}\right]$, B= $\left[\begin{array}{cc}1& 5\\ 6& 11\end{array}\right]$ & $3A+5B+2X=0$ then X =

• A. $\left[\begin{array}{cc}16& -14\\ 21& -32\end{array}\right]$
• B. $\left[\begin{array}{cc}16& 14\\ -21& -32\end{array}\right]$
• C. $\left[\begin{array}{cc}-16& -14\\ -21& -32\end{array}\right]$
• D. $\left[\begin{array}{cc}16& 14\\ 21& 32\end{array}\right]$

Answer 1

The correct option is C $\left[\begin{array}{cc}-16& -14\\ -21& -32\end{array}\right]$

$\Rightarrow 3A+5B+2X=0$
$\Rightarrow 3\left[\begin{array}{cc}9& 1\\ 4& 3\end{array}\right]+5\left[\begin{array}{cc}1& 5\\ 6& 11\end{array}\right]+2\left[\begin{array}{cc}a& b\\ c& d\end{array}\right]=0$
$\Rightarrow \left[\begin{array}{cc}27& 3\\ 12& 9\end{array}\right]+\left[\begin{array}{cc}5& 25\\ 30& 55\end{array}\right]+2\left[\begin{array}{cc}a& b\\ c& d\end{array}\right]=0$
$\Rightarrow \left[\begin{array}{cc}32& 28\\ 42& 64\end{array}\right]+2\left[\begin{array}{cc}a& b\\ c& d\end{array}\right]=0$
$\Rightarrow \left[\begin{array}{cc}a& b\\ c& d\end{array}\right]=\left[\begin{array}{cc}-16& -14\\ -21& -32\end{array}\right]$