Question

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

Answer 1

Given: $A={\left[\begin{array}{cc}1& 5\\ 7& 12\end{array}\right]}_{2\times 2}\&B={\left[\begin{array}{cc}9& 1\\ 7& 8\end{array}\right]}_{2\times 2}$

Also, $3A+5B+2C=0$

As addition of matrices is possible if the matrices are of same order.

Let $C={\left[\begin{array}{cc}a& b\\ c& d\end{array}\right]}_{2\times 2}$

So. from (1), we get

$\left[\begin{array}{cc}3& 15\\ 21& 16\end{array}\right]+\left[\begin{array}{cc}45& 5\\ 35& 40\end{array}\right]+\left[\begin{array}{cc}2a& 2b\\ 2c& 2d\end{array}\right]=\left[\begin{array}{cc}0& 0\\ 0& 0\end{array}\right]$

$\Rightarrow \left[\begin{array}{cc}48+2a& 20+2b\\ 56+2c& 76+2d\end{array}\right]=\left[\begin{array}{cc}0& 0\\ 0& 0\end{array}\right]$

By comparing corresponding elements, we get

$2a+48=0\Rightarrow a=-24$

$20+2b=0\Rightarrow b=-10$

$56+2c=0\Rightarrow c=-28$

$76+2d=0\Rightarrow d=-38$

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