Question

Given $A=\left[\begin{array}{cc}3& 6\\ -2& -8\end{array}\right]$ and $B=\left[\begin{array}{cc}2& 16\end{array}\right]$, find the matrix $X$ such that $XA=B$.

• A. $\left[\begin{array}{cc}-\frac{4}{3}& -3\end{array}\right]$
• B. $\left[\begin{array}{cc}\frac{4}{3}& 3\end{array}\right]$
• C. $\left[\begin{array}{cc}\frac{4}{3}& -3\end{array}\right]$
• D. $\left[\begin{array}{cc}-\frac{4}{3}& 3\end{array}\right]$

Answer 1

The correct option is A $\left[\begin{array}{cc}-\frac{4}{3}& -3\end{array}\right]$

$XA=B$

$X_{1\times 2}{\left[\begin{array}{cc}3& 6\\ -2& -8\end{array}\right]}_{2\times 2}={\left[\begin{array}{cc}2& 16\end{array}\right]}_{1\times 2}$

$X$ should be of order $1\times 2$

Let $X$ be $\left[\begin{array}{cc}a& b\end{array}\right]$

$\therefore \left[\begin{array}{cc}a& b\end{array}\right]\left[\begin{array}{cc}3& 6\\ -2& -8\end{array}\right]=\left[\begin{array}{cc}2& 16\end{array}\right]\Rightarrow \left[\begin{array}{cc}3a-2b& 6a-8b\end{array}\right]=\left[\begin{array}{cc}2& 16\end{array}\right]\therefore 3a-2b=2⟶\left(1\right)\times 26a-8b=16⟶\left(2\right)\times 1\therefore 6a-4b=46a-8b=16\_\_\_\_\_\_\_\_\_\_\_4b=-12\Rightarrow b=-3\therefore 3a+6=2a=\frac{-4}{3}\therefore X=\left[\begin{array}{cc}\frac{-4}{3}& -3\end{array}\right]$