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

Hint: According to the question we find the unknown matrix X such that it satisfies the given equation $XA=B$. Now, firstly we need to identify the order of the matrix X using the product rule property of the matrices and then continue to solve for X in order to get X.

Complete step by step answer:
In the given question we are given two matrices and we need to find the third matrix such that the product of two matrices is equal to the third matrix.
Now, we know that order of matrix A is $2\times 2$ and order or matrix B is $1\times 2$ and B is equal to the product of XA. Therefore, for this product to occur we must have the number of columns of X equal to the number of rows of A and also since the number of rows of B is 1 therefore, the number of rows of matrix X is 1.
Therefore, we now know that the order of X is $1\times 2$ .
Now, product of XA is
$\begin{align}
  & XA=\left( \begin{matrix}
   x & y \\
\end{matrix} \right)\left( \begin{matrix}
   3 & 6 \\
   -2 & -8 \\
\end{matrix} \right) \\
 & \Rightarrow \left( \begin{matrix}
   3x-2y & 6x-8y \\
\end{matrix} \right) \\
\end{align}$
And now this is equal to B which is $B=\left( \begin{matrix}
   2 & 16 \\
\end{matrix} \right)$.
Now, we get two equations which are
$\begin{align}
  & 3x-2y=2 \\
 & 6x-8y=16 \\
\end{align}$
Now, dividing second equation by 2 we get $3x-4y=8$ and now subtracting first equation from this we get $y=-3$ and now putting this value in equation one we get $x=-\dfrac{4}{3}$ .
Therefore, the X matrix attained is $X=\left( \begin{matrix}
   -\dfrac{4}{3} & -3 \\
\end{matrix} \right)$ .

Note: In such type of question, we need to make sure that we need to find the product of X and A in order to make it equal to B. So, X should satisfy that, therefore for that we will have to make use of the fact that how matrices are multiplied and what kind of order it should be of both the matrices. Also, we need to be careful while calculating and should also be sure about the matrix’s multiplication.