(a) If $A = \left[ {\begin{array}{*{20}{c}}
3&x\\
0&1
\end{array}} \right]$ and \[B = \left[ {\begin{array}{*{20}{c}}
9&{ - 16}\\
0&{ - y}
\end{array}} \right]\] , then find x and y when ${A^2} = B$
(b) Find x and y if $\left[ {\begin{array}{*{20}{c}}
{ - 2}&0\\
3&1
\end{array}} \right]\left[ {\begin{array}{*{20}{c}}
{ - 1}\\
{2x}
\end{array}} \right] + 3\left[ {\begin{array}{*{20}{c}}
{ - 2}\\
1
\end{array}} \right] = 2\left[ {\begin{array}{*{20}{c}}
y\\
3
\end{array}} \right]$

Hint:For solving these questions we need to know the concept of matrix multiplication which is mention below:-
Suppose we have to multiply two matrices which is given below,
$A = \left[ {\begin{array}{*{20}{c}}
a&b\\
c&d
\end{array}} \right]\left[ {\begin{array}{*{20}{c}}
w&x\\
y&z
\end{array}} \right]$
Now the rule is multiply elements of row one with column one and two and then row two with column one and two.
 $ \Rightarrow A = \left[ {\begin{array}{*{20}{c}}
{aw + by}&{ax + bz}\\
{cw + dy}&{cx + dz}
\end{array}} \right]$

Complete step-by-step answer:
We will first multiply matrix A with itself and then equate the resultant with matrix B as it is given in the question that ${A^2} = B$
$ \Rightarrow {A^2} = \left[ {\begin{array}{*{20}{c}}
3&x\\
0&1
\end{array}} \right]\left[ {\begin{array}{*{20}{c}}
3&x\\
0&1
\end{array}} \right]$
Now we will do multiplication as per the rule $\left[ {\begin{array}{*{20}{c}}
{aw + by}&{ax + bz}\\
{cw + dy}&{cx + dz}
\end{array}} \right]$
$ \Rightarrow {A^2} = \left[ {\begin{array}{*{20}{c}}
{9 + 0}&{3x + x}\\
{0 + 0}&{0 + 1}
\end{array}} \right]$
$ \Rightarrow {A^2} = \left[ {\begin{array}{*{20}{c}}
9&{4x}\\
0&1
\end{array}} \right]$
Now we will equate this matrix with the B matrix.
$ \Rightarrow \left[ {\begin{array}{*{20}{c}}
9&{ - 16}\\
0&{ - y}
\end{array}} \right] = \left[ {\begin{array}{*{20}{c}}
9&{4x}\\
0&1
\end{array}} \right]$
By comparing the elements we get that -y=1 and -16=4x
$ \Rightarrow x = \dfrac{{ - 16}}{4} = - 4$ and $y = - 1$

b) Now we use again product rule in given equation $\left[ {\begin{array}{*{20}{c}}
{ - 2}&0\\
3&1
\end{array}} \right]\left[ {\begin{array}{*{20}{c}}
{ - 1}\\
{2x}
\end{array}} \right] + 3\left[ {\begin{array}{*{20}{c}}
{ - 2}\\
1
\end{array}} \right] = 2\left[ {\begin{array}{*{20}{c}}
y\\
3
\end{array}} \right]$\[ \Rightarrow \left[ {\begin{array}{*{20}{c}}
2\\
{ - 3 + 2x}
\end{array}} \right] + 3\left[ {\begin{array}{*{20}{c}}
{ - 2}\\
1
\end{array}} \right] = 2\left[ {\begin{array}{*{20}{c}}
y\\
3
\end{array}} \right]\]
Now we will multiply 3 inside and 2 inside the matrices
\[ \Rightarrow \left[ {\begin{array}{*{20}{c}}
2\\
{ - 3 + 2x}
\end{array}} \right] + \left[ {\begin{array}{*{20}{c}}
{ - 6}\\
3
\end{array}} \right] = \left[ {\begin{array}{*{20}{c}}
{2y}\\
6
\end{array}} \right]\]
Now we will simply add the two matrices.

\[ \Rightarrow \left[ {\begin{array}{*{20}{c}}
{ - 4}\\
{ - 3 + 2x + 3}
\end{array}} \right] = \left[ {\begin{array}{*{20}{c}}
{2y}\\
6
\end{array}} \right]\]
Now we will cancel -3 and 3
\[ \Rightarrow \left[ {\begin{array}{*{20}{c}}
{ - 4}\\
{2x}
\end{array}} \right] = \left[ {\begin{array}{*{20}{c}}
{2y}\\
6
\end{array}} \right]\]
Now we will compare the elements and will get two equations.

$ \Rightarrow 2y = - 4$ and $2x = 6$
After solving we will get y=-2 and x=3

Note:Students may find difficulty in determining which order will get after multiplication of two matrices if the order are not same so here is trick if one matrix order is $2 \times 2$ where 2 are the rows and 2 are the columns and other’s order is $2 \times 1$ then eliminate the middle similar 2 and left outer $2 \times 1$ will be the order of resultant product. Remember that similar column numbers should be there.