Solve the matrix equation \[\left[ \begin{align}
  & \begin{matrix}
   5 & 4 \\
\end{matrix} \\
 & \begin{matrix}
   1 & 1 \\
\end{matrix} \\
\end{align} \right]X\] = \[\left[ \begin{align}
  & \begin{matrix}
   1 & -2 \\
\end{matrix} \\
 & \begin{matrix}
   1 & 3 \\
\end{matrix} \\
\end{align} \right]\] , where X is a \[2\times 2\] matrix.

Hint: Assume a matrix X, \[X=\left[ \begin{align}
  & \begin{matrix}
   {{a}_{1}} & {{b}_{1}} \\
\end{matrix} \\
 & \begin{matrix}
   {{c}_{1}} & {{d}_{1}} \\
\end{matrix} \\
\end{align} \right]\] . Now, replace X by matrix X in the equation \[\left[ \begin{align}
  & \begin{matrix}
   5 & 4 \\
\end{matrix} \\
 & \begin{matrix}
   1 & 1 \\
\end{matrix} \\
\end{align} \right]X\] = \[\left[ \begin{align}
  & \begin{matrix}
   1 & -2 \\
\end{matrix} \\
 & \begin{matrix}
   1 & 3 \\
\end{matrix} \\
\end{align} \right]\] . Now, compare the elements of rows of the matrix in the LHS with the corresponding elements of the matrix in the RHS. Now, we have four equations and we have four variables. Solve it further and get the values of \[{{a}_{1}}\] , \[{{b}_{1}}\] , \[{{c}_{1}}\] , and \[{{d}_{1}}\] .

Complete step-by-step answer:
According to the question, it is given that
\[\left[ \begin{align}
  & \begin{matrix}
   5 & 4 \\
\end{matrix} \\
 & \begin{matrix}
   1 & 1 \\
\end{matrix} \\
\end{align} \right]X\] = \[\left[ \begin{align}
  & \begin{matrix}
   1 & -2 \\
\end{matrix} \\
 & \begin{matrix}
   1 & 3 \\
\end{matrix} \\
\end{align} \right]\]………………………(1)
Here, X is a \[2\times 2\] matrix.
Let us assume a \[2\times 2\] matrix be X and the elements of the matrix X be
\[X=\left[ \begin{align}
  & \begin{matrix}
   {{a}_{1}} & {{b}_{1}} \\
\end{matrix} \\
 & \begin{matrix}
   {{c}_{1}} & {{d}_{1}} \\
\end{matrix} \\
\end{align} \right]\] ………………………(2)
Putting the matrix X in equation (1), we get
\[\left[ \begin{align}
  & \begin{matrix}
   5 & 4 \\
\end{matrix} \\
 & \begin{matrix}
   1 & 1 \\
\end{matrix} \\
\end{align} \right]\]\[\left[ \begin{align}
  & \begin{matrix}
   {{a}_{1}} & {{b}_{1}} \\
\end{matrix} \\
 & \begin{matrix}
   {{c}_{1}} & {{d}_{1}} \\
\end{matrix} \\
\end{align} \right]\] = \[\left[ \begin{align}
  & \begin{matrix}
   1 & -2 \\
\end{matrix} \\
 & \begin{matrix}
   1 & 3 \\
\end{matrix} \\
\end{align} \right]\] …………………………(3)
Now, we have to multiply the matrix \[\left[ \begin{align}
  & \begin{matrix}
   {{a}_{1}} & {{b}_{1}} \\
\end{matrix} \\
 & \begin{matrix}
   {{c}_{1}} & {{d}_{1}} \\
\end{matrix} \\
\end{align} \right]\] and \[\left[ \begin{align}
  & \begin{matrix}
   5 & 4 \\
\end{matrix} \\
 & \begin{matrix}
   1 & 1 \\
\end{matrix} \\
\end{align} \right]\] .
On multiplying, we get
\[\left[ \begin{align}
  & \begin{matrix}
   5 & 4 \\
\end{matrix} \\
 & \begin{matrix}
   1 & 1 \\
\end{matrix} \\
\end{align} \right]\]\[\left[ \begin{align}
  & \begin{matrix}
   {{a}_{1}} & {{b}_{1}} \\
\end{matrix} \\
 & \begin{matrix}
   {{c}_{1}} & {{d}_{1}} \\
\end{matrix} \\
\end{align} \right]\]
\[=\left[ \begin{align}
  & \begin{matrix}
   5{{a}_{1}}+4{{c}_{1}} & 5{{b}_{1}}+4{{d}_{1}} \\
\end{matrix} \\
 & \begin{matrix}
   {{a}_{1}}+{{c}_{1}} & \,\,\,\,\,\,\,\,{{b}_{1}}+{{d}_{1}} \\
\end{matrix} \\
\end{align} \right]\] …………………….(4)
From equation (3), we have \[\left[ \begin{align}
  & \begin{matrix}
   5 & 4 \\
\end{matrix} \\
 & \begin{matrix}
   1 & 1 \\
\end{matrix} \\
\end{align} \right]\]\[\left[ \begin{align}
  & \begin{matrix}
   {{a}_{1}} & {{b}_{1}} \\
\end{matrix} \\
 & \begin{matrix}
   {{c}_{1}} & {{d}_{1}} \\
\end{matrix} \\
\end{align} \right]\] = \[\left[ \begin{align}
  & \begin{matrix}
   1 & -2 \\
\end{matrix} \\
 & \begin{matrix}
   1 & 3 \\
\end{matrix} \\
\end{align} \right]\] and from equation (4), we have
\[\left[ \begin{align}
  & \begin{matrix}
   1 & -2 \\
\end{matrix} \\
 & \begin{matrix}
   1 & 3 \\
\end{matrix} \\
\end{align} \right]\] = \[\left[ \begin{align}
  & \begin{matrix}
   5{{a}_{1}}+4{{c}_{1}} & 5{{b}_{1}}+4{{d}_{1}} \\
\end{matrix} \\
 & \begin{matrix}
   {{a}_{1}}+{{c}_{1}} & \,\,\,\,\,\,\,\,{{b}_{1}}+{{d}_{1}} \\
\end{matrix} \\
\end{align} \right]\] ……………………..(5)
We have to compare the elements of each row and columns of LHS with the elements of each column.
On comparing, we get
\[5{{a}_{1}}+4{{c}_{1}}=1\] ……………………..(6)
\[{{a}_{1}}+{{c}_{1}}=1\] …………………..(7)
\[5{{b}_{1}}+4{{d}_{1}}=-2\] ………………………(8)
\[{{b}_{1}}+{{d}_{1}}=3\] ……………………(9)
Multiplying equation (7) by 4 and then deducing equation (6), we get
\[5{{a}_{1}}+4{{c}_{1}}=1\]
\[4{{a}_{1}}+4{{c}_{1}}=4\]
After deducting these two equations, we get \[{{a}_{1}}=-3\] .
Now, putting the value of \[{{a}_{1}}\] in equation (7), we get
\[{{a}_{1}}+{{c}_{1}}=1\]
\[\begin{align}
  & \Rightarrow -3+{{c}_{1}}=1 \\
 & \Rightarrow {{c}_{1}}=1+3 \\
 & \Rightarrow {{c}_{1}}=4 \\
\end{align}\]
Multiplying equation (9) by 4 and then deducing equation (8), we get
\[5{{b}_{1}}+4{{d}_{1}}=-2\]
\[4{{b}_{1}}+4{{d}_{1}}=12\]
After deducting these two equations, we get \[{{b}_{1}}=-14\] .
Now, putting the value of \[{{b}_{1}}\] in equation (9), we get
\[{{b}_{1}}+{{d}_{1}}=3\]
\[\begin{align}
  & \Rightarrow -14+{{d}_{1}}=3 \\
 & \Rightarrow {{d}_{1}}=14+3 \\
 & \Rightarrow {{d}_{1}}=17 \\
\end{align}\]
Now, we have got the values of each element of rows and columns of the matrix X.
 Hence, the matrix X is, \[X=\left[ \begin{align}
  & \begin{matrix}
   -3 & 17 \\
\end{matrix} \\
 & \begin{matrix}
   4 & -14 \\
\end{matrix} \\
\end{align} \right]\] .

Note: To solve this question, one may think to solve the equation \[\left[ \begin{align}
  & \begin{matrix}
   5 & 4 \\
\end{matrix} \\
 & \begin{matrix}
   1 & 1 \\
\end{matrix} \\
\end{align} \right]X\] = \[\left[ \begin{align}
  & \begin{matrix}
   1 & -2 \\
\end{matrix} \\
 & \begin{matrix}
   1 & 3 \\
\end{matrix} \\
\end{align} \right]\] and after solving we will get \[X=\dfrac{\left[ \begin{align}
  & \begin{matrix}
   1 & -2 \\
\end{matrix} \\
 & \begin{matrix}
   1 & 3 \\
\end{matrix} \\
\end{align} \right]}{\left[ \begin{align}
  & \begin{matrix}
   5 & 4 \\
\end{matrix} \\
 & \begin{matrix}
   1 & 1 \\
\end{matrix} \\
\end{align} \right]}\] . This equation is complex to be solved and we don’t have any direct method to solve the division of two matrices. So, we cannot approach this question through this method.