Question

If $\left[ \begin{matrix}
   \text{ }\!\!~\!\!\text{ }3 & 1 \\
   4 & 1 \\
\end{matrix} \right]X=\left[ \begin{matrix}
   5 & -1 \\
   2 & 3 \\
\end{matrix} \right]$, then $X =$
A . $\left[ \begin{matrix}
   \text{ }\!\!~\!\!\text{ -}3 & 4 \\
   14 & -13 \\
\end{matrix} \right]$
B $\left[ \begin{matrix}
   \text{ }\!\!~\!\!\text{ }3 & -4 \\
   -14 & 13 \\
\end{matrix} \right]$
C $\left[ \begin{matrix}
   \text{ }\!\!~\!\!\text{ }3 & 4 \\
   14 & 13 \\
\end{matrix} \right]$
D \[\left[ \begin{matrix}
   \text{ }\!\!~\!\!\text{ }-3 & 4 \\
   -14 & 13 \\
\end{matrix} \right]\]

Answer 1

Hint: To solve this question we will take each of the options as the value of $X$ and substitute in the given equation and see if both sides are equal or not, that is if $L.H.S=R.H.S$. We will take each of the options and them. And the option which will prove $L.H.S=R.H.S$ by substituting as the value of $X$ in the equation then will be the correct answer.

Complete step by step Solution:
We will check the first option,
$\left[ {\begin{array}{*{20}{c}}
  3&1 \\
  4&1
\end{array}} \right]X = \left[ {\begin{array}{*{20}{c}}
  5&{ - 1} \\
  2&3
\end{array}} \right]$
Let $X = \left[ {\begin{array}{*{20}{c}}
  { - 3}&4 \\
  {14}&{ - 13}
\end{array}} \right]$
\[\left[ {\begin{array}{*{20}{c}}
  3&1 \\
  4&1
\end{array}} \right]\left[ {\begin{array}{*{20}{c}}
  { - 3}&4 \\
  {14}&{ - 13}
\end{array}} \right] = \left[ {\begin{array}{*{20}{c}}
  5&{ - 1} \\
  2&3
\end{array}} \right]\]
After multiplication, we will get
$\left[ {\begin{array}{*{20}{c}}
  {3( - 3) + 14}&{3(4) + ( - 13)} \\
  {4( - 3) + 14}&{4(4) + ( - 13)}
\end{array}} \right] = \left[ {\begin{array}{*{20}{c}}
  5&{ - 1} \\
  2&3
\end{array}} \right]$
$\left[ {\begin{array}{*{20}{c}}
  { - 9 + 14}&{12 - 13} \\
  { - 12 + 14}&{16 - 13}
\end{array}} \right] = \left[ {\begin{array}{*{20}{c}}
  5&{ - 1} \\
  2&3
\end{array}} \right]$
After solving, we get
$\left[ {\begin{array}{*{20}{c}}
  5&{ - 1} \\
  2&3
\end{array}} \right] = \left[ {\begin{array}{*{20}{c}}
  5&{ - 1} \\
  2&3
\end{array}} \right]$
LHS=RHS
Hence, $X = \left[ {\begin{array}{*{20}{c}}
  { - 3}&4 \\
  {14}&{ - 13}
\end{array}} \right]$
Therefore, option A is correct.
We will now take the second option $\left[ \begin{matrix}
   \text{ }\!\!~\!\!\text{ }3 & -4 \\
   -14 & 13 \\
\end{matrix} \right]$.and substitute in the equation in place of $X$.
$\left[ \begin{matrix}
   \text{ }\!\!~\!\!\text{ }3 & 1 \\
   4 & 1 \\
\end{matrix} \right]X=\left[ \begin{matrix}
   5 & -1 \\
   2 & 3 \\
\end{matrix} \right]$
$\left[ \begin{matrix}
   \text{ }\!\!~\!\!\text{ }3 & 1 \\
   4 & 1 \\
\end{matrix} \right]\left[ \begin{matrix}
   \text{ }\!\!~\!\!\text{ }3 & -4 \\
   -14 & 13 \\
\end{matrix} \right]=\left[ \begin{matrix}
   5 & -1 \\
   2 & 3 \\
\end{matrix} \right]$
We will now check if $L.H.S=R.H.S$.
$\left[ \begin{matrix}
   \text{ }\!\!~\!\!\text{ }3\times 3+1\times (-14) & \text{ }\!\!~\!\!\text{ }3\times (-4)+1\times 13 \\
   4\times 3+1\times (-14) & \text{ }\!\!~\!\!\text{ }4\times (-4)+1\times 13 \\
\end{matrix} \right]=\left[ \begin{matrix}
   5 & -1 \\
   2 & 3 \\
\end{matrix} \right]$
$\left[ \begin{matrix}
   -5 & 1 \\
   -2 & -3 \\
\end{matrix} \right]\ne \left[ \begin{matrix}
   5 & -1 \\
   2 & 3 \\
\end{matrix} \right]$
So the second option is incorrect.
Now we will check third option $\left[ \begin{matrix}
   \text{ }\!\!~\!\!\text{ }3 & 4 \\
   14 & 13 \\
\end{matrix} \right]$ and substitute in the equation in place of $X$.
$\left[ \begin{matrix}
   \text{ }\!\!~\!\!\text{ }3 & 1 \\
   4 & 1 \\
\end{matrix} \right]X=\left[ \begin{matrix}
   5 & -1 \\
   2 & 3 \\
\end{matrix} \right]$
$\left[ \begin{matrix}
   \text{ }\!\!~\!\!\text{ }3 & 1 \\
   4 & 1 \\
\end{matrix} \right]\left[ \begin{matrix}
   \text{ }\!\!~\!\!\text{ }3 & 4 \\
   14 & 13 \\
\end{matrix} \right]=\left[ \begin{matrix}
   5 & -1 \\
   2 & 3 \\
\end{matrix} \right]$
We will now check if $L.H.S=R.H.S$.
$\left[ \begin{matrix}
   \text{ }\!\!~\!\!\text{ }3\times 3+1\times 14 & \text{ }\!\!~\!\!\text{ }3\times 4+1\times 13 \\
   4\times 3+1\times 14 & \text{ }\!\!~\!\!\text{ }4\times 4+1\times 13 \\
\end{matrix} \right]=\left[ \begin{matrix}
   5 & -1 \\
   2 & 3 \\
\end{matrix} \right]$
$\left[ \begin{matrix}
   23 & 25 \\
   26 & 29 \\
\end{matrix} \right]\ne \left[ \begin{matrix}
   5 & -1 \\
   2 & 3 \\
\end{matrix} \right]$
Hence the third option is also incorrect.
Now we will check fourth option \[\left[ \begin{matrix}
   \text{ }\!\!~\!\!\text{ }-3 & 4 \\
   -14 & 13 \\
\end{matrix} \right]\] and substitute in the equation in place of $X$.
$\left[ \begin{matrix}
   \text{ }\!\!~\!\!\text{ }3 & 1 \\
   4 & 1 \\
\end{matrix} \right]X=\left[ \begin{matrix}
   5 & -1 \\
   2 & 3 \\
\end{matrix} \right]$
$\left[ \begin{matrix}
   \text{ }\!\!~\!\!\text{ }3 & 1 \\
   4 & 1 \\
\end{matrix} \right]\left[ \begin{matrix}
   \text{ }\!\!~\!\!\text{ }-3 & 4 \\
   -14 & 13 \\
\end{matrix} \right]=\left[ \begin{matrix}
   5 & -1 \\
   2 & 3 \\
\end{matrix} \right]$
We will now check if $L.H.S=R.H.S$.
$\left[ \begin{matrix}
   \text{ }\!\!~\!\!\text{ }3\times (-3)+1\times (-14) & \text{ }\!\!~\!\!\text{ }3\times 4+1\times 13 \\
   4\times (-3)+1\times (-14) & \text{ }\!\!~\!\!\text{ }4\times 4+1\times 13 \\
\end{matrix} \right]=\left[ \begin{matrix}
   5 & -1 \\
   2 & 3 \\
\end{matrix} \right]$
$\left[ \begin{matrix}
   -23 & 25 \\
   -26 & 29 \\
\end{matrix} \right]\ne \left[ \begin{matrix}
   5 & -1 \\
   2 & 3 \\
\end{matrix} \right]$
Therefore, the fourth option is also incorrect.

Therefore, the correct option is (A).

Note: Before doing multiplication in matrices we need to check first the orders of the matrices then do calculations carefully. To multiply we will take each element of the column of the first matrix and then multiply with each element of rows of the other matrix and then add them .
After we checked the first option and as it was equal on both sides that is $L.H.S=R.H.S$ , it was not necessary to check all the other remaining options.