MCQ
If $\left[ {\begin{array}{*{20}{c}}3&1\\4&1\end{array}} \right]X = \left[ {\begin{array}{*{20}{c}}5&{ - 1}\\2&3\end{array}} \right],$then $X =$
  • $\left[ {\begin{array}{*{20}{c}}{ - 3}&4\\{14}&{ - 13}\end{array}} \right]$
  • B
    $\left[ {\begin{array}{*{20}{c}}3&{ - 4}\\{ - 14}&{13}\end{array}} \right]$
  • C
    $\left[ {\begin{array}{*{20}{c}}3&4\\{14}&{13}\end{array}} \right]$
  • D
    $\left[ {\begin{array}{*{20}{c}}{ - 3}&4\\{ - 14}&{13}\end{array}} \right]$

Answer

Correct option: A.
$\left[ {\begin{array}{*{20}{c}}{ - 3}&4\\{14}&{ - 13}\end{array}} \right]$
a
(a)$\left[ {\begin{array}{*{20}{c}}3&1\\4&1\end{array}} \right]X = \left[ {\begin{array}{*{20}{c}}5&{ - 1}\\2&{\,\,3}\end{array}} \right] \Rightarrow X = \left[ {\begin{array}{*{20}{c}}{ - 3}&4\\{14}&{ - 13}\end{array}} \right]$
Since $\left[ {\begin{array}{*{20}{c}}3&1\\4&1\end{array}} \right]\,\,\left[ {\begin{array}{*{20}{c}}{ - 3}&4\\{14}&{ - \,13}\end{array}} \right] = \left[ {\begin{array}{*{20}{c}}5&{ - 1}\\2&3\end{array}} \right]$.

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