3 and 4 .Determinants and Matrices
easy

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 $X =$

A

$\left[ {\begin{array}{*{20}{c}}{ - 3}&4\\{14}&{ - 13}\end{array}} \right]$

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

$\left[ {\begin{array}{*{20}{c}}{ - 3}&4\\{ - 14}&{13}\end{array}} \right]$

Solution

(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 $\left[ {\begin{array}{*{20}{c}}3&1\\4&1\end{array}} \right]\,\,\left[ {\begin{array}{*{20}{c}}{ – 3}&4\\{14}&{ – \,13}\end{array}} \right] = \left[ {\begin{array}{*{20}{c}}5&{ – 1}\\2&3\end{array}} \right]$.

Standard 12
Mathematics

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