3 and 4 .Determinants and Matrices
easy

If $A = \left[ {\begin{array}{*{20}{c}}1&{ - 2}\\5&3\end{array}} \right]$, then $A + {A^T}$equals

A

$\left[ {\begin{array}{*{20}{c}}2&3\\3&6\end{array}} \right]$

B

$\left[ {\begin{array}{*{20}{c}}2&{ - 4}\\{10}&6\end{array}} \right]$

C

$\left[ {\begin{array}{*{20}{c}}2&4\\{ - 10}&6\end{array}} \right]$

D

None of these

Solution

(a) $A = \left[ {\begin{array}{*{20}{c}}1&{ – 2}\\5&3\end{array}} \right],\,$${A^T} = \left[ {\begin{array}{*{20}{c}}1&5\\{ – 2}&3\end{array}} \right],\,A + {A^T} = \left[ {\begin{array}{*{20}{c}}2&3\\3&6\end{array}} \right]$.

Standard 12
Mathematics

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