MCQ
If $A = \left[ {\begin{array}{*{20}{c}}1&{ - 2}&1\\2&1&3\end{array}} \right]$ and $B = \left[ {\begin{array}{*{20}{c}}2&1\\3&2\\1&1\end{array}} \right]$, then ${(AB)^T} = $
  • A
    $\left[ {\begin{array}{*{20}{c}}{ - 3}&{ - 2}\\{10}&7\end{array}} \right]$
  • $\left[ {\begin{array}{*{20}{c}}{ - 3}&{10}\\{ - 2}&7\end{array}} \right]$
  • C
    $\left[ {\begin{array}{*{20}{c}}{ - 3}&{10}\\7&{ - 2}\end{array}} \right]$
  • D
    $\left[ {\begin{array}{*{20}{c}}3&{10}\\2&7\end{array}} \right]$

Answer

Correct option: B.
$\left[ {\begin{array}{*{20}{c}}{ - 3}&{10}\\{ - 2}&7\end{array}} \right]$
b
(b) $AB = \left[ {\begin{array}{*{20}{c}}1&{ - 2}&1\\2&1&3\end{array}} \right]\left[ {\begin{array}{*{20}{c}}2&1\\3&2\\1&1\end{array}} \right]$

$AB = \left[ {\begin{array}{*{20}{c}}{ - 3}&{ - 2}\\{10}&7\end{array}} \right] \Rightarrow {(AB)^T} = \left[ {\begin{array}{*{20}{c}}{ - 3}&{10}\\{ - 2}&7\end{array}} \right]$.

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