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3 and 4 . determinant and metrices — Maths STD 12 Science — Question

Gujarat BoardEnglish MediumSTD 12 ScienceMaths3 and 4 . determinant and metrices1 Mark
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 ${(AB)^T}$ is equal to
  • 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
    $\left( {\begin{array}{*{20}{c}}{ - 3}&7\\{10}&2\end{array}} \right)$
  • D
    None of these
✓

Answer

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

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

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