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STD 12 - 3 and 4 . determinant and metrices — Mathematics STD 12 Science — Question

CBSE BoardEnglish MediumSTD 12 ScienceMathematicsSTD 12 - 3 and 4 . determinant and metrices1 Mark
MCQ
$\left[ {\begin{array}{*{20}{c}}7&1&2\\9&2&1\end{array}} \right]\,\left[ \begin{array}{l}3\\4\\5\end{array} \right] + 2\left[ \begin{array}{l}4\\2\end{array} \right]$ is equal to
  • ✓
    $\left[ \begin{array}{l}43\\44\end{array} \right]$
  • B
    $\left[ \begin{array}{l}43\\45\end{array} \right]$
  • C
    $\left[ \begin{array}{l}45\\44\end{array} \right]$
  • D
    $\left[ \begin{array}{l}44\\45\end{array} \right]$
✓

Answer

Correct option: A.
$\left[ \begin{array}{l}43\\44\end{array} \right]$
a
(a) $\left[ {\begin{array}{*{20}{c}}7&1&2\\9&2&1\end{array}} \right]\,\left[ {\begin{array}{*{20}{c}}3\\4\\5\end{array}} \right]\, = \,\,\left[ {\begin{array}{*{20}{c}}{35}\\{40}\end{array}} \right]$;

$\therefore$ $\left[ {\begin{array}{*{20}{c}}{35}\\{40}\end{array}} \right]\, + \,\left[ {\begin{array}{*{20}{c}}8\\4\end{array}} \right]\, = \,\left[ {\begin{array}{*{20}{c}}{43}\\{44}\end{array}} \right]$.

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