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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
If $A = \left[ {\begin{array}{*{20}{c}}3&{ - 5}\\{ - 4}&2\end{array}} \right],$then ${A^2} - 5A = $
  • A
    $I$
  • ✓
    $14\ I$
  • C
    $0$
  • D
    None of these
✓

Answer

Correct option: B.
$14\ I$
b
(b) ${A^2} = A.A = \left[ {\begin{array}{*{20}{c}}3&{ - 5}\\{ - 4}&2\end{array}} \right]{\rm{ }}\left[ {\begin{array}{*{20}{c}}3&{ - 5}\\{ - 4}&2\end{array}} \right]$

$ \Rightarrow \,{A^2} = \left[ {\begin{array}{*{20}{c}}{29}&{ - 25}\\{ - 20}&{24}\end{array}} \right]$ and $5A = \left[ {\begin{array}{*{20}{c}}{15}&{ - 25}\\{ - 20}&{10}\end{array}} \right]$

$\therefore$ ${A^2} - 5A = 14\left[ {\begin{array}{*{20}{c}}1&0\\0&1\end{array}} \right] = 14I$.

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