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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 $A = \left( {\begin{array}{*{20}{c}}3&2\\0&1\end{array}} \right)$, then ${({A^{ - 1}})^3}$is =
  • ✓
    $\frac{1}{{27}}\left( {\begin{array}{*{20}{c}}1&{ - 26}\\0&{27}\end{array}} \right)$
  • B
    $\frac{1}{{27}}\left( {\begin{array}{*{20}{c}}{ - 1}&{26}\\0&{27}\end{array}} \right)$
  • C
    $\frac{1}{{27}}\left( {\begin{array}{*{20}{c}}1&{ - 26}\\0&{ - 27}\end{array}} \right)$
  • D
    $\frac{1}{{27}}\left( {\begin{array}{*{20}{c}}{ - 1}&{ - 26}\\0&{ - 27}\end{array}} \right)$
✓

Answer

Correct option: A.
$\frac{1}{{27}}\left( {\begin{array}{*{20}{c}}1&{ - 26}\\0&{27}\end{array}} \right)$
a
(a) $|A| = 3,\,AdjA = \left( {\begin{array}{*{20}{c}}1&{ - 2}\\0&3\end{array}} \right)$; $\therefore $ ${A^{ - 1}} = \frac{1}{3}\,\left( {\begin{array}{*{20}{c}}1&{ - 2}\\0&3\end{array}} \right)$

$ \Rightarrow $ ${({A^{ - 1}})^3} = \frac{1}{{27}}\,{\left( {\begin{array}{*{20}{c}}1&{ - 2}\\0&3\end{array}} \right)^3} = \frac{1}{{27}}\,\left( {\begin{array}{*{20}{c}}1&{ - 26}\\0&{27}\end{array}} \right)$.

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