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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 matrix $A = \left[ {\begin{array}{*{20}{c}}0&{ - 1}\\1&0\end{array}} \right]$, then ${A^{16}} = $
  • A
    $\left[ {\begin{array}{*{20}{c}}0&{ - 1}\\1&0\end{array}} \right]$
  • B
    $\left[ {\begin{array}{*{20}{c}}0&1\\1&0\end{array}} \right]$
  • C
    $\left[ {\begin{array}{*{20}{c}}{ - 1}&0\\0&1\end{array}} \right]$
  • ✓
    $\left[ {\begin{array}{*{20}{c}}1&0\\0&1\end{array}} \right]$
✓

Answer

Correct option: D.
$\left[ {\begin{array}{*{20}{c}}1&0\\0&1\end{array}} \right]$
d
(d) Given, Matrix $A = \,\left[ {\begin{array}{*{20}{c}}0&{ - 1}\\1&0\end{array}} \right]$.
We know that ${A^2} = A.A = \left[ {\begin{array}{*{20}{c}}0&{ - 1}\\1&0\end{array}} \right]\,\left[ {\begin{array}{*{20}{c}}0&{ - 1}\\1&0\end{array}} \right] = \left[ {\begin{array}{*{20}{c}}{ - 1}&0\\0&{ - 1}\end{array}} \right].$

Therefore

${A^{16}} = {({A^2})^8} = {\left[ {\begin{array}{*{20}{c}}{ - 1}&0\\0&{ - 1}\end{array}} \right]^8} = \left[ {\begin{array}{*{20}{c}}{{{( - 1)}^8}}&0\\0&{{{( - 1)}^8}}\end{array}} \right]$$ = \left[ {\begin{array}{*{20}{c}}1&0\\0&1\end{array}} \right]$.

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