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

Answer

Correct option: A.
$\left[ {\begin{array}{*{20}{c}}1&n\\0&1\end{array}} \right]$
a
(a) ${A^2} = \left[ {\begin{array}{*{20}{c}}1&1\\0&1\end{array}} \right]\,\left[ {\begin{array}{*{20}{c}}1&1\\0&1\end{array}} \right] = \left[ {\begin{array}{*{20}{c}}1&2\\0&1\end{array}} \right],$

and ${A^3} = {A^2}.A = \left[ {\begin{array}{*{20}{c}}1&2\\0&1\end{array}} \right]\,\left[ {\begin{array}{*{20}{c}}1&1\\0&1\end{array}} \right] = \left[ {\begin{array}{*{20}{c}}1&3\\0&1\end{array}} \right]$

==> ${A^n} = {A^{n - 1}}.A = \left[ {\begin{array}{*{20}{c}}1&{n - 1}\\0&1\end{array}} \right]\,\left[ {\begin{array}{*{20}{c}}1&1\\0&1\end{array}} \right] = \left[ {\begin{array}{*{20}{c}}1&n\\0&1\end{array}} \right]$.

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