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STD 12 - 3 and 4 . determinant and metrices — JEE STD 11 Science — Question

Gujarat BoardEnglish MediumSTD 11 ScienceJEESTD 12 - 3 and 4 . determinant and metrices4 Marks
MCQ
If $A = \left[ {\begin{array}{*{20}{c}}a&0&0\\0&b&0\\0&0&c\end{array}} \right]$, then ${A^n} = $
  • A
    $\left[ {\begin{array}{*{20}{c}}{na}&0&0\\0&{nb}&0\\0&0&{nc}\end{array}} \right]$
  • B
    $\left[ {\begin{array}{*{20}{c}}a&0&0\\0&b&0\\0&0&c\end{array}} \right]$
  • ✓
    $\left[ {\begin{array}{*{20}{c}}{{a^n}}&0&0\\0&{{b^n}}&0\\0&0&{{c^n}}\end{array}} \right]$
  • D
    None of these
✓

Answer

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

And ${A^3} = \left[ {\begin{array}{*{20}{c}}{{a^3}}&0&0\\0&{{b^3}}&0\\0&0&{{c^3}}\end{array}} \right]$,....

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

$ = \left[ {\begin{array}{*{20}{c}}{{a^n}}&0&0\\0&{{b^n}}&0\\0&0&{{c^n}}\end{array}} \right]$.

Note: Students should remember this question as a formula.

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