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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 $A = \left[ {\begin{array}{*{20}{c}}1&0\\1&1\end{array}} \right]$ and $I = \left[ {\begin{array}{*{20}{c}}1&0\\0&1\end{array}} \right]$, then which one of the following holds for all $n \ge 1$, (by the principal of mathematical induction)
  • A
    ${A^n} = nA + (n - 1)I$
  • B
    ${A^n} = {2^{n - 1}}A + (n - 1)I$
  • ✓
    ${A^n} = nA - (n - 1)I$
  • D
    ${A^n} = {2^{n - 1}}A - (n - 1)I$
✓

Answer

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

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

$\therefore $ ${A^n} = \left[ {\begin{array}{*{20}{c}}1&0\\n&1\end{array}} \right]$ $;$ $nA = \left[ {\begin{array}{*{20}{c}}n&0\\n&n\end{array}} \right],(n - 1)I = \left[ {\begin{array}{*{20}{c}}{n - 1}&0\\0&{n - 1}\end{array}} \right]$

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

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