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3 and 4 .Determinants and Matrices
normal
$A = \left[ {\begin{array}{*{20}{c}}1&0\\1&1\end{array}} \right]$ and $I = \left[ {\begin{array}{*{20}{c}}
1&0\\0&1\end{array}} \right]$ , then which of the following holds for all $n \geq 2, n \in N$ ?
A
${A^n} = {2^{n - 1}}A + \left( {n - 1} \right)I$
B
${A^n} = nA + \left( {n - 1} \right)I$
C
${A^n} = {2^{n - 1}}A - \left( {n - 1} \right)I$
D
${A^n} = nA - \left( {n - 1} \right)I$
Solution
$A^2 = \left[ {\begin{array}{*{20}{c}}
1&0 \\
1&1
\end{array}} \right]\left[ {\begin{array}{*{20}{c}}
1&0 \\
1&1
\end{array}} \right] = \left[ {\begin{array}{*{20}{c}}
1&0 \\
2&1
\end{array}} \right]$
it satisfies only option $(D)$
Standard 12
Mathematics
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