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If $A = \left[ {\begin{array}{*{20}{c}}
1&1\\
1&1
\end{array}} \right]$ and $\det ({A^n} - I) = 1 - {\lambda ^n}\,,\,n \in N$ then $\lambda $ is
$1$
$2$
$3$
$4$
Solution
$A=\left[\begin{array}{ll}{1} & {1} \\ {1} & {1}\end{array}\right]$
$A^{2}=\left[\begin{array}{ll}{1} & {1} \\ {1} & {1}\end{array}\right]\left[\begin{array}{ll}{1} & {1} \\ {1} & {1}\end{array}\right]$
$A^{2}=\left[\begin{array}{ll}{2} & {2} \\ {2} & {2}\end{array}\right]=2 \mathrm{A} \Rightarrow \mathrm{A}^{3}=2 \mathrm{A}^{2}$
$=2^{2} \mathrm{A}$
Similarly $\mathrm{A}^{4}=2^{3} \mathrm{A}$ and so on
So $\mathrm{A}^{\mathrm{n}}=2^{\mathrm{n}-1} \mathrm{A}$
$\Rightarrow A^{n}=2^{n-1}\left[\begin{array}{ll}{1} & {1} \\ {1} & {1}\end{array}\right]$
${{\rm{A}}^{\rm{n}}} – {\rm{I}} = \left[ {\begin{array}{*{20}{c}}
{{2^{{\rm{n}} – 1}} – 1}&{{2^{{\rm{n}} – 1}}}\\
{{2^{{\rm{n}} – 1}}}&{{2^{{\rm{n}} – 1}} – 1}
\end{array}} \right]$
$\left| {{{\rm{A}}^{\rm{n}}} – {\rm{I}}} \right| = {\left( {{2^{{\rm{n}} – 1}} – 1} \right)^2} – {\left( {{2^{{\rm{n}} – 1}}} \right)^2}$
$=1-2^{\mathrm{n}} $
$ \Rightarrow \lambda =2$
