If A = left[ {begin{array}{*{20}{c}}
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Question

$A=\left[\begin{array}{ll}{1} & {1} \ {1} & {1}\end{array}ight]$

$A^{2}=\left[\begin{array}{ll}{1} & {1} \ {1} & {1}\end{array}

Vedclass · Accepted Answer

$2$

Vedclass · Answer

$A=\left[\begin{array}{ll}{1} & {1} \ {1} & {1}\end{array}ight]$

$A^{2}=\left[\begin{array}{ll}{1} & {1} \ {1} & {1}\end{array}ight]\left[\begin{array}{ll}{1} & {1} \ {1} & {1}\end{array}ight]$

$A^{2}=\left[\begin{array}{ll}{2} & {2} \ {2} & {2}\end{array}ight]=2 \mathrm{A} \Rightarrow \mathrm{A}^{3}=2 \mathrm{A}^{2}$

$=2^{2} \mathrm{A}$

Similarly $\mathrm{A}^{4}=2^{3} \mathrm{A}$ and so on

So $\mathrm{A}^{\mathrm{n}}=2^{\mathrm{n}-1} \mathrm{A}$

$\Rightarrow A^{n}=2^{n-1}\left[\begin{array}{ll}{1} & {1} \ {1} & {1}\end{array}ight]$

${{m{A}}^{m{n}}} - {m{I}} = \left[ {\begin{array}{*{20}{c}}
{{2^{{m{n}} - 1}} - 1}&{{2^{{m{n}} - 1}}}\
{{2^{{m{n}} - 1}}}&{{2^{{m{n}} - 1}} - 1}
\end{array}} ight]$

$\left| {{{m{A}}^{m{n}}} - {m{I}}} ight| = {\left( {{2^{{m{n}} - 1}} - 1} ight)^2} - {\left( {{2^{{m{n}} - 1}}} ight)^2}$

$=1-2^{\mathrm{n}} $

$ \Rightarrow \lambda =2$

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

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