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

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

$A^{2}=\left[\begin{array}{ll}{2} & {2} \\ {2} & {2}\end{array}\right]=2 \mathrm{A} \Rightarrow \mathrm{A}^{3}=2 \mathrm{A}^{2}$

$=2^{2} \mathrm{A}$

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$