$A^{2}=\left[\begin{array}{ll}{1} & {1} \\ {1} & {1}\end{array}\right]\left[\begin{array}{ll}{1} & {1} \\ {1} & {1}\end{array}\right]$
$A^{2}=\left[\begin{array}{ll}{2} & {2} \\ {2} & {2}\end{array}\right]=2 \mathrm{A} \Rightarrow \mathrm{A}^{3}=2 \mathrm{A}^{2}$
$=2^{2} \mathrm{A}$
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$
Generate a complete, print-ready paper with questions like this in minutes — across 16+ boards, with answer keys.
The set of points where $f$ is not differentiable is
| Column $I$ | Column $II$ |
| $(A)$ If $\vec{a}=\hat{j}+\sqrt{3} \hat{k}, \vec{b}=-\hat{j}+\sqrt{3} \hat{k}$ and $\vec{c}=2 \sqrt{3} \hat{k}$ form a triangle, then the internal angle of the triangle between $\vec{a}$ and $\vec{b}$ is | $(p)$ $\frac{\pi}{6}$ |
| $(B)$ If $\int_a^b(f(x)-3 x) d x=a^2-b^2$, then the value of $f\left(\frac{\pi}{6}\right)$ is | $(q)$ $\frac{2 \pi}{3}$ |
| $(C)$ The value of $\frac{\pi^2}{\ln 3} \int_{1 / 6}^{5 / 6} \sec (\pi x) d x$ is | $(r)$ $\frac{\pi}{3}$ |
| $(D)$ The maximum value of $\left|\operatorname{Arg}\left(\frac{1}{1-z}\right)\right|$ for $|z|=1, \quad z \neq 1$ is given by | $(s)$ $\pi$ |
| $(t)$ $\frac{\pi}{2}$ |
$(a+b)-\left(\frac{a^{2}+b^{2}}{2}\right)+\left(\frac{a^{3}+b^{3}}{3}\right)-\left(\frac{a^{4}+b^{4}}{4}\right)+\ldots$ is ..... .