If A = left[ {begin{array}{*{20}{c}}0&1 1&0end{array}} right], then {A^4} =

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3 and 4 .Determinants and Matrices

easy

If $A = \left[ {\begin{array}{*{20}{c}}0&1 \\ 1&0\end{array}} \right],$then ${A^4}$=

$\left[ {\begin{array}{*{20}{c}}1&0 \\ 0&1\end{array}} \right]$

$\left[ {\begin{array}{*{20}{c}}1&1 \\ 0&0\end{array}} \right]$

$\left[ {\begin{array}{*{20}{c}}0&0 \\ 1&1\end{array}} \right]$

$\left[ {\begin{array}{*{20}{c}}0&1 \\1&0\end{array}} \right]$

Solution

(a) We have $A = \left[ {\begin{array}{*{20}{c}}0&1\\1&0\end{array}} \right]$

$\therefore$ ${A^2} = \left[ {\begin{array}{*{20}{c}}0&1\\1&0\end{array}} \right]\,\,\left[ {\begin{array}{*{20}{c}}0&1\\1&0\end{array}} \right] = \left[ {\begin{array}{*{20}{c}}1&0\\0&1\end{array}} \right] = {I_2}$

$\therefore$ ${A^4} = {A^2}.{A^2} = {I_2}.{I_2} = {I_2} = \left[ {\begin{array}{*{20}{c}}1&0\\0&1\end{array}} \right]$.

Standard 12

Mathematics

Let $A=\left[\begin{array}{lll}1 & a & a \\ 0 & 1 & b \\ 0 & 0 & 1\end{array}\right], a, b \in R$. If for some $n \in N$, $A ^{ n }=\left[\begin{array}{ccc}1 & 48 & 2160 \\ 0 & 1 & 96 \\ 0 & 0 & 1\end{array}\right]$ then $n + a + b$ is equal to $\dots\dots$

hard

(JEE MAIN-2022)

Compute the following : $\left[\begin{array}{cc}a & b \\ -b & a\end{array}\right]+\left[\begin{array}{ll}a & b \\ b & a\end{array}\right]$

easy

For the matrix $A=\left[\begin{array}{ll}3 & 2 \\ 1 & 1\end{array}\right]$. find the number $a$ and $b$ such that $A^{2}+a A+b I=0$

medium

Let $P$ be an $m \times m$ matrix such that $P^2=P$. Then, $(I+P)^n$ equals

normal

(KVPY-2011)

If $A$ and $B$ are square matrices of order $n × n$, then ${(A – B)^2}$ is equal to