If matrix A = left[ {begin{array}{*{20}{c}}0&{ - 1}1&0end{array}} right] , then {A^{16}} =

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3 and 4 .Determinants and Matrices

easy

If matrix $A = \left[ {\begin{array}{*{20}{c}}0&{ - 1}\\1&0\end{array}} \right]$, then ${A^{16}} = $

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

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

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

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

Solution

(d) Given, Matrix $A = \,\left[ {\begin{array}{*{20}{c}}0&{ – 1}\\1&0\end{array}} \right]$.
We know that ${A^2} = A.A = \left[ {\begin{array}{*{20}{c}}0&{ – 1}\\1&0\end{array}} \right]\,\left[ {\begin{array}{*{20}{c}}0&{ – 1}\\1&0\end{array}} \right] = \left[ {\begin{array}{*{20}{c}}{ – 1}&0\\0&{ – 1}\end{array}} \right].$

Therefore

${A^{16}} = {({A^2})^8} = {\left[ {\begin{array}{*{20}{c}}{ – 1}&0\\0&{ – 1}\end{array}} \right]^8} = \left[ {\begin{array}{*{20}{c}}{{{( – 1)}^8}}&0\\0&{{{( – 1)}^8}}\end{array}} \right]$$ = \left[ {\begin{array}{*{20}{c}}1&0\\0&1\end{array}} \right]$.

Standard 12

Mathematics

Let $A$ be a $2 \times 2$ matrix with $\operatorname{det}(A)=-1$ and det $(( A + I )(\operatorname{Adj}( A )+ I ))=4$. Then the sum of the diagonal elements of $A$ can be.

hard

(JEE MAIN-2022)

If $A = \left[ {\begin{array}{*{20}{c}}2&{ – 3}\\{ – 4}&1\end{array}} \right],$ then $adj\;\left( {3{A^2} + 12A} \right) = $ . . . .

medium

(JEE MAIN-2017)

A manufacturer produces three products $x,\, y,\, z$ which he sells in two markets. Annual sales are indicated below:

Market $x$ $y$ $z$

$I$ $10,000$ $2,000$ $18,000$

$II$ $6,000$ $20,000$ $8,000$

If the unit costs of the above three commodities are $\mathrm{Rs} $. $2.00, $ $\mathrm{Rs} $. $1.00$ and $50$ paise respectively. Find the gross profit.

hard

$\cos \theta \left[ {\begin{array}{{20}{c}}{\cos \theta }&{\sin \theta }\\{ – \sin \theta }&{\cos \theta }\end{array}} \right] + \sin \theta \left[ {\begin{array}{{20}{c}}{\sin \theta }&{ – \cos \theta }\\{\cos \theta }&{\sin \theta }\end{array}} \right] = $

easy

$A = \left[ {\begin{array}{{20}{c}}1&0\\1&1\end{array}} \right]$ and $I = \left[ {\begin{array}{{20}{c}}
1&0\\0&1\end{array}} \right]$ , then which of the following holds for all $n \geq 2, n \in N$ ?

normal

Market	$x$	$y$	$z$
$I$	$10,000$	$2,000$	$18,000$
$II$	$6,000$	$20,000$	$8,000$

If matrix $A = \left[ {\begin{array}{*{20}{c}}0&{ - 1}\\1&0\end{array}} \right]$, then ${A^{16}} = $

Solution

Similar Questions

Let $A$ be a $2 \times 2$ matrix with $\operatorname{det}(A)=-1$ and det $(( A + I )(\operatorname{Adj}( A )+ I ))=4$. Then the sum of the diagonal elements of $A$ can be.

If $A = \left[ {\begin{array}{*{20}{c}}2&{ – 3}\\{ – 4}&1\end{array}} \right],$ then $adj\;\left( {3{A^2} + 12A} \right) = $ . . . .

$\cos \theta \left[ {\begin{array}{*{20}{c}}{\cos \theta }&{\sin \theta }\\{ – \sin \theta }&{\cos \theta }\end{array}} \right] + \sin \theta \left[ {\begin{array}{*{20}{c}}{\sin \theta }&{ – \cos \theta }\\{\cos \theta }&{\sin \theta }\end{array}} \right] = $

$A = \left[ {\begin{array}{*{20}{c}}1&0\\1&1\end{array}} \right]$ and $I = \left[ {\begin{array}{*{20}{c}} 1&0\\0&1\end{array}} \right]$ , then which of the following holds for all $n \geq 2, n \in N$ ?

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$\cos \theta \left[ {\begin{array}{{20}{c}}{\cos \theta }&{\sin \theta }\\{ – \sin \theta }&{\cos \theta }\end{array}} \right] + \sin \theta \left[ {\begin{array}{{20}{c}}{\sin \theta }&{ – \cos \theta }\\{\cos \theta }&{\sin \theta }\end{array}} \right] = $

$A = \left[ {\begin{array}{{20}{c}}1&0\\1&1\end{array}} \right]$ and $I = \left[ {\begin{array}{{20}{c}}
1&0\\0&1\end{array}} \right]$ , then which of the following holds for all $n \geq 2, n \in N$ ?