Matrix A satisfies A^2 = 2A - I where I is identity matrix then for n ge 2 , A^n is equal to (n ∈

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3 and 4 .Determinants and Matrices

normal

Matrix $A$ satisfies $A^2 = 2A - I$ where $I$ is the identity matrix then for $n \ge 2$, $A^n$ is equal to $(n \in N)$

$nA - I$

$2^{n - 1}A - (n - 1)I$

$nA - (n - 1)I$

$2^{n - 1}A - I$

Solution

$A^2 = 2A – I$ ==>$A^3 = 2A^2 – IA$
                                  $= 2(2A – I) – A$
                            $A^3 = 3A – 2I$
                             $A^4 = 3A^2 – 2A$
                                    $= 3(2A – I) – 2A$
                             $A^4 = 4A – 3I$
                             $A^5 = 5A – 4I$

$\vdots$
$A^n = nA – (n – 1)I$

Standard 12

Mathematics

$AB = 0$, if and only if

normal

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

easy

Let $S = \left\{ {\left( {\begin{array}{*{20}{c}}
{{a_{11}}}&{{a_{12}}}\\
{{a_{21}}}&{{a_{22}}}
\end{array}} \right):{a_{ij}} \in \left\{ {0,1,2} \right\},{a_{11}} = {a_{22}}} \right\}$ Then the number of non-singular matrices in the set $S$ is

hard

(JEE MAIN-2013)

$\left[ \begin{array}{l}\,\,\,1\\ – 1\\\,\,\,2\end{array} \right]\,\,[2{\rm{ }}\,\,1{\rm{ }} – 1]$ =

easy

If $A=\left[\begin{array}{cc}\alpha & \beta \\ \gamma & -\alpha\end{array}\right]$ is such that $A^{2}=I$, then

medium