3 and 4 .Determinants and Matrices
hard

અહી $S =\left\{ m \in Z : A ^{ m ^2}+ A ^{ m }=3 I - A ^{-6}\right\}$ કે જ્યાં $A=\left[\begin{array}{cc}2 & -1 \\ 1 & 0\end{array}\right]$ હોય તો $n(S)$ ની કિમંત મેળવો.

A$1$
B$2$
C$3$
D$4$
(JEE MAIN-2025)

Solution

$A=\left[\begin{array}{cc}2 & -1 \\ 1 & 0\end{array}\right]$
$A^2=\left[\begin{array}{ll}3 & -2 \\ 2 & -1\end{array}\right], A^3=\left[\begin{array}{ll}4 & -3 \\ 3 & -2\end{array}\right], A^4=\left[\begin{array}{ll}5 & -4 \\ 4 & -3\end{array}\right]$
and so on
$A^6=\left[\begin{array}{ll}7 & -6 \\ 6 & -5\end{array}\right]$
$A^{ m }=\left[\begin{array}{cc} m +1 & – m \\ m & – m -1\end{array}\right]$
$A^{m^2}=\left[\begin{array}{cc}m^2+1 & -m^2 \\ m^2 & -\left(m^2-1\right)\end{array}\right]$
$A ^{ m ^2}+ A ^{ m }=3 I – A ^{-6}$
$\left[\begin{array}{cc} m +1 & – m ^2 \\ m^2 & -\left( m ^2-1\right)\end{array}\right]+\left[\begin{array}{cc} m +1 & – m \\ m & -( m -1)\end{array}\right]$
$=3\left[\begin{array}{ll}1 & 0 \\ 0 & 1\end{array}\right]-\left[\begin{array}{ll}-5 & 6 \\ -6 & 7\end{array}\right]$
$=\left[\begin{array}{ll}8 & -6 \\ 6 & -4\end{array}\right]$
$= m ^2+1+ m +1=8$
$= m ^2+ m -6=0 \Rightarrow m=-3,2$
$n ( s )=2$
Standard 12
Mathematics

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