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Let $A=\left[\begin{array}{lll}0 & 1 & 2 \\ a & 0 & 3 \\ 1 & c & 0\end{array}\right]$, where $a, c \in R$. If $A^3=A$ and the positive value of a belongs to the interval ( $n -1, n$ ], where $n \in N$, then $n$ is equal to $..........$.
$4$
$2$
$6$
$8$
Solution
$A=\left[\begin{array}{lll}0 & 1 & 2 \\ a & 0 & 3 \\ 1 & c & 0\end{array}\right]$
$A ^3= A$
$A ^2=\left[\begin{array}{lll}0 & 1 & 2 \\ a & 0 & 3 \\ 1 & c & 0\end{array}\right]\left[\begin{array}{lll}0 & 1 & 2 \\ a & 0 & 3 \\ 1 & c & 0\end{array}\right]$
$A ^2=\left[\begin{array}{ccc} a +2 & 2 c & 3 \\ 3 & a +3 c & 2 a \\ ac & 1 & 2+3 c \end{array}\right]$
$A ^3=\left[\begin{array}{ccc} a +2 & 2 c & 3 \\ 3 & a+3 c & 2 a \\ ac & a & 2+3 c \end{array}\right]\left[\begin{array}{ccc}0 & 1 & 2 \\ a & 0 & 3 \\ 1 & c & 0\end{array}\right]$
$A^3=\left[\begin{array}{ccc}2 a c+3 & a+2+3 c & 2 a+4+6 c \\ a(a+3 c)+2 a & 3+2 a c & 6+3 a+9 c \\ a+2+3 c & a c+c(2+3 c) & 2 a c+3\end{array}\right]$
Given $A ^3=A$
$2 ac +3=0 \ldots(1) \text { and } a+2+3 c=1$
$a +1+3 c =0$
$a +1-\frac{9}{2 a}=0$
$2 a ^2+2 a -9=0$
$f (1) < 0, f (2) > 0$
$a \in(1,2]$
$n=2$
