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(A) Let $A = \begin{bmatrix} 0 & i \ 1 & 0 \end{bmatrix}$ and $X = \begin{bmatrix} a & b \ c & d \end{bmatrix}$.The given condition is $A^n X = X$ f

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$11$

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(A) Let $A = \begin{bmatrix} 0 & i \ 1 & 0 \end{bmatrix}$ and $X = \begin{bmatrix} a & b \ c & d \end{bmatrix}$.The given condition is $A^n X = X$ for all $a, b, c, d \in R$.Since $X$ can be any $2 	imes 2$ matrix,we can choose $X = I$ (the identity matrix),which implies $A^n = I$.Now,calculate powers of $A$:$A^2 = \begin{bmatrix} 0 & i \ 1 & 0 \end{bmatrix} \begin{bmatrix} 0 & i \ 1 & 0 \end{bmatrix} = \begin{bmatrix} i & 0 \ 0 & i \end{bmatrix} = iI$.$A^4 = (A^2)^2 = (iI)^2 = i^2 I = -I$.$A^8 = (A^4)^2 = (-I)^2 = I$.Thus,$A^n = I$ if and only if $n$ is a multiple of $8$.We need to find the number of $2$-digit numbers $n$ such that $n$ is a multiple of $8$.The $2$-digit multiples of $8$ are $16, 24, 32, 40, 48, 56, 64, 72, 80, 88, 96$.Counting these,we have $11$ such numbers.

Let $S=\{n \in N \mid \begin{bmatrix} 0 & i \\ 1 & 0 \end{bmatrix}^{n} \begin{bmatrix} a & b \\ c & d \end{bmatrix} = \begin{bmatrix} a & b \\ c & d \end{bmatrix} \forall a, b, c, d \in R \}$,where $i=\sqrt{-1}$. Then the number of $2$-digit numbers in the set $S$ is $......$

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