Let A = ( cases 1 + i & 1 - i & 0 cases ) where i = - 1 . Then, the...

We have A = \left[\begin{cases}1 + i & 1 \\ - i & 0\end{cases}\right]

{A}^{2} = A \cdot A = \left[\begin{cases}1 + i & 1 \\ - i & 0\end{cases}\right]\left[\begin{cases}1 + i & 1 \\ - i & 0\end{cases}\right] = \left[\begin{cases}i & 1 + i \\ - i + 1 & - i\end{cases}\right]

{A}^{3} = {A}^{2} \cdot A = \left[\begin{cases}i & 1 + i \\ 1 - i & - i\end{cases}\right]\left[\begin{cases}1 + i & 1 \\ - i & 0\end{cases}\right] = \left[\begin{cases}0 & i \\ 1 & 1 - i\end{cases}\right]

{A}^{4} = {A}^{3} \cdot A = \left[\begin{cases}0 & i \\ 1 & 1 - i\end{cases}\right]\left[\begin{cases}1 + i & 1 \\ - i & 0\end{cases}\right]\left[\begin{cases}1 & 0 \\ 0 & 1\end{cases}\right] = I

$∴ {A}^{4} = I$

So, ${A}^{5} = {A}^{4} \cdot A = I \cdot A = A$

${A}^{6} = {A}^{4} \cdot {A}^{2} = I \cdot {A}^{2} = {A}^{2}$ and so on

$∵ {A}^{1} = {A}^{5} = {A}^{9} = . . . . . = {A}^{97} = A$

Hence, possible values of $n$, such that ${A}^{n} = A$

$= \left\{1, 5, 9, \ldots , 97\right\}$

Clearly, above sequence is in A.P. where

$a = 1, d = 4 \text{ and } {t}_{n} = 97$ $⇒ a + \left(n - 1\right)d = 97$

$⇒ 1 + \left(n - 1\right)4 = 97$ $⇒ n = 25$

$∴$ The number of elements in the given set $= 25$.