Let A = [ cases 0 & 1 & 2 a & 0 & 3 1 & c & 0 cases ] , where a, c...

Given,

A = \left[\begin{cases}0 & 1 & 2 \\ a & 0 & 3 \\ 1 & c & 0\end{cases}\right]

⇒ {A}^{2} = \left[\begin{cases}0 & 1 & 2 \\ a & 0 & 3 \\ 1 & c & 0\end{cases}\right]\left[\begin{cases}0 & 1 & 2 \\ a & 0 & 3 \\ 1 & c & 0\end{cases}\right]

⇒ {A}^{2} = \left[\begin{cases}a + 2 & 2c & 3 \\ 3 & a + 3c & 2a \\ ca & 1 & 2 + 3c\end{cases}\right]

Now finding,

{A}^{3} = \left[\begin{cases}a + 2 & 2c & 3 \\ 3 & a + 3c & 2a \\ ca & 1 & 2 + 3c\end{cases}\right]\left[\begin{cases}0 & 1 & 2 \\ a & 0 & 3 \\ 1 & c & 0\end{cases}\right]

⇒ {A}^{3} = \left[\begin{cases}2ca + 3 & \alpha + 2 + 3c & 2a + 4 + 6c \\ {a}^{2} + 3ca + 2a & 3 + 2ac & 6 + 3a + 9c \\ a + 2 + 3c & ca + 2c + 3{c}^{2} & 2ca + 3\end{cases}\right]

Now equating ${A}^{3} = A$ and comparing both side we get,

$2ca + 3 = 0 ⇒ c = \dfrac{- 3}{2a}$

And $a + 2 + 3c = 1$

$⇒ a + 2 + 3\left(\dfrac{- 3}{2a}\right) = 1$

$⇒ a + 1 - \dfrac{9}{2a} = 0$

$⇒ 2{a}^{2} + 2a - 9 = 0$

$⇒ a = \dfrac{- 2 + \sqrt{4 + 4 \times 9 \times 2}}{4} = \dfrac{- 2 + \sqrt{76}}{4} \approx \dfrac{6 . 7}{4} \approx 1 . 4$

Hence, $a \in ( 1, 2 ]$

So, on comparing with $a \in ( n - 1, n ]$ we get $n = 2$