Let A = [ 0 1 2 a 0 3 1 c 0 ] , where a , c. If A 3 = A and the positive value of a belongs to the interval ( n - 1 , n ] , where n, then n is equal to. [2023]

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Solution

Q.
$Let A = [\begin{matrix} 0 & 1 & 2 \\ a & 0 & 3 \\ 1 & c & 0 \end{matrix}], where a, c \in ℝ . If A^{3} = A and the positive value of a belongs to the interval$ $(n - 1, n],$

$where n \in ℕ, then n is equal to$ _________ . [2023]

Ans.
(2)

We have,

$A = [\begin{matrix} 0 & 1 & 2 \\ a & 0 & 3 \\ 1 & c & 0 \end{matrix}] and A^{3} = A$

Now, $A^{2} = [\begin{matrix} 0 & 1 & 2 \\ a & 0 & 3 \\ 1 & c & 0 \end{matrix}] [\begin{matrix} 0 & 1 & 2 \\ a & 0 & 3 \\ 1 & c & 0 \end{matrix}] \begin{matrix} \end{matrix}$

$\Rightarrow A^{2} = [\begin{matrix} a + 2 & 2 c & 3 \\ 3 & a + 3 c & 2 a \\ a c & 1 & 2 + 3 c \end{matrix}] \begin{matrix} \end{matrix}$

and $A^{3} = [\begin{matrix} a + 2 & 2 c & 3 \\ 3 & a + 3 c & 2 a \\ a c & 1 & 2 + 3 c \end{matrix}] [\begin{matrix} 0 & 1 & 2 \\ a & 0 & 3 \\ 1 & c & 0 \end{matrix}]$

$\Rightarrow A^{3} = [\begin{matrix} 2 a c + 3 & a + 2 + 3 c & 2 a + 4 + 6 c \\ a (a + 3 c) + 2 a & 3 + 2 a c & 6 + 3 a + 9 c \\ a + 2 + 3 c & a c + c (2 + 3 c) & 2 a c + 3 \end{matrix}]$

Given that $A^{3} = A$ ,

$2 a c + 3 = 0 ...(i)$

and $a + 2 + 3 c = 1 \Rightarrow a + 1 + 3 c = 0 ...(ii)$

$\Rightarrow a + 1 - \frac{9}{2 a} = 0$ [From (i)]

$\Rightarrow 2 a^{2} + 2 a - 9 = 0$

Since $f (1) < 0, f (2) > 0$

$∴ a \in (1, 2]$ $[∵ f (a) = 2 a^{2} + 2 a - 9]$

Now, $n - 1 = 1 \Rightarrow n = 2$

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