Let $A=\begin{bmatrix} 0 & 1 & 2 \\ a & 0 & 3 \\ 1 & c & 0 \end{bmatrix}$, where $a, c \in \mathbb{R}$. If…
Let $A=\begin{bmatrix} 0 & 1 & 2 \\ a & 0 & 3 \\ 1 & c & 0 \end{bmatrix}$, where $a, c \in \mathbb{R}$. If $A^3 = A$ and the positive value of $a$ belongs to the interval $(n-1, n]$, where $n \in \mathbb{N}$, then $n$ is equal to ____.
Solution
Given,
$A = \begin{bmatrix} 0 & 1 & 2 \\ a & 0 & 3 \\ 1 & c & 0 \end{bmatrix}$
$\Rightarrow A^2 = \begin{bmatrix} 0 & 1 & 2 \\ a & 0 & 3 \\ 1 & c & 0 \end{bmatrix} \begin{bmatrix} 0 & 1 & 2 \\ a & 0 & 3 \\ 1 & c & 0 \end{bmatrix}$
$\Rightarrow A^2 = \begin{bmatrix} a+2 & 2c & 3 \\ 3 & a+3c & 2a \\ ca & 1 & 2+3c \end{bmatrix}$
Now finding,
$A^3 = \begin{bmatrix} a+2 & 2c & 3 \\ 3 & a+3c & 2a \\ ca & 1 & 2+3c \end{bmatrix} \begin{bmatrix} 0 & 1 & 2 \\ a & 0 & 3 \\ 1 & c & 0 \end{bmatrix}$
$\Rightarrow A^3 = \begin{bmatrix} 2ca+3 & \alpha+2+3c & 2a+4+6c \\ a^2+3ca+2a & 3+2ac & 6+3a+9c \\ a+2+3c & ca+2c+3c^2 & 2ca+3 \end{bmatrix}$
Now equating $A^3 = A$ and comparing both sides we get,
$2ca+3 = 0 \Rightarrow c = \frac{-3}{2a}$
And $a+2+3c = 1$
$\Rightarrow a+2+3\left(\frac{-3}{2a}\right) = 1$
$\Rightarrow a+1-\frac{9}{2a} = 0$
$\Rightarrow 2a^2+2a-9 = 0$
$\Rightarrow a = \frac{-2 \pm \sqrt{4+4 \times 9 \times 2}}{4} = \frac{-2 \pm \sqrt{76}}{4} \approx \frac{6.7}{4} \approx 1.4$
Hence, $a \in (1,2]$
So, on comparing with $a \in (n-1,n]$ we get $n=2$