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Question:

If A = $\left[\begin{array}{ccc}1& 0& 0\\ 0& 0& 1\\ 0& 1& 0\end{array}\right]$, then the absolute value of the determinant of (A⁹ + A⁶ + A³ + A) is __________.

Explanation:

Given, A = $\left[\begin{array}{ccc}1& 0& 0\\ 0& 0& 1\\ 0& 1& 0\end{array}\right]$ 

Now, A × A = $\left[\begin{array}{ccc}1& 0& 0\\ 0& 0& 1\\ 0& 1& 0\end{array}\right]$ × $\left[\begin{array}{ccc}1& 0& 0\\ 0& 0& 1\\ 0& 1& 0\end{array}\right]$ = $\left[\begin{array}{ccc}1\times 1+0\times 0+0\times 0& 1\times 0+0\times 0+0\times 1& 1\times 0+0\times 1+0\times 0\\ 0\times 1+0\times 0+1\times 0& 0\times 0+0\times 0+1\times 1& 0\times 0+0\times 1+1\times 0\\ 0\times 1+1\times 0+0\times 0& 0\times 0+1\times 0+0\times 1& 0\times 0+1\times 1+0\times 0\end{array}\right]$ = $\left[\begin{array}{ccc}1& 0& 0\\ 0& 1& 0\\ 0& 0& 1\end{array}\right]$ = I

⇒ A³ = A² × A = I × A = A

⇒ A⁶ = A³ × A³ = A × A = I

⇒ A⁹ = A³ × A³ × A³ = A × A × A = A² × A = I × A = A

Now, (A⁹ + A⁶ + A³ + A)

= A + I + A + A + A

= 3A + I

= $\left[\begin{array}{ccc}3& 0& 0\\ 0& 0& 3\\ 0& 3& 0\end{array}\right]$ + $\left[\begin{array}{ccc}1& 0& 0\\ 0& 1& 0\\ 0& 0& 1\end{array}\right]$

= $\left[\begin{array}{ccc}4& 0& 0\\ 0& 1& 3\\ 0& 3& 1\end{array}\right]$

∴ The absolute value of the determinant of (A⁹ + A⁶ + A³ + A) = absolute value of $\left[\begin{array}{ccc}4& 0& 0\\ 0& 1& 3\\ 0& 3& 1\end{array}\right]$

= |4 × (1 × 1 - 3 × 3)| = |4 × (1 - 9)| = 32.

Hence, 32.