Solve this

Question:

Let $A=\left[\begin{array}{lll}a & 0 & 0 \\ 0 & a & 0 \\ 0 & 0 & a\end{array}\right]$, then $A^{n}$ is equal to

(a) $\left[\begin{array}{ccc}a^{n} & 0 & 0 \\ 0 & a^{n} & 0 \\ 0 & 0 & a\end{array}\right]$

(b) $\left[\begin{array}{ccc}a^{n} & 0 & 0 \\ 0 & a & 0 \\ 0 & 0 & a\end{array}\right]$

(c) $\left[\begin{array}{ccc}a^{n} & 0 & 0 \\ 0 & a^{n} & 0 \\ 0 & 0 & a^{n}\end{array}\right]$

(d) $\left[\begin{array}{ccc}n a & 0 & 0 \\ 0 & n a & 0 \\ 0 & 0 & n a\end{array}\right]$

Solution:

(c) $\left[\begin{array}{ccc}a^{n} & 0 & 0 \\ 0 & a^{n} & 0 \\ 0 & 0 & a^{n}\end{array}\right]$

Here,

$A=\left[\begin{array}{lll}a & 0 & 0 \\ 0 & a & 0 \\ 0 & 0 & a\end{array}\right]$

$\Rightarrow A^{2}=\left[\begin{array}{lll}a & 0 & 0 \\ 0 & a & 0 \\ 0 & 0 & a\end{array}\right]\left[\begin{array}{lll}a & 0 & 0 \\ 0 & a & 0 \\ 0 & 0 & a\end{array}\right]=\left[\begin{array}{ccc}a^{2} & 0 & 0 \\ 0 & a^{2} & 0 \\ 0 & 0 & a^{2}\end{array}\right]$

$\Rightarrow A^{3}=\left[\begin{array}{ccc}a^{2} & 0 & 0 \\ 0 & a^{2} & 0 \\ 0 & 0 & a^{2}\end{array}\right]\left[\begin{array}{lll}a & 0 & 0 \\ 0 & a & 0 \\ 0 & 0 & a\end{array}\right]=\left[\begin{array}{ccc}a^{3} & 0 & 0 \\ 0 & a^{3} & 0 \\ 0 & 0 & a^{3}\end{array}\right]$

This pattern is applicable on all natural numbers.

$\therefore A^{n}=\left[\begin{array}{ccc}a^{n} & 0 & 0 \\ 0 & a^{n} & 0 \\ 0 & 0 & a^{n}\end{array}\right]$