Let \( A=\begin{bmatrix}2 & 2 & 2\\0 & 2 & 2\\0 & 0 & 2\end{bmatrix} \) , then \( A^6 \) is equal to:
Answer & explanation
Correct answer: option 2
The correct answer is Option (2) → \( 2^6\begin{bmatrix}1 & 6 & 21\\0 & 1 & 6\\0 & 0 & 1\end{bmatrix} \)
Every entry of A is a multiple of 2, so A = 2B, and B splits into I and a matrix N:
\( \displaystyle B=\begin{bmatrix}1 & 1 & 1\\0 & 1 & 1\\0 & 0 & 1\end{bmatrix}=I+N \)
\( \displaystyle N=\begin{bmatrix}0 & 1 & 1\\0 & 0 & 1\\0 & 0 & 0\end{bmatrix} \)
\( \displaystyle N^2=\begin{bmatrix}0 & 0 & 1\\0 & 0 & 0\\0 & 0 & 0\end{bmatrix} \)
\( \displaystyle N^3=O \)
Since I and N commute, the binomial theorem applies, and all terms from \( N^3 \) onwards vanish:
\( \displaystyle B^6=(I+N)^6=I+{}^{6}C_{1}N+{}^{6}C_{2}N^2 \)
\( \displaystyle =I+6N+15N^2 \)
\( \displaystyle =\begin{bmatrix}1 & 6 & 6+15\\0 & 1 & 6\\0 & 0 & 1\end{bmatrix} \)
\( \displaystyle =\begin{bmatrix}1 & 6 & 21\\0 & 1 & 6\\0 & 0 & 1\end{bmatrix} \)
\( \displaystyle A^6=2^6B^6 \)
\( \displaystyle =2^6\begin{bmatrix}1 & 6 & 21\\0 & 1 & 6\\0 & 0 & 1\end{bmatrix} \)
Why the other options are incorrect
Option (1) \( \begin{bmatrix}1 & 6 & 21\\0 & 1 & 6\\0 & 0 & 1\end{bmatrix} \) — Incorrect. This is \( B^6 \); the factor \( 2^6 \) from \( A=2B \) is missing.
Option (3) \( \begin{bmatrix}1 & 4 & 12\\0 & 1 & 4\\0 & 0 & 1\end{bmatrix} \) — Incorrect. This is \( B^4 \), not \( B^6 \), and the factor of 2 is missing.
Option (4) \( 2^4\begin{bmatrix}1 & 4 & 12\\0 & 1 & 4\\0 & 0 & 1\end{bmatrix} \) — Incorrect. This is \( A^4=2^4B^4 \), not \( A^6 \).