If \( A=\begin{bmatrix}1 & 1\\0 & 1\end{bmatrix} \), then \( A^{26} \) is equal to
Answer & explanation
Correct answer: option 3
The correct answer is Option (3) → \( \begin{bmatrix}1 & 26\\0 & 1\end{bmatrix} \)
The first few powers of A show a pattern:
\( \displaystyle A^2=\begin{bmatrix}1 & 1\\0 & 1\end{bmatrix}\begin{bmatrix}1 & 1\\0 & 1\end{bmatrix} \)
\( \displaystyle =\begin{bmatrix}1 & 2\\0 & 1\end{bmatrix} \)
\( \displaystyle A^3=A^2A \)
\( \displaystyle =\begin{bmatrix}1 & 2\\0 & 1\end{bmatrix}\begin{bmatrix}1 & 1\\0 & 1\end{bmatrix} \)
\( \displaystyle =\begin{bmatrix}1 & 3\\0 & 1\end{bmatrix} \)
Each multiplication by A adds 1 to the top-right entry, so \( A^n=\begin{bmatrix}1 & n\\0 & 1\end{bmatrix} \).
\( \displaystyle A^{26}=\begin{bmatrix}1 & 26\\0 & 1\end{bmatrix} \)
Why the other options are incorrect
Option (1) \( \begin{bmatrix}1 & 1\\0 & 1\end{bmatrix} \) — Incorrect. This is A itself; the top-right entry of \( A^{26} \) is 26, not 1.
Option (2) \( \begin{bmatrix}26 & 26\\0 & 26\end{bmatrix} \) — Incorrect. This is 26A, not \( A^{26} \); a power of a matrix is not found by multiplying each entry by 26.
Option (4) \( \begin{bmatrix}1 & 0\\0 & 1\end{bmatrix} \) — Incorrect. This is the identity matrix; the top-right entry of \( A^{26} \) is 26, not 0.