If \(A = \begin{bmatrix}1&2\\0&3\end{bmatrix}\), then for every natural number n, \(A^n\) is equal to
Answer & explanation
Correct answer: option 3
The correct answer is Option (3) → \(\begin{bmatrix}1&3^n - 1\\0&3^n\end{bmatrix}\)
Working out the first few powers shows the pattern:
\( \displaystyle A^2 = \begin{bmatrix}1&2\\0&3\end{bmatrix}\begin{bmatrix}1&2\\0&3\end{bmatrix} = \begin{bmatrix}1&2 + 6\\0&9\end{bmatrix} = \begin{bmatrix}1&8\\0&9\end{bmatrix} \)
\( \displaystyle A^3 = \begin{bmatrix}1&8\\0&9\end{bmatrix}\begin{bmatrix}1&2\\0&3\end{bmatrix} = \begin{bmatrix}1&2 + 24\\0&27\end{bmatrix} = \begin{bmatrix}1&26\\0&27\end{bmatrix} \)
The elements 2, 8, 26 are \(3 - 1\), \(9 - 1\), \(27 - 1\). The pattern carries on from each power to the next:
\( \displaystyle \begin{bmatrix}1&3^n - 1\\0&3^n\end{bmatrix}\begin{bmatrix}1&2\\0&3\end{bmatrix} = \begin{bmatrix}1&2 + 3(3^n - 1)\\0&3^{n+1}\end{bmatrix} = \begin{bmatrix}1&3^{n+1} - 1\\0&3^{n+1}\end{bmatrix} \)
Hence \(A^n = \begin{bmatrix}1&3^n - 1\\0&3^n\end{bmatrix}\) for every natural number n.
Why the other options are incorrect
Option (1) \(\begin{bmatrix}1&2n\\0&3^n\end{bmatrix}\) — Incorrect. For \(n = 2\) it gives 4 in place of 8.
Option (2) \(\begin{bmatrix}1&2 \cdot 3^{n-1}\\0&3^n\end{bmatrix}\) — Incorrect. For \(n = 2\) it gives 6 in place of 8.
Option (4) \(\begin{bmatrix}1&\displaystyle\frac{3^n - 1}{2}\\0&3^n\end{bmatrix}\) — Incorrect. For \(n = 1\) it gives 1 in place of 2.