If \(A = \begin{bmatrix}3&1\\-4&-1\end{bmatrix}\), then \(A^{50}\) is equal to
Answer & explanation
Correct answer: option 2
The correct answer is Option (2) → \(\begin{bmatrix}101&50\\-200&-99\end{bmatrix}\)
Write A as I + N, where
\( \displaystyle N = A - I = \begin{bmatrix}2&1\\-4&-2\end{bmatrix} \)
\( \displaystyle N^2 = \begin{bmatrix}4 - 4&2 - 2\\-8 + 8&-4 + 4\end{bmatrix} = \begin{bmatrix}0&0\\0&0\end{bmatrix} \)
Since IN = NI = N and \(N^2 = O\), each multiplication by A = I + N adds one more N:
\(A^2 = (I + N)(I + N) = I + 2N + N^2 = I + 2N\)
\(A^3 = (I + 2N)(I + N) = I + 3N + 2N^2 = I + 3N\)
Continuing in the same way,
\(A^{50} = I + 50N\)
\( \displaystyle I + 50N = \begin{bmatrix}1 + 100&50\\-200&1 - 100\end{bmatrix} = \begin{bmatrix}101&50\\-200&-99\end{bmatrix} \)
Why the other options are incorrect
Option (1) \(\begin{bmatrix}101&50\\-200&-101\end{bmatrix}\) — Incorrect. The last element is \(1 + 50(-2) = -99\), not \(-101\).
Option (3) \(\begin{bmatrix}100&50\\-200&-100\end{bmatrix}\) — Incorrect. This is 50N alone; the I term has been left out.
Option (4) \(\begin{bmatrix}101&-50\\200&-99\end{bmatrix}\) — Incorrect. The off-diagonal elements have the wrong signs; they are 50(1) = 50 and 50(−4) = −200.