If \(A = \begin{bmatrix}0&1\\-1&0\end{bmatrix}\), then \(A + A^2 + A^3 + \dots + A^{2026}\) is equal to
Answer & explanation
Correct answer: option 3
The correct answer is Option (3) → \(\begin{bmatrix}-1&1\\-1&-1\end{bmatrix}\)
Squaring A:
\( \displaystyle A^2 = \begin{bmatrix}0 - 1&0 + 0\\0 + 0&-1 + 0\end{bmatrix} = -I \)
So the powers repeat in a cycle of four:
\(A^3 = A^2A = -A\)
\(A^4 = (A^2)^2 = I\)
Any four consecutive powers add to zero:
\(A^{4m+1} + A^{4m+2} + A^{4m+3} + A^{4m+4} = A - I - A + I = O\)
Since \(2026 = 4 \times 506 + 2\), the first 2024 powers add to O, and only \(A^{2025} = A\) and \(A^{2026} = A^2 = -I\) remain:
\( \displaystyle A + A^2 + \dots + A^{2026} = A - I = \begin{bmatrix}-1&1\\-1&-1\end{bmatrix} \)
Why the other options are incorrect
Option (1) \(\begin{bmatrix}0&0\\0&0\end{bmatrix}\) — Incorrect. The sum is O only when the number of terms is a multiple of 4; 2026 leaves two terms over.
Option (2) \(\begin{bmatrix}1&1\\-1&1\end{bmatrix}\) — Incorrect. This is \(A + I\), which takes \(A^2\) as I instead of −I.
Option (4) \(\begin{bmatrix}-1&-1\\1&-1\end{bmatrix}\) — Incorrect. This is \(A^2 + A^3 = -I - A\), the wrong two left-over powers.