If \(A = \begin{bmatrix}2&-3\\1&4\end{bmatrix}\), then \(A^2\) is equal to
Answer & explanation
Correct answer: option 2
The correct answer is Option (2) → \(\begin{bmatrix}1&-18\\6&13\end{bmatrix}\)
\(A^2 = A \times A\), and each element is a row of A times a column of A:
\(\text{row 1, column 1}: 2(2) + (-3)(1) = 1\)
\(\text{row 1, column 2}: 2(-3) + (-3)(4) = -18\)
\(\text{row 2, column 1}: 1(2) + 4(1) = 6\)
\(\text{row 2, column 2}: 1(-3) + 4(4) = 13\)
Hence
\( \displaystyle A^2 = \begin{bmatrix}1&-18\\6&13\end{bmatrix} \)
Why the other options are incorrect
Option (1) \(\begin{bmatrix}4&9\\1&16\end{bmatrix}\) — Incorrect. This squares each element separately, which is not matrix multiplication.
Option (3) \(\begin{bmatrix}1&6\\-18&13\end{bmatrix}\) — Incorrect. This is the transpose of \(A^2\); the off-diagonal elements are in the wrong places.
Option (4) \(\begin{bmatrix}7&-18\\6&13\end{bmatrix}\) — Incorrect. Its first element takes \((-3)(1)\) as \(+3\); the correct value is \(4 - 3 = 1\).