If \(A = \begin{bmatrix}x&1\\0&y\end{bmatrix}\), where x and y are real numbers, and \(A^2 = \begin{bmatrix}4&5\\0&9\end{bmatrix}\), then the value of xy is
Answer & explanation
Correct answer: option 1
The correct answer is Option (1) → 6
Multiplying row by column:
\( \displaystyle A^2 = \begin{bmatrix}x^2 + 0&x + y\\0&0 + y^2\end{bmatrix} = \begin{bmatrix}x^2&x + y\\0&y^2\end{bmatrix} \)
Comparing with the given \(A^2\),
\(x^2 = 4\), so \(x = 2\) or \(x = -2\)
\(y^2 = 9\), so \(y = 3\) or \(y = -3\)
\(x + y = 5\)
Of the four pairs, only \(x = 2,\ y = 3\) gives \(x + y = 5\). Hence
\(xy = 2 \times 3 = 6\)
Why the other options are incorrect
Option (2) \(-6\) — Incorrect. This needs \(x = -2,\ y = 3\) or \(x = 2,\ y = -3\), where \(x + y\) is 1 or −1, not 5.
Option (3) 13 — Incorrect. This is \(x^2 + y^2\), not xy.
Option (4) 5 — Incorrect. This is \(x + y\), not xy.