If \(\begin{bmatrix}2&x\end{bmatrix}\begin{bmatrix}1&3\\-2&x\end{bmatrix} = \begin{bmatrix}-4&15\end{bmatrix}\), then the value of x is
Answer & explanation
Correct answer: option 1
The correct answer is Option (1) → 3
Multiplying the row matrix by each column of the \(2 \times 2\) matrix:
\(\begin{bmatrix}2(1) + x(-2)&2(3) + x(x)\end{bmatrix} = \begin{bmatrix}2 - 2x&6 + x^2\end{bmatrix}\)
Equating the corresponding elements with \(\begin{bmatrix}-4&15\end{bmatrix}\):
\(2 - 2x = -4\), so \(x = 3\)
\(6 + x^2 = 15\), so \(x = 3\) or \(x = -3\)
x must satisfy both equations, so \(x = 3\).
Why the other options are incorrect
Option (2) −3 — Incorrect. It makes the first element \(2 + 6 = 8\), not −4.
Option (3) 3 or −3 — Incorrect. These are the roots of the second equation only; −3 fails the first.
Option (4) 2 — Incorrect. It gives the product \(\begin{bmatrix}-2&10\end{bmatrix}\).