If A is a square matrix of order 2 such that \(A\begin{bmatrix}1\\1\end{bmatrix} = \begin{bmatrix}3\\5\end{bmatrix}\) and \(A\begin{bmatrix}1\\-1\end{bmatrix} = \begin{bmatrix}1\\-1\end{bmatrix}\), then A is equal to
Answer & explanation
Correct answer: option 2
The correct answer is Option (2) → \(\begin{bmatrix}2&1\\2&3\end{bmatrix}\)
Let \(A = \begin{bmatrix}a&b\\c&d\end{bmatrix}\). Multiplying out the two given products and equating corresponding elements:
\(a + b = 3,\quad c + d = 5\)
\(a - b = 1,\quad c - d = -1\)
Adding and subtracting each pair:
\(2a = 4\), so \(a = 2\), and \(b = 3 - 2 = 1\)
\(2c = 4\), so \(c = 2\), and \(d = 5 - 2 = 3\)
Hence
\( \displaystyle A = \begin{bmatrix}2&1\\2&3\end{bmatrix} \)
Why the other options are incorrect
Option (1) \(\begin{bmatrix}1&2\\3&2\end{bmatrix}\) — Incorrect. It gives the first product correctly, but its product with \(\begin{bmatrix}1\\-1\end{bmatrix}\) is \(\begin{bmatrix}-1\\1\end{bmatrix}\).
Option (3) \(\begin{bmatrix}2&2\\1&3\end{bmatrix}\) — Incorrect. Its product with \(\begin{bmatrix}1\\1\end{bmatrix}\) is \(\begin{bmatrix}4\\4\end{bmatrix}\).
Option (4) \(\begin{bmatrix}3&0\\1&4\end{bmatrix}\) — Incorrect. It gives the first product correctly, but its product with \(\begin{bmatrix}1\\-1\end{bmatrix}\) is \(\begin{bmatrix}3\\-3\end{bmatrix}\).