If A is a square matrix of order 2 such that \(A + A' = \begin{bmatrix}4&3\\3&6\end{bmatrix}\) and \(A - A' = \begin{bmatrix}0&-1\\1&0\end{bmatrix}\), then A is equal to
Answer & explanation
Correct answer: option 3
The correct answer is Option (3) → \(\begin{bmatrix}2&1\\2&3\end{bmatrix}\)
Adding the two equations removes A':
\( \displaystyle 2A = \begin{bmatrix}4 + 0&3 + (-1)\\3 + 1&6 + 0\end{bmatrix} = \begin{bmatrix}4&2\\4&6\end{bmatrix} \)
\( \displaystyle A = \frac{1}{2}\begin{bmatrix}4&2\\4&6\end{bmatrix} = \begin{bmatrix}2&1\\2&3\end{bmatrix} \)
Check: \(A' = \begin{bmatrix}2&2\\1&3\end{bmatrix}\), and \(A + A'\) and \(A - A'\) are the given matrices.
Why the other options are incorrect
Option (1) \(\begin{bmatrix}2&2\\1&3\end{bmatrix}\) — Incorrect. This is A', obtained by subtracting the equations the wrong way; for it \(A - A' = \begin{bmatrix}0&1\\-1&0\end{bmatrix}\).
Option (2) \(\begin{bmatrix}4&2\\4&6\end{bmatrix}\) — Incorrect. This is 2A; it was not halved.
Option (4) \(\begin{bmatrix}2&\frac{3}{2}\\\frac{3}{2}&3\end{bmatrix}\) — Incorrect. This is only \(\displaystyle\frac{1}{2}(A + A')\); for it \(A - A' = O\), not the given matrix.