The sum of a matrix \( A \) and its transpose is \( \begin{bmatrix}6 & -1\\-1 & 4\end{bmatrix} \).
Which of the following statements is true?
Answer & explanation
Correct answer: option 3
The correct answer is Option (3) → \( A=\begin{bmatrix}3 & b\\-1-b & 2\end{bmatrix} \), \( b \) is any real number.
Let \( A=\begin{bmatrix}p & b\\c & q\end{bmatrix} \). Then
\( \displaystyle A+A^T=\begin{bmatrix}2p & b+c\\b+c & 2q\end{bmatrix} \)
\( \displaystyle =\begin{bmatrix}6 & -1\\-1 & 4\end{bmatrix} \)
So 2p = 6, 2q = 4 and b + c = −1, i.e. p = 3, q = 2 and c = −1 − b, with b free.
\( \displaystyle A=\begin{bmatrix}3 & b\\-1-b & 2\end{bmatrix} \)
Why the other options are incorrect
Option (1) \( A=\begin{bmatrix}3 & 0\\-1 & -2\end{bmatrix} \) — Incorrect. The diagonal entry is −2 instead of 2, so \( A+A^T \) has −4, not 4, in the last place.
Option (2) \( A=\begin{bmatrix}2 & b\\-1-b & 3\end{bmatrix} \), \( b \) is any real number. — Incorrect. The diagonal entries are the wrong way round: \( A+A^T \) would have 4 and 6 on its diagonal, not 6 and 4.
Option (4) \( A=\begin{bmatrix}3 & 5\\-6 & -2\end{bmatrix} \) — Incorrect. The diagonal entry is −2 instead of 2, so \( A+A^T \) has −4, not 4, in the last place.