If \( A=\begin{bmatrix}a & b\\b & a\end{bmatrix} \) and \( A^2=\begin{bmatrix}\alpha & \beta\\\beta & \alpha\end{bmatrix} \), then the value(s) of \( (a-b) \) is/are
Answer & explanation
Correct answer: option 3
The correct answer is Option (3) → \( \pm\sqrt{\alpha-\beta} \)
\( \displaystyle A^2=\begin{bmatrix}a^2+b^2 & 2ab\\2ab & a^2+b^2\end{bmatrix} \)
Comparing entries,
\( \alpha=a^2+b^2,\ \beta=2ab \)
\( (a-b)^2=\alpha-\beta \)
\( a-b=\pm\sqrt{\alpha-\beta} \)
Both signs are needed, since a − b can be positive or negative.
Why the other options are incorrect
Option (1) \( \sqrt{\alpha-\beta} \) — Incorrect. a − b can also be negative (for example a = 1, b = 2), so both signs are needed.
Option (2) \( -\sqrt{\alpha+\beta} \) — Incorrect. \( \alpha+\beta \) equals \( (a+b)^2 \), not \( (a-b)^2 \).
Option (4) \( \sqrt{\alpha+\beta} \) — Incorrect. \( \alpha+\beta \) equals \( (a+b)^2 \), not \( (a-b)^2 \).