Question

If $A=\begin{bmatrix} a & b \\ b & a \end{bmatrix}$ and $A^{2}=\begin{bmatrix} \alpha & \beta \\ \beta & a \end{bmatrix}$, then $(a-b)$ is:

• A. $\sqrt{\alpha-\beta}$
• B. $-\sqrt{\alpha+\beta}$
• C. $\pm\sqrt{\alpha-\beta}$
• D. $\sqrt{\alpha+\beta}$

Hint:

Always compute $A^2$ explicitly before comparing matrices.

Answer 1

The Correct Option is C

Concept: Compute $A^2$ using matrix multiplication.

Step 1: {Multiply matrices.}
\[ A^2= \begin{bmatrix} a^2+b^2 & ab+ba \\ ab+ba & a^2+b^2 \end{bmatrix} = \begin{bmatrix} a^2+b^2 & 2ab \\ 2ab & a^2+b^2 \end{bmatrix} \]

Step 2: {Compare with given matrix.}
\[ \alpha=a^2+b^2,\quad \beta=2ab \]

Step 3: {Compute $(a-b)^2$.}
\[ (a-b)^2=a^2+b^2-2ab \]

Step 4: {Substitute values.}
\[ (a-b)^2=\alpha-\beta \]

Step 5: {Take square root.}
\[ a-b=\pm\sqrt{\alpha-\beta} \]