Question

If $A = \begin{pmatrix} 0 & 1 \\ -1 & 0 \end{pmatrix}$ and $(\alpha I + \beta A)^2 = A$, where $I$ is $2 \times 2$ unit matrix, then $\alpha^2 - \beta^2 =$:

• A. $2$
• B. $-2$
• C. $-1$
• D. $1$
• E. $0$

Hint:

If $A^2 = -I$, treat it like $i^2 = -1$ for simplification.

Answer 1

Answer: (E)

Concept:
• $A^2 = -I$ for given matrix
• Expand $(\alpha I + \beta A)^2$

Step 1: Compute $A^2$
\[ A^2 = \begin{pmatrix} 0 & 1 -1 & 0 \end{pmatrix}^2 = \begin{pmatrix} -1 & 0 0 & -1 \end{pmatrix} = -I \]

Step 2: Expand expression
\[ (\alpha I + \beta A)^2 = \alpha^2 I + 2\alpha\beta A + \beta^2 A^2 \] \[ = \alpha^2 I + 2\alpha\beta A - \beta^2 I \] \[ = (\alpha^2 - \beta^2)I + 2\alpha\beta A \]

Step 3: Compare with RHS
\[ (\alpha^2 - \beta^2)I + 2\alpha\beta A = A \] Equating coefficients: \[ \alpha^2 - \beta^2 = 0, \quad 2\alpha\beta = 1 \] Final Conclusion:
\[ \alpha^2 - \beta^2 = 0 \]