Question

If
\[ A= \begin{bmatrix} \alpha^2 & 5 \\ 5 & -\alpha \end{bmatrix} \] and
\[ \det(A^{10})=1024, \] then \(\alpha=\)

• A. \(-2\)
• B. \(-1\)
• C. \(-3\)
• D. \(0\)

Hint:

Remember the important property: \(\det(A^n)=(\det A)^n\). It simplifies many determinant-based matrix problems.

Answer 1

The Correct Option is C

Step 1: Use the determinant property.
For any square matrix \(A\),
\[ \det(A^n)=(\det A)^n \]
Therefore,
\[ \det(A^{10})=(\det A)^{10} \]
Given that
\[ \det(A^{10})=1024 \] we get
\[ (\det A)^{10}=1024 \]
Since
\[ 1024=2^{10}, \] therefore,
\[ \det A=\pm 2 \]

Step 2: Find \(\det A\).
Given matrix is
\[ A= \begin{bmatrix} \alpha^2 & 5 \\ 5 & -\alpha \end{bmatrix} \]
Its determinant is
\[ \det A= \alpha^2(-\alpha)-5\cdot 5 \]
\[ \det A=-\alpha^3-25 \]

Step 3: Equate determinant values.
We have
\[ -\alpha^3-25=\pm 2 \]
Case 1:
\[ -\alpha^3-25=2 \] \[ -\alpha^3=27 \] \[ \alpha^3=-27 \] \[ \alpha=-3 \]
Case 2:
\[ -\alpha^3-25=-2 \] \[ -\alpha^3=23 \] \[ \alpha^3=-23 \] which is not among the given options.
Hence,
\[ \alpha=-3 \]

Step 4: Verification.
Substitute \(\alpha=-3\):
\[ \det A=-(-3)^3-25 \] \[ =27-25 \] \[ =2 \]
Therefore,
\[ \det(A^{10})=2^{10}=1024 \] which satisfies the given condition.

Step 5: Final conclusion.
Hence,
\[ \boxed{-3} \]