Question

Let $A$ be a $3 \times 3$ matrix such that $A^2 = I$. If $\det(A) = -1$ and the sum of eigenvalues is $1$, find the set of eigenvalues of $A$.

• A. \( 1, 1, 1 \)
• B. \( 1, 1, -1 \)
• C. \( -1, -1, 1 \)
• D. \( -1, -1, -1 \)

Hint:

For matrices satisfying \(A^2=I\), eigenvalues are always restricted to \(\pm1\). Use determinant and trace conditions together to uniquely identify them.

Answer 1

The Correct Option is B

Concept: An involutory matrix satisfies \(A^2 = I\). Hence, every eigenvalue \(\lambda\) of \(A\) satisfies: \[ \lambda^2 = 1 \] Therefore, the eigenvalues can only be \(1\) or \(-1\).

Step 1: Use the determinant condition.
The determinant of a matrix equals the product of its eigenvalues: \[ \lambda_1\lambda_2\lambda_3 = -1 \] Thus, an odd number of eigenvalues must be negative.

Step 2: Use the trace condition.
The sum of eigenvalues is: \[ \lambda_1+\lambda_2+\lambda_3 = 1 \]

Step 3: Check possible combinations.
Possible eigenvalue sets are: \[ (1,1,-1), \quad (-1,-1,-1) \] For \((1,1,-1)\): \[ 1+1-1 = 1 \] and \[ 1\cdot1\cdot(-1)=-1 \] Both conditions are satisfied.