Question

Let the matrix \( A = \begin{bmatrix} 1 & 4 \\ 2 & -1 \end{bmatrix} \).
Statement-I: \( A^2 = 9I \)
Statement-II: The eigen values of \( A \) are \( -3 \) and \( 3 \)

• A. Both statement-I and statement-II are true
• B. Both statement-I and statement-II are false
• C. Statement-I is true, but statement-II is false
• D. Statement-I is false, but statement-II is true

Hint:

By the Cayley-Hamilton Theorem, if a matrix satisfies \( A^2 = 9I \), its matrix characteristic equation is \( \lambda^2 - 9 = 0 \). Solving this directly yields eigenvalues \( \lambda = \pm 3 \) instantly without needing separate determinant expansion calculations!

Answer 1

The Correct Option is A

Step 1: Find the trace and determinant of \( A \).
For \( A = \begin{bmatrix} 1 & 4 \\ 2 & -1 \end{bmatrix} \), the trace is \( \text{Tr}(A) = 1 + (-1) = 0 \) and the determinant is \( \det(A) = (1)(-1) - (4)(2) = -1 - 8 = -9 \).

Step 2: Write the characteristic equation.
For a \( 2 \times 2 \) matrix, the characteristic equation is \( \lambda^2 - \text{Tr}(A)\lambda + \det(A) = 0 \). Plugging in our values: \[ \lambda^2 - 0\cdot\lambda + (-9) = 0 \quad \Rightarrow \quad \lambda^2 = 9 \quad \Rightarrow \quad \lambda = \pm 3 \] This already settles Statement-II: the eigenvalues are \( 3 \) and \( -3 \), so it is true.

Step 3: Use Cayley-Hamilton to get \( A^2 \) without multiplying matrices.
By the Cayley-Hamilton theorem, \( A \) itself satisfies its own characteristic equation: \[ A^2 - 0\cdot A - 9I = 0 \quad \Rightarrow \quad A^2 = 9I \] This confirms Statement-I is also true, using the same trace and determinant equation instead of multiplying the matrix out by hand.

Since both statements check out from the same characteristic equation, the answer is option (A).