Question

If \( A = \begin{pmatrix} 1 & 5 \\ 0 & 2 \end{pmatrix} \), then which of the following equations is satisfied by matrix \( A \)?

• A. \( A^2 - 2A + 2I = 0 \)
• B. \( A^2 - 3A + 2I = 0 \)
• C. \( A^2 - 5A + 2I = 0 \)
• D. \( 2A^2 - A + I = 0 \)
• E. \( A^2 + 3A + 2I = 0 \)

Hint:

For a \( 2 \times 2 \) matrix, don't waste time calculating \( A^2 \) manually. Use \( A^2 - (\text{sum of diagonals})A + (\text{determinant})I = 0 \) to find the result in seconds.

Answer 1

The Correct Option is B

Concept: According to the Cayley-Hamilton Theorem, every square matrix satisfies its own characteristic equation. The characteristic equation is given by \( \det(A - \lambda I) = 0 \). For a \( 2 \times 2 \) matrix \( A \), this equation is \( \lambda^2 - \text{tr}(A)\lambda + \det(A) = 0 \).

Step 1: Finding the trace and determinant of matrix \( A \).
The trace (\( \text{tr} \)) is the sum of the diagonal elements: \[ \text{tr}(A) = 1 + 2 = 3 \] The determinant (\( \det \)) is calculated as: \[ \det(A) = (1 \times 2) - (5 \times 0) = 2 \]

Step 2: Forming the characteristic equation and applying the theorem.
Substituting the values into the formula \( \lambda^2 - \text{tr}(A)\lambda + \det(A) = 0 \): \[ \lambda^2 - 3\lambda + 2 = 0 \] By Cayley-Hamilton Theorem, replacing \( \lambda \) with \( A \) and the constant term with \( 2I \): \[ A^2 - 3A + 2I = 0 \]