Question

If the matrix \[ A = \begin{pmatrix} 2 & 0 & -1 \\ 5 & 1 & 0 \\ 0 & 1 & 3 \end{pmatrix} \] satisfies the matrix equation:

• A. \( A^3 - 6A^2 + 11A - I = 0 \)
• B. \( A^3 + 6A^2 - 11A + I = 0 \)
• C. \( A^3 + 6A^2 + 11A + I = 0 \)
• D. \( A^3 - 6A^2 - 11A - I = 0 \)

Hint:

Using the formula \( \lambda^3 - (\text{tr}(A))\lambda^2 + (\text{sum of diagonal cofactors})\lambda - \det(A) = 0 \) is a much faster way to find the characteristic equation for a 3x3 matrix than expanding \( \det(A - \lambda I) \) directly, reducing the chance of algebraic errors.

Answer 1

The Correct Option is A

Step 1: Understanding the Concept:
By the Cayley–Hamilton theorem, every square matrix satisfies its own characteristic equation. Thus, to find the matrix equation satisfied by \(A\), we must determine its characteristic polynomial.
Step 2: Key Formula or Approach:
The characteristic polynomial of a 3×3 matrix \(A\) is: \[ \lambda^3 - (\text{tr}(A))\lambda^2 + (\text{sum of principal minors of order 2})\lambda - \det(A) = 0 \] where: - \(\text{tr}(A)\) = trace of \(A\) = sum of diagonal elements, - principal minors of order 2 are cofactors of the diagonal elements, - \(\det(A)\) is the determinant of \(A\).
Step 3: Detailed Explanation:
Let \[ A = \begin{pmatrix} 2 & 0 & -1 \\ 5 & 1 & 0 \\ 0 & 1 & 3 \end{pmatrix} \] 1. Trace of A:
\[ \text{tr}(A) = 2 + 1 + 3 = 6 \] 2. Sum of cofactors of diagonal elements:
- \(M_{11} = \det\begin{pmatrix} 1 & 0 \\ 1 & 3 \end{pmatrix} = 3\) - \(M_{22} = \det\begin{pmatrix} 2 & -1 \\ 0 & 3 \end{pmatrix} = 6\) - \(M_{33} = \det\begin{pmatrix} 2 & 0 \\ 5 & 1 \end{pmatrix} = 2\) Sum = \(3 + 6 + 2 = 11\) 3. Determinant of A:
\[ \det(A) = 2 \begin{vmatrix} 1 & 0 \\ 1 & 3 \end{vmatrix} - 0 \begin{vmatrix} 5 & 0 \\ 0 & 3 \end{vmatrix} + (-1) \begin{vmatrix} 5 & 1 \\ 0 & 1 \end{vmatrix} \] \[ = 2(3) - 0 + (-1)(5) = 6 - 5 = 1 \] 4. Characteristic Equation:
\[ \lambda^3 - 6\lambda^2 + 11\lambda - 1 = 0 \] 5. Apply Cayley–Hamilton Theorem:
Replacing \(\lambda\) with \(A\): \[ A^3 - 6A^2 + 11A - I = 0 \] Step 4: Final Answer:
The matrix \(A\) satisfies the equation: \[ \boxed{A^3 - 6A^2 + 11A - I = 0} \]