Step 1: Conceptual Foundation:
The Cayley–Hamilton theorem states that every square matrix is a root of its characteristic polynomial. Therefore, to find the equation a matrix \(A\) satisfies, we first compute its characteristic polynomial.
Step 2: Defining Characteristic Polynomial:
For a 3×3 matrix \(A\), the characteristic polynomial is given by: \[\lambda^3 - (\text{tr}(A))\lambda^2 + (\text{sum of principal minors of order 2})\lambda - \det(A) = 0\] where: - \(\text{tr}(A)\) is the trace of \(A\), the sum of its diagonal elements. - The principal minors of order 2 are the cofactors of the diagonal elements. - \(\det(A)\) is the determinant of \(A\).
Step 3: Application to a Specific Matrix:
Consider the matrix\[A = \begin{pmatrix} 2 & 0 & -1 \\ 5 & 1 & 0 \\ 0 & 1 & 3 \end{pmatrix}\]1. Calculation of Trace:
\[\text{tr}(A) = 2 + 1 + 3 = 6\]2. Calculation of Sum of Cofactors of Diagonal Elements:
- Cofactor of \(a_{11}\): \(M_{11} = \det\begin{pmatrix} 1 & 0 \\ 1 & 3 \end{pmatrix} = 3\) - Cofactor of \(a_{22}\): \(M_{22} = \det\begin{pmatrix} 2 & -1 \\ 0 & 3 \end{pmatrix} = 6\) - Cofactor of \(a_{33}\): \(M_{33} = \det\begin{pmatrix} 2 & 0 \\ 5 & 1 \end{pmatrix} = 2\) Sum of cofactors = \(3 + 6 + 2 = 11\)3. Calculation of Determinant:
\[\det(A) = 2 \begin{vmatrix} 1 & 0 \\ 1 & 3 \end{vmatrix} - 0 \begin{vmatrix} 5 & 0 \\ 0 & 3 \end{vmatrix} + (-1) \begin{vmatrix} 5 & 1 \\ 0 & 1 \end{vmatrix}\] \[= 2(3) - 0 + (-1)(5) = 6 - 5 = 1\]4. Formulation of Characteristic Equation:
Substituting the calculated values into the characteristic polynomial formula: \[\lambda^3 - 6\lambda^2 + 11\lambda - 1 = 0\]5. Application of Cayley–Hamilton Theorem:
According to the theorem, the matrix \(A\) satisfies its characteristic equation. Replacing \(\lambda\) with \(A\) and the constant term \(1\) with the identity matrix \(I\): \[A^3 - 6A^2 + 11A - I = 0\]Step 4: Conclusion:
The matrix \(A\) satisfies the following equation: \[\boxed{A^3 - 6A^2 + 11A - I = 0}\]