The Cayley-Hamilton theorem states that every square matrix satisfies its own characteristic equation. For a $3 \times 3$ matrix $A$, this means if you compute the characteristic polynomial of $A$, say $p(\lambda) = \det(\lambda I - A)$, and substitute $A$ itself into the polynomial, then the result will be the zero matrix.

Example:

Consider the following $3 \times 3$ matrix $A$:

2 & 1 & 0 \\ 0 & 3 & 1 \\ 1 & 0 & 2 \end{pmatrix}$$ **Step 1: Find the characteristic polynomial** The characteristic polynomial of $$A$$ is given by: $$p(\lambda) = \det(\lambda I - A)$$ Where $$I$$ is the identity matrix. So, we calculate $$\lambda I - A$$: $$\lambda I - A = \begin{pmatrix} \lambda - 2 & -1 & 0 \\ 0 & \lambda - 3 & -1 \\ -1 & 0 & \lambda - 2 \end{pmatrix}$$ Now compute the determinant: $$p(\lambda) = \det\begin{pmatrix} \lambda - 2 & -1 & 0 \\ 0 & \lambda - 3 & -1 \\ -1 & 0 & \lambda - 2 \end{pmatrix}$$ Expanding this determinant yields: $$p(\lambda) = (\lambda - 2)\left[(\lambda - 3)(\lambda - 2)\right] - (-1)(-1)(\lambda - 2)$$ $$= (\lambda - 2)\left[(\lambda - 3)(\lambda - 2) - 1\right]$$ $$= (\lambda - 2)\left[\lambda^2 - 5\lambda + 6 - 1\right]$$ $$= (\lambda - 2)(\lambda^2 - 5\lambda + 5)$$ So, the characteristic polynomial is: $$p(\lambda) = (\lambda - 2)(\lambda^2 - 5\lambda + 5)$$ Expanding it further: $$p(\lambda) = \lambda^3 - 7\lambda^2 + 15\lambda - 10$$ **Step 2: Apply the Cayley-Hamilton theorem** The Cayley-Hamilton theorem says that the matrix $$A$$ satisfies its own characteristic polynomial. So, substituting $$A$$ into the polynomial $$p(\lambda)$$, we should get the zero matrix: $$p(A) = A^3 - 7A^2 + 15A - 10I = 0$$ You can now verify this by computing $$A^2$$ and $$A^3$$, and substituting them into the above equation, showing that indeed the result will be the zero matrix. Would you like me to go further with the calculations or provide more details? --- Here are 5 questions to deepen your understanding: 1. How do you find the characteristic polynomial of a matrix in general? 2. Why does the Cayley-Hamilton theorem hold for any square matrix? 3. Can the Cayley-Hamilton theorem be applied to non-invertible matrices? 4. What is the significance of the identity matrix in the Cayley-Hamilton theorem? 5. How can the Cayley-Hamilton theorem be used to compute matrix powers? **Tip:** The Cayley-Hamilton theorem can simplify matrix computations, especially when dealing with high powers of a matrix, by expressing powers of the matrix in terms of lower powers.