Q.9. Verify Cayley-Hamilton theorem for the matrix A = [[1, 1, 1], [1, 2, 1], [3, 2, 3]]

The Cayley-Hamilton theorem states that every square matrix satisfies its own characteristic equation. To verify the theorem for the given matrix \( A \), we need to follow these steps:

Given:
\[
A = \begin{bmatrix} 
1 & 1 & 1 \\ 
1 & 2 & 1 \\ 
3 & 2 & 3 
\end{bmatrix}
\]

### Step 1: Find the Characteristic Polynomial of \( A \)
The characteristic polynomial of matrix \( A \) is given by:
\[
p(\lambda) = \det(A - \lambda I)
\]
where \( I \) is the identity matrix of the same order as \( A \) (in this case, \( 3 \times 3 \)), and \( \lambda \) is the eigenvalue variable.

First, compute \( A - \lambda I \):

\[
A - \lambda I = \begin{bmatrix}
1 & 1 & 1 \\
1 & 2 & 1 \\
3 & 2 & 3
\end{bmatrix}
- \lambda \begin{bmatrix}
1 & 0 & 0 \\
0 & 1 & 0 \\
0 & 0 & 1
\end{bmatrix}
= \begin{bmatrix}
1 - \lambda & 1 & 1 \\
1 & 2 - \lambda & 1 \\
3 & 2 & 3 - \lambda
\end{bmatrix}
\]

Now, compute the determinant of \( A - \lambda I \):

\[
\det(A - \lambda I) = \det \begin{bmatrix}
1 - \lambda & 1 & 1 \\
1 & 2 - \lambda & 1 \\
3 & 2 & 3 - \lambda
\end{bmatrix}
\]

Let's calculate this determinant. I'll perform the computation step by step.

### Step 2: Solve the Characteristic Polynomial

Let me calculate the determinant and then solve for the characteristic polynomial.The characteristic polynomial of the matrix \( A \) is:

\[
p(\lambda) = \lambda(-\lambda^2 + 6\lambda - 5) = -\lambda^3 + 6\lambda^2 - 5\lambda
\]

### Step 3: Apply the Cayley-Hamilton Theorem
According to the Cayley-Hamilton theorem, the matrix \( A \) must satisfy its characteristic equation, which is obtained by replacing \( \lambda \) in the characteristic polynomial with \( A \):

\[
p(A) = -A^3 + 6A^2 - 5A = 0
\]

Now, let's calculate \( A^3 \), \( A^2 \), and then verify the equation \( -A^3 + 6A^2 - 5A = 0 \).

I'll compute these matrices step by step.After computing the expression \( -A^3 + 6A^2 - 5A \), the result is the zero matrix, confirming that:

\[
-A^3 + 6A^2 - 5A = 0
\]

### Conclusion:
The matrix \( A \) satisfies its characteristic equation, thus verifying the Cayley-Hamilton theorem for the given matrix.

---

Would you like more details on any step or further clarification? Here are some related questions you may find useful:

1. How is the characteristic polynomial derived for larger matrices?
2. What are the applications of the Cayley-Hamilton theorem in solving linear systems?
3. Can the Cayley-Hamilton theorem be used to compute matrix functions, such as exponentials?
4. How can we compute the inverse of a matrix using the Cayley-Hamilton theorem?
5. What is the significance of eigenvalues in verifying the Cayley-Hamilton theorem?

**Tip:** The Cayley-Hamilton theorem is a powerful tool that simplifies matrix computations by reducing higher powers of matrices to linear combinations of lower powers.

Verify Cayley-Hamilton theorem for the matrix A = [[1, 1, 1], [1, 2, 1], [3, 2, 3]]

Explore how to verify the Cayley-Hamilton theorem for the matrix A = [[1, 1, 1], [1, 2, 1], [3, 2, 3]]. This detailed solution involves finding the characteristic polynomial and proving that the matrix satisfies its own characteristic equation. Suitable for undergraduate students studying linear algebra.

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