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To verify **Cayley-Hamilton Theorem** for the given matrix \( A = \begin{pmatrix} 2 & 5 \\ 1 & -3 \end{pmatrix} \), we follow these steps:

### Cayley-Hamilton Theorem:
This theorem states that every square matrix satisfies its own characteristic equation.

### Step 1: Find the characteristic equation
The characteristic equation of a 2x2 matrix \( A \) is given by:
\[
\text{det}(A - \lambda I) = 0
\]
where \( I \) is the identity matrix and \( \lambda \) is the eigenvalue.

For \( A = \begin{pmatrix} 2 & 5 \\ 1 & -3 \end{pmatrix} \), the matrix \( A - \lambda I \) is:
\[
A - \lambda I = \begin{pmatrix} 2 - \lambda & 5 \\ 1 & -3 - \lambda \end{pmatrix}
\]
The determinant of this matrix is:
\[
\text{det}(A - \lambda I) = (2 - \lambda)(-3 - \lambda) - (5)(1)
\]
Expanding this:
\[
(2 - \lambda)(-3 - \lambda) = 2(-3 - \lambda) - \lambda(-3 - \lambda)
= -6 - 2\lambda + 3\lambda + \lambda^2 = \lambda^2 + \lambda - 6
\]
Now subtract the remaining term:
\[
\lambda^2 + \lambda - 6 - 5 = \lambda^2 + \lambda - 11
\]
Thus, the characteristic equation is:
\[
\lambda^2 + \lambda - 11 = 0
\]

### Step 2: Apply the characteristic equation to matrix \( A \)
The Cayley-Hamilton theorem states that the matrix \( A \) satisfies its characteristic equation, i.e., if we substitute \( A \) into the equation, we should get the zero matrix:
\[
A^2 + A - 11I = 0
\]

First, compute \( A^2 \):
\[
A^2 = \begin{pmatrix} 2 & 5 \\ 1 & -3 \end{pmatrix} \times \begin{pmatrix} 2 & 5 \\ 1 & -3 \end{pmatrix}
= \begin{pmatrix} 2(2) + 5(1) & 2(5) + 5(-3) \\ 1(2) + (-3)(1) & 1(5) + (-3)(-3) \end{pmatrix}
= \begin{pmatrix} 4 + 5 & 10 - 15 \\ 2 - 3 & 5 + 9 \end{pmatrix}
= \begin{pmatrix} 9 & -5 \\ -1 & 14 \end{pmatrix}
\]

Now, calculate \( A + A \):
\[
A + A = 2A = \begin{pmatrix} 2(2) & 2(5) \\ 2(1) & 2(-3) \end{pmatrix} = \begin{pmatrix} 4 & 10 \\ 2 & -6 \end{pmatrix}
\]

Finally, calculate \( 11I \) (where \( I \) is the identity matrix):
\[
11I = 11 \times \begin{pmatrix} 1 & 0 \\ 0 & 1 \end{pmatrix} = \begin{pmatrix} 11 & 0 \\ 0 & 11 \end{pmatrix}
\]

Now, check the equation \( A^2 + A - 11I \):
\[
A^2 + A - 11I = \begin{pmatrix} 9 & -5 \\ -1 & 14 \end{pmatrix} + \begin{pmatrix} 4 & 10 \\ 2 & -6 \end{pmatrix} - \begin{pmatrix} 11 & 0 \\ 0 & 11 \end{pmatrix}
= \begin{pmatrix} 9 + 4 - 11 & -5 + 10 - 0 \\ -1 + 2 - 0 & 14 - 6 - 11 \end{pmatrix}
= \begin{pmatrix} 2 & 5 \\ 1 & -3 \end{pmatrix} - \begin{pmatrix} 2 & 5 \\ 1 & -3 \end{pmatrix}
= \begin{pmatrix} 0 & 0 \\ 0 & 0 \end{pmatrix}
\]

Thus, \( A^2 + A - 11I = 0 \), verifying Cayley-Hamilton Theorem for this matrix.

Would you like further details or have any questions?

### Related questions:
1. How can we derive the characteristic equation for higher-dimensional matrices?
2. How do eigenvalues relate to the Cayley-Hamilton theorem?
3. What are the practical applications of Cayley-Hamilton theorem?
4. How does the Cayley-Hamilton theorem simplify matrix exponentiation?
5. Can the Cayley-Hamilton theorem be applied to non-square matrices?

**Tip:** Verifying Cayley-Hamilton involves solving the characteristic equation and showing that the matrix satisfies it.

Verify Cayley's Hamilton Theorem for the matrix [2  5; 1  -3]

Learn how to verify Cayley-Hamilton's theorem for the matrix [2 5; 1 -3] through a detailed step-by-step solution. This solution covers key linear algebra concepts such as matrix multiplication and the characteristic equation. Suitable for university-level students.

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