Example 4: Find the eigen values of A and hence find n A ( n is a positive integer), 
given that 
1 2
4 3
A
 
    
and also find 3 A .
Solution: The characteristic equation of A is A I    0
  
2
1 2 i e S S . ., 0.      1 2 S S    4, 5. The characteristic equation of A is 
2
     4 5 0. By Cayley Hamilton theorem  
2 A A I     4 5 0. 1
The eigen value of A are   1, 5. When
n
 is divided by
2
    4 5, let the quotient be Q  and the remainder be a b   .
Then        
2
4 5 2 n
           Q a b
Put  1in (2),  1 3   
n
     a b
Put   5in (2), 5 5 4    
n
a b   
Solving (3) and (4) we get, 
 
5 1 5 5 1     &
6 6
n n n n
a b
   


To solve this problem, we need to find the eigenvalues of the matrix \( A \) and then use them to find \( A^n \) for any positive integer \( n \), and specifically \( A^3 \).

Given the matrix:

\[ A = \begin{pmatrix} 1 & 2 \\ 4 & 3 \end{pmatrix} \]

### Finding Eigenvalues of \( A \)

The eigenvalues of a matrix are found by solving the characteristic equation \( \det(A - \lambda I) = 0 \).

1. **Form the matrix \( A - \lambda I \)**:

\[ A - \lambda I = \begin{pmatrix} 1 & 2 \\ 4 & 3 \end{pmatrix} - \lambda \begin{pmatrix} 1 & 0 \\ 0 & 1 \end{pmatrix} = \begin{pmatrix} 1 - \lambda & 2 \\ 4 & 3 - \lambda \end{pmatrix} \]

2. **Find the determinant**:

\[ \det(A - \lambda I) = (1 - \lambda)(3 - \lambda) - (4 \cdot 2) = (1 - \lambda)(3 - \lambda) - 8 \]

\[ = \lambda^2 - 4\lambda - 5 \]

3. **Solve the characteristic equation**:

\[ \lambda^2 - 4\lambda - 5 = 0 \]

Using the quadratic formula \( \lambda = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \):

\[ \lambda = \frac{4 \pm \sqrt{16 + 20}}{2} = \frac{4 \pm \sqrt{36}}{2} = \frac{4 \pm 6}{2} \]

So the eigenvalues are:

\[ \lambda_1 = 5 \quad \text{and} \quad \lambda_2 = -1 \]

### Finding \( A^n \) using Eigenvalues

By the Cayley-Hamilton theorem, any square matrix satisfies its own characteristic equation. The characteristic equation of \( A \) is:

\[ A^2 - 4A - 5I = 0 \]

So,

\[ A^2 = 4A + 5I \]

For \( A^n \):

\[ A^3 = A \cdot A^2 = A \cdot (4A + 5I) = 4A^2 + 5A \]

Using \( A^2 = 4A + 5I \):

\[ A^3 = 4(4A + 5I) + 5A = 16A + 20I + 5A = 21A + 20I \]

### Summary

1. The eigenvalues of \( A \) are \( \lambda_1 = 5 \) and \( \lambda_2 = -1 \).
2. Using the Cayley-Hamilton theorem, \( A^3 \) is found as:

\[ A^3 = 21A + 20I \]

So, if we need to compute any higher power \( A^n \), we can recursively apply the relation derived from the Cayley-Hamilton theorem.

Would you like a more detailed explanation or do you have any questions about this solution?

### Further Questions
1. How can we find the eigenvectors of \( A \)?
2. What is the significance of the eigenvalues in this context?
3. How can we verify the eigenvalues by solving \( A \mathbf{v} = \lambda \mathbf{v} \)?
4. What are the properties of matrices that allow the Cayley-Hamilton theorem to work?
5. How can we use the diagonalization of \( A \) to find \( A^n \)?
6. Can we generalize this approach to non-square matrices?
7. What is the role of the identity matrix \( I \) in matrix computations?
8. How does matrix multiplication differ from scalar multiplication in this context?

### Tip
To solve higher powers of matrices efficiently, consider diagonalization if the matrix is diagonalizable, which simplifies the computation of matrix powers.

Learn how to find eigenvalues and compute matrix powers using matrix A = [[1, 2], [4, 3]]. Explore the characteristic equation, Cayley-Hamilton theorem, and detailed steps to find A^n for any positive integer n. Suitable for advanced high school students and beyond.

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