Verify Cayley–Hamilton theorem
A= |1 2 3|
|2 -1 1|
|3 1 1|
The Cayley–Hamilton theorem states that every square matrix satisfies its own characteristic equation. In other words, if A is an n × n matrix and p(λ) = det(A - λI) is the characteristic polynomial of A, then by substituting A for λ in the polynomial, we should get the zero matrix:
p(A) = 0
where 0 is the zero matrix of the same size as A.
Let's apply the Cayley–Hamilton theorem to the matrix A that you provided:
```
A = | 1 2 3 |
| 2 -1 1 |
| 3 1 1 |
```
Step 1: Calculate the characteristic polynomial p(λ):
```
p(λ) = det(A - λI)
```
Substitute λ for the identity matrix I:
```
A - λI = | 1-λ 2 3 |
| 2 -1-λ 1 |
| 3 1 1-λ |
```
Calculate the determinant:
```
p(λ) = (1-λ)[(-1-λ)(1-λ) - (1)(3)] - (2)[(2)(1-λ) - (3)(1)] + (3)[(2)(1) - (-1-λ)(3)]
p(λ) = (1-λ)[λ² + 2λ - 4] - (2)[-2λ + 2] + (3)[2λ + 3]
p(λ) = λ³ - 3λ² - 3λ + 7
```
Step 2: Substitute A into the characteristic polynomial:
```
p(A) = A³ - 3A² - 3A + 7I
```
Step 3: Substitute the matrix A into the polynomial:
```
p(A) = | 1 2 3 |³ - 3| 1 2 3 |² - 3| 1 2 3 | + 7| 1 0 0 |
| 2 -1 1 | | 2 -1 1 | | 2 -1 1 | | 0 1 0 |
| 3 1 1 | | 3 1 1 | | 3 1 1 | | 0 0 1 |
Calculating each term:
A³ = | 13 -4 2 |
| 2 -1 16 |
| 14 -2 9 |
3A² = 3 * | 1 2 3 |² = | 14 -2 10 |
| 2 -1 1 | | 4 -7 5 |
| 3 1 1 | | 23 1 11 |
3A = 3 * | 1 2 3 | = | 3 6 9 |
| 2 -1 1 | | 6 -3 3 |
| 3 1 1 | | 9 3 3 |
7I = | 7 0 0 |
| 0 7 0 |
| 0 0 7 |
Now calculate p(A) by subtracting the matrices:
p(A) = A³ - 3A² - 3A + 7I = | 13 -4 2 | - | 14 -2 10 | - | 3 6 9 | + | 7 0 0 |
| 2 -1 16 | | 4 -7 5 | | 6 -3 3 | | 0 7 0 |
| 14 -2 9 | | 23 1 11 | | 9 3 3 | | 0 0 7 |
p(A) = | 13 -4 2 | - | 14 -2 10 | - | 3 6 9 | + | 7 0 0 |
| 2 -1 16 | | 4 -7 5 | | 6 -3 3 | | 0 7 0 |
| 14 -2 9 | | 23 1 11 | | 9 3 3 | | 0 0 7 |
Performing matrix operations and simplifying:
p(A) = | -3 -2 -8 |
| -2 6 2 |
| 8 14 -4 |
This is not the zero matrix, which means p(A) ≠ 0.
Since p(A) ≠ 0, the Cayley–Hamilton theorem does not hold for the given matrix A. In other words, the matrix A does not satisfy its own characteristic equation.
Determine the value of λsuch that the following systems of linear equations
x+y+z=1
2x+3y+λ=3
x+λy+3z=2
To determine the value of λ that makes the system of linear equations consistent, we need to solve for the variables x, y, and z. We can do this by using techniques such as Gaussian elimination or matrix operations. Let's set up the augmented matrix for the system:
```
[ 1 1 1 | 1 ]
[ 2 3 λ | 3 ]
[ 1 λ 3 | 2 ]
```
Now, let's perform row operations to bring the matrix into row-echelon form. The goal is to get zeros below the diagonal in the first two columns. We'll start with row 2 and row 3:
1. Subtract Row 1 from Row 2: Row 2 = Row 2 - Row 1
2. Subtract Row 1 from Row 3: Row 3 = Row 3 - Row 1
The matrix after these operations becomes:
```
[ 1 1 1 | 1 ]
[ 1 2 λ-1 | 2 ]
[ 0 λ-1 2 | 1 ]
```
Now, let's consider the case where λ ≠ 1. In this case, we can proceed with more row operations:
3. Divide Row 2 by (λ - 1): Row 2 = Row 2 / (λ - 1)
4. Subtract λ-1 times Row 2 from Row 3: Row 3 = Row 3 - (λ - 1) * Row 2
The matrix now becomes:
```
[ 1 1 1 | 1 ]
[ 0 1 1 | 2 / (λ - 1) ]
[ 0 0 2 - λ | 1 - (λ - 1) * (2 / (λ - 1)) ]
```
For the system to have a unique solution (i.e., consistent), the third row should not have a free variable. This means that the determinant of the coefficient matrix must be nonzero. In other words, (2 - λ) ≠ 0, which leads to λ ≠ 2.
Therefore, if λ ≠ 1 and λ ≠ 2, the system of linear equations will have a unique solution, making it consistent.
In summary, the values of λ that make the system of linear equations consistent are all real numbers except for λ = 1 and λ = 2.
