To find the eigenvalues and corresponding eigenvectors of a matrix, we need to solve the characteristic equation and then find the eigenvectors associated with each eigenvalue.
Given matrix A:
```
A = |4 1|
|3 2|
```
The characteristic equation is given by:
```
det(A - λI) = 0
```
Where λ (lambda) is the eigenvalue we're trying to find, and I is the identity matrix of the same size as A.
Let's proceed to calculate the eigenvalues:
```
A - λI = |4-λ 1|
| 3 2-λ|
det(A - λI) = (4-λ)(2-λ) - 3*1
det(A - λI) = λ^2 - 6λ + 5
```
Now we set the determinant equal to zero and solve for λ:
```
λ^2 - 6λ + 5 = 0
Factoring the quadratic equation:
(λ - 5)(λ - 1) = 0
So, the eigenvalues are λ = 5 and λ = 1.
```
Now let's find the eigenvectors associated with each eigenvalue.
For λ = 5:
```
A - 5I = |-1 1|
| 3 -3|
Solving the system (A - 5I)v = 0 gives us:
-1x + y = 0
3x - 3y = 0
From the first equation, we get x = y.
From the second equation, we get x = y.
So, an eigenvector corresponding to λ = 5 is [x, y] = [1, 1].
```
For λ = 1:
```
A - I = |3 1|
|3 1|
Solving the system (A - I)v = 0 gives us:
3x + y = 0
3x + y = 0
From the first equation, we get y = -3x.
From the second equation, we get y = -3x.
So, an eigenvector corresponding to λ = 1 is [x, y] = [1, -3].
```
To summarize:
Eigenvalue λ = 5 with eigenvector [1, 1].
Eigenvalue λ = 1 with eigenvector [1, -3].
Find the eigen value and corresponding eigen vector of it the matrix
A=|1 0 - 2|
|0 0 0|
|-2 0 4|
To find the eigenvalues and corresponding eigenvectors of the given matrix A, we'll follow a similar procedure as before. We need to solve the characteristic equation and find the eigenvectors associated with each eigenvalue.
Given matrix A:
```
A = | 1 0 -2 |
| 0 0 0 |
|-2 0 4 |
```
The characteristic equation is given by:
```
det(A - λI) = 0
```
Where λ (lambda) is the eigenvalue we're trying to find, and I is the identity matrix of the same size as A.
Let's proceed to calculate the eigenvalues:
```
A - λI = |1-λ 0 -2 |
| 0 -λ 0 |
|-2 0 4-λ |
det(A - λI) = (1-λ)(-λ)(4-λ) - 0 - (-2*0)
det(A - λI) = -λ(λ^2 - 5λ + 4)
```
Now we set the determinant equal to zero and solve for λ:
```
-λ(λ^2 - 5λ + 4) = 0
So, the eigenvalues are λ = 0, λ = 1, and λ = 4.
```
Now let's find the eigenvectors associated with each eigenvalue.
For λ = 0:
```
A - 0I = | 1 0 -2 |
| 0 0 0 |
|-2 0 4 |
Solving the system (A - 0I)v = 0 gives us:
x - 2z = 0
0 = 0
-2x + 4z = 0
From the first equation, we get x = 2z.
From the third equation, we get x = 2z.
So, an eigenvector corresponding to λ = 0 is [x, y, z] = [2z, y, z]. Let's denote this as [2z, y, z] = z * [2, 0, 1].
```
For λ = 1:
```
A - I = | 0 0 -2 |
| 0 -1 0 |
|-2 0 3 |
Solving the system (A - I)v = 0 gives us:
-2z = 0
-y = 0
-2z + 3z = 0
From the first equation, we get z = 0.
From the second equation, we get y = 0.
From the third equation, we get z = 0.
So, an eigenvector corresponding to λ = 1 is [x, y, z] = [0, 0, 0], which is not a valid eigenvector.
```
For λ = 4:
```
A - 4I = |-3 0 -2 |
| 0 -4 0 |
|-2 0 0 |
Solving the system (A - 4I)v = 0 gives us:
-3x - 2z = 0
-4y = 0
-2x = 0
From the first equation, we get x = -2z.
From the second equation, we get y = 0.
From the third equation, we get x = 0.
So, an eigenvector corresponding to λ = 4 is [x, y, z] = [-2z, 0, z], which can be denoted as z * [-2, 0, 1].
```
To summarize:
Eigenvalue λ = 0 with eigenvector [2, 0, 1].
Eigenvalue λ = 4 with eigenvector [-2, 0, 1].
Eigenvalue λ = 1 does not have a valid eigenvector.
