Question

6. Find the Eigen values Eigen vector of the matrix
311
151
113​

Answer 1

Step-by-step explanation:

this is the way to solve this

Answer 2

The eigenvalues and eigenvectors of the matrix are:

Eigenvalues: λ1 = 412, λ2 = 600

Eigenvectors: v1 = [0 1], v2 = [1 -289/151]

To find the eigenvalues and eigenvectors of a matrix, we can use the following method:

Start by finding the characteristic equation of the matrix by det(A - λI) = 0.

For the matrix

[311 151]

[113 311]

The characteristic equation is:

| 311 - λ 151 |

$| 113 311 - λ | = (311-λ)^2 - 151*113 = λ^2 - 622λ + 100844 = 0$

Solve the characteristic equation to find the eigenvalues, which are the roots of the equation.

$λ^2 - 622λ + 100844 = 0$

(λ-412)(λ-600) = 0

So the eigenvalues are λ1 = 412 and λ2 = 600

Once we have the eigenvalues, we can find the corresponding eigenvectors by solving the equation (A - λI)v = 0

For λ1 = 412, we have:

$(A - λ1I)v = [311-412 151] [v1] = [ -101 151] [v1] = 0$

So v1 = [0 1] is an eigenvector for λ1 = 412

For λ2 = 600, we have:

$(A - λ2I)v = [311-600 151] [v2] = [-289 151] [v2] = 0$

So v2 = [1 -289/151] is an eigenvector for λ2 = 600

So the eigenvalues and eigenvectors of the matrix are:

Eigenvalues: λ1 = 412, λ2 = 600

Eigenvectors: v1 = [0 1], v2 = [1 -289/151]

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