Question

find eigenvalue and corresponding eigenvectors of the matrix A=[2 0 0][0 3 4][0 -4 -9]​

Answer 1

Answer:

:

Eigenvector (X) that corresponding to Eigenvalue (λ) satisfies the equation AX = λX.

Determining eigenvalues first by characteristic Equation

Given matrix A = [2−2 −16]

The characteristic equation for the given matrix is

|A - λI| = 0

λ λ

M=|2−λ−2−16−λ|=0

⇒ (2 - λ) × (6 - λ) - (-2)(-1) = 0

⇒ 12 - 2 λ - 6 λ + λ2 - 2 = 0

⇒ λ2 - 8λ + 10 = 0

λ

λ=−8±√64−402=−8±√242=4±√6

⇒ λ = 4 + √6 , 4 - √6

∴ Eigen values are 4 + √6 & 4 - √6

Smallest Eigen Value = 4 - √6 = 1.55

Now Determining Eigenvectors corresponding to the eigenvalue λ = 1.55.

For λ = 1.55 The corresponding Eigenvector is

[2−1.55−2−16−1.55][x1x2]=[00]

[0.45−2−14.45][x1x2]=[00]

0.45 x1 - 2 x2 = 0 -- (1)

-x1 + 4.45 x2 = 0 -- (2)

Solving (1) and (2) we get.

x1 = 2 & x2 = 0.45

∴ The eigenvector corresponding to 1.55 is [20.45]

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