Question

Find the eigenvalues and eigenvectors of the matrix [

8 −6 2

−6 7 −4

2 −4 3

]​

Answer 1

Answer:

Solve linear equations (Sor method) - relaxation method 8x+y+z=8,2x+4y+z=4,x+3y+5z=5. 60. Solve linear equations - Gauss Jacobi 7y+2x=11,3x-y=5.

Answer 2

Concept:

A matrix is a set array of numbers that are being arranged in rows and columns in a rectangular array.

Given:

Matrix,

           8 −6 2

A =     −6 7 −4

            2 −4 3

Find:

We are asked to find the eigenvalues and eigenvectors of the given matrix.

Solution:

We have,

Matrix,

           8 −6 2

A =     −6 7 −4

            2 −4 3 ,

So,

So to find eigenvalues;

Eigenvalues = [(determinent of matrix) - (λI)]

i.e. [A - λI] = 0

So,

[   8 −6 2 ]          [ λ 0 0 ]

[ −6 7 −4 ]    -     [ 0 λ 0 ]     = 0

[   2 −4 3 ]          [ 0 0 λ ]

We get,

[ (8 - λ)    −6          2   ]          

[ −6       (7 - λ)     −4    ]        = 0

[   2        −4       (3 - λ) ]

Now,

Solve for λ ,

We get,

[(168 - 45λ + 3λ² - 56λ + 15λ² - λ³) + 48 + 48 ] - [ 28 - 4λ + 128 - 16λ + 108 - 36λ] = 0

Simplify,

(8 - λ) [(7 - λ) × (3-λ) - (-4) × (-4)] - [(-6) [(-6) × (3-λ) - (-4)×2)] +2 [(-6) × (-4) - (7-λ) × 2)] = 0

Simplify more,

[(8 - λ) (21 - 10λ + λ²) - 16] - [6 ( -18 + 6λ) - (-8)] + 2 [(24 - (14 - 2λ)] = 0

[(8 - λ) (5- 10λ + λ²)] + 6 (-10 + 6λ) + 2 (10 + 2λ) = 0

Now, making it more simple,

(40 - 85λ + 18λ² - λ³) + ( -60 + 36λ) + (20 +4 λ) = 0

40 - 85λ + 18λ² - λ³  -60 + 36λ + 20 +4 λ = 0

Rewrite it,

- λ³ + 18λ²  - 85λ + 40λ + 60 - 60 = 0

We get,

( -λ³ + 18λ² - 45λ) = 0

Now, making fctors,

We get,

-λ (λ - 3) (λ - 15)=0

So, the factors i.e. eigenvalues and eigenvectors are (0, 3, 15).

Hence we can say that the eigenvalues and eigenvectors are (0, 3, 15).

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