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Find eigenvectors [[8,-6,2],[-6,7,-4],[2,-4,3]] [ Calculator, Method and examples ]

Solution:
Your problem `->` eigenvectors [[8,-6,2],[-6,7,-4],[2,-4,3]]


`|A-lamdaI|=0`

 `(8-lamda)`  `-6`  `2` 
 `-6`  `(7-lamda)`  `-4` 
 `2`  `-4`  `(3-lamda)` 
 = 0


`:.(8-lamda)((7-lamda) × (3-lamda) - (-4) × (-4))-(-6)((-6) × (3-lamda) - (-4) × 2)+2((-6) × (-4) - (7-lamda) × 2)=0`

`:.(8-lamda)((21-10lamda+lamda^2)-16)+6((-18+6lamda)-(-8))+2(24-(14-2lamda))=0`

`:.(8-lamda)(5-10lamda+lamda^2)+6(-10+6lamda)+2(10+2lamda)=0`

`:. (40-85lamda+18lamda^2-lamda^3)+(-60+36lamda)+(20+4lamda)=0`

`:.(-lamda^3+18lamda^2-45lamda)=0`

`:.-lamda(lamda-3)(lamda-15)=0`

`:.lamda=0 or(lamda-3)=0 or(lamda-15)=0 `

`:.` The eigenvalues of the matrix `A` are given by `lamda=0,3,15`,

1. Eigenvectors for `lamda=0`




1. Eigenvectors for `lamda=0`

`A-lamdaI = `
`8``-6``2`
`-6``7``-4`
`2``-4``3`
 - `0` 
`1``0``0`
`0``1``0`
`0``0``1`


 = 
`8``-6``2`
`-6``7``-4`
`2``-4``3`


Now, reduce this matrix
`R_1 larr R_1-:8`

 = 
 `1` `1=8-:8`
`R_1 larr R_1-:8`
 `-3/4` `-3/4=-6-:8`
`R_1 larr R_1-:8`
 `1/4` `1/4=2-:8`
`R_1 larr R_1-:8`
`-6``7``-4`
`2``-4``3`


`R_2 larr R_2+6xx R_1`

 = 
`1``-3/4``1/4`
 `0` `0=-6+6xx1`
`R_2 larr R_2+6xx R_1`
 `5/2` `5/2=7+6xx-3/4`
`R_2 larr R_2+6xx R_1`
 `-5/2` `-5/2=-4+6xx1/4`
`R_2 larr R_2+6xx R_1`
`2``-4``3`


`R_3 larr R_3-2xx R_1`

 = 
`1``-3/4``1/4`
`0``5/2``-5/2`
 `0` `0=2-2xx1`
`R_3 larr R_3-2xx R_1`
 `-5/2` `-5/2=-4-2xx-3/4`
`R_3 larr R_3-2xx R_1`
 `5/2` `5/2=3-2xx1/4`
`R_3 larr R_3-2xx R_1`


`R_2 larr R_2-:5/2`

 = 
`1``-3/4``1/4`
 `0` `0=0-:5/2`
`R_2 larr R_2-:5/2`
 `1` `1=5/2-:5/2`
`R_2 larr R_2-:5/2`
 `-1` `-1=-5/2-:5/2`
`R_2 larr R_2-:5/2`
`0``-5/2``5/2`


`R_1 larr R_1+(3/4)xx R_2`

 = 
 `1` `1=1+(3/4)xx0`
`R_1 larr R_1+(3/4)xx R_2`
 `0` `0=-3/4+(3/4)xx1`
`R_1 larr R_1+(3/4)xx R_2`
 `-1/2` `-1/2=1/4+(3/4)xx-1`
`R_1 larr R_1+(3/4)xx R_2`
`0``1``-1`
`0``-5/2``5/2`


`R_3 larr R_3+(5/2)xx R_2`

 = 
`1``0``-1/2`
`0``1``-1`
 `0` `0=0+(5/2)xx0`
`R_3 larr R_3+(5/2)xx R_2`
 `0` `0=-5/2+(5/2)xx1`
`R_3 larr R_3+(5/2)xx R_2`
 `0` `0=5/2+(5/2)xx-1`
`R_3 larr R_3+(5/2)xx R_2`


The system associated with the eigenvalue `lamda=0`

`(A-0I)`
`x_1`
`x_2`
`x_3`
 = 
`1``0``-1/2`
`0``1``-1`
`0``0``0`
 
`x_1`
`x_2`
`x_3`
 = 
`0`
`0`
`0`


`=>x_1-(1/2)x_3=0,x_2-x_3=0`

`=>x_1=(1/2)x_3,x_2=x_3`

`:.` eigenvectors corresponding to the eigenvalue `lamda=0` is

`v=`
`(1/2)x_3`
`x_3`
`x_3`


Let `x_3=1`

`v_1=`
`1/2`
`1`
`1`
`v_1=`
`1/2`
`1`
`1`


2. Eigenvectors for `lamda=3`




2. Eigenvectors for `lamda=3`

`A-lamdaI = `
`8``-6``2`
`-6``7``-4`
`2``-4``3`
 - `3` 
`1``0``0`
`0``1``0`
`0``0``1`


 = 
`8``-6``2`
`-6``7``-4`
`2``-4``3`
 - 
`3``0``0`
`0``3``0`
`0``0``3`


 = 
`5``-6``2`
`-6``4``-4`
`2``-4``0`


Now, reduce this matrix
interchanging rows `R_1 harr R_2`

 = 
`-6``4``-4`
`5``-6``2`
`2``-4``0`


`R_1 larr R_1-:-6`

 = 
 `1` `1=-6-:-6`
`R_1 larr R_1-:-6`
 `-2/3` `-2/3=4-:-6`
`R_1 larr R_1-:-6`
 `2/3` `2/3=-4-:-6`
`R_1 larr R_1-:-6`
`5``-6``2`
`2``-4``0`







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