Let matrix `A=[(4,6,6),(1,3,2),(-1,-4,-3)],` Find the non-zero column vector X such that `AX= lambdaX` for some scalar `lambda.`

Explanation

The characteristic equation is `|A-lambdaI|=0`
`rArr" " [(4-lambda,6,6),(1,3-lambda,2),(-1,-4,-3-lambda)]=0`
`rArr " " lambda^(3)-4lambda^(2)-lambda+4=0`
or `(lambda+1)(lambda-1)(lambda-4)=0`
The eigen values are `lambda=-1,1,4`
If `lambda=-1`, we get `5x+6y+6z=0,x+4y+2z=0`
and `-x-4y-2z=0`
Giving `(x)/(6)=(y)/(2)=(z)/(-7),X=[(,6),(,2),(,-7)]`
If `lambda =4`, we get `0.x+6y+6z=0,x-y+2z=0`
and `-x-4y-7=0`
Giving `(x)/(3)=(y)/(1)=(3)/(-1),X=[(,0),(,1),(,-1)]`
Hence, vector are X `X=[(,6),(,2),(,-7)][(,0),(,1),(,-1)][(,3),(,1),(,-1)]`