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after i plug in the value, i have the matrix
|dw:1334524320293:dw|

multiply second row by -1/2 and add to 3rd row...
|dw:1334524531744:dw|

so now i have these equations
v1+v2+2v3=0
-2v2+2v3=0
-6v3=0

but that's what you wanted |A-I(Lambda)| =0

just put any value, for example v3=a and find the other

you should get a singular matrix the last row should be 0

-2v2+2a=0
v2=a
v1+v2+2v3=0
v1+a+2a=0
v1 = -3a
eigenvector (1,1,3)

it doesn't have to be all 0

|A-I(Lambda)| =0

remember you are solving
det(A- lambda*I)= 0

i was taught the bottom left triangle has to be 0s.

so it would be the bottom left corner, and the values right above and to the right

the determinant has to be 0

i've checked and checked to see if i made a mistake...i will look again

what is the original matrix?

i think your eigenvalue is wrong

ugh idiot i did do it wrong. value is right, i will rework this

just trying to help