Can there be a different Eigen vector for a particular Eigen value?

Question

Please see the photo.

Here, my answer came $k$ $\begin{bmatrix} 1\\ 1\\ -1 \end{bmatrix}$

But their answer is given : $k$ $\begin{bmatrix} 1\\ -1\\ 1 \end{bmatrix}$ My answer is correct?? Please tell--

THE ACTUAL MATRIX]

Answer 1

Sure. An eigenvalue can be associated with multiple eigenvectors.

For example: Let $$A=\pmatrix{1&0&0\\0&1&0\\0&0&1}.$$

The eigenvalue $\lambda=1$ has all non-zero vectors as eigenvectors.

Answer 2

For $\lambda=1$ we have: $$(A-I)v=0$$ $$\pmatrix{0&1&1\\-1&-2&-1\\0&0&0}\pmatrix {x\\y\\z}=\pmatrix {0\\0\\0}$$ This gives us two equations: $$y=-z$$ $$-x-2y-z=0 \implies x=z$$ Therefore we have: $$\implies v=(x,y,z)=x(1,-1,1)$$ You certainly made a sign mistake somewhere.