$A$ is a 3×3 matrix with eigenvalues 0,3,6 and corresponding eigenvectors $u$,$v$,$w$. [closed]
Solution 1:
Since $u,v,w$ correspond to different eigenvalues, they are linearly independent, and hence form a basis for $\mathbb{R^3}$. If $$ A(au+bv+cw)=u $$ or, equivalently, $$ a0 + 3bv + 6cw = u \mbox{,}$$ then $u,v,w$ are linearly dependent, a contradiction. Thus, $Ax=u$ has no solution.
I think you are able to solve the other question yourself now.