Find $X$ such that there exists a scalar $c$. Let $A = \begin{bmatrix}5 & 0 & 0 \\1 & 5 & 0 \\ 0 & 1 & 5\end{bmatrix}$.
For which $X$ does there exist a scalar $c$ such that $AX = cX$?
Do they mean $X$ is the null matrix?
 A: The question is asking about a very important concept in linear algebra called eigenvalues (and eigenvectors). Eigen is the German word for 'proper', and by definition, an eigenpair  is a non-zero vector $X$ and a scalar $c$ (which can be zero) that solves the equation $AX = cX$. Finding eigenpairs is a huge part of linear algebra, and if you haven't learned about it yet, you probably will soon. 
In general for a 3x3 matrix the method is pretty simple. You rewrite the equation as $AX - cX = 0$, and then factor out $X$ to get $(A - cI)X = 0$ (where $I$ is the 3x3 identity, so $cI$ is just a matrix with all zeros except $c$'s on the diagonal). Since $X$ is nonzero, you now need to find the values of $c$ for which $A-cI$ has a nontrivial null space, which is often done by finding $c$ that make $\det (A-cI) = 0$. This boils down to solving a cubic polynomial equation. 
In your case this is quite easy: $\det(A-cI) = (5-c)^3$, so the only eigenvalue is $c = 5$. If you want to check that this works, try to find the null space of $A-cI$. It should be three dimensional.
