If you really want to do it in terms of matrices, yes, "2" is a triple eigenvalue.
To find an eigenvector, we have first
\(\displaystyle \begin{bmatrix}2 & 1 & 3 \\ 0 & 2 & -1\\ 0 & 0 & 2\end{bmatrix}\begin{bmatrix}x \\ y \\ z\end{bmatrix}= \begin{bmatrix}2x+ y+ 3z \\ 2y- z \\ 2z\end{bmatrix}= \begin{bmatrix}2x \\ 2y \\ 2z\end{bmatrix}\)
which gives the three equations 2x+ y+ 3z= 2x, 2y- z= 2y, 2z= 2z. The last equation is, of course, true for any z so we are left with two equations in terms of three variables. The first is y+ 3z= 0 and the second, -z= 0. Since z= 0, y= -3z= 0. There is no equation left for x so x can be anything- any eigenvector, is of the form <x, 0, 0>= x<1, 0, 0>. Now, that's only a single vector so the "geometric multiplicity" of eigenvalue 2 is only 1 while its "algebraic multiplicity" is 3.
We can't "diagonalize" this matrix, the best we can do is write it in "Jordan Normal form". That is, if, in addition to the single eigenvector, we find two "generalized eigenvectors", and form a matrix B with them as columns, we can write \(\displaystyle B^{-1}AB\) as \(\displaystyle \begin{bmatrix}2 & 1 & 0 \\ 0 & 2 & 1 \\ 0 & 0 & 2\end{bmatrix}\). But, in fact, that is not really any simpler to solve than the method I showed in the previous post so, frankly, I wouldn't bother.