So I took linear algebra a few years ago, so I'm a little rusty on how to do ref and rref, but I imagine it can be used to solve this. I just don't understand my output.
Here is the system of equations:
$x'=x+2y-z$
$y'=x+z$
$z'=4x-4y+5z$
I put this into a calculator and treated $x'$, $y'$, and $z'$ as $0$ so my matrix looked like this:
$$\left[\begin{matrix}1 & 2 & -1 & 0\\ 1 & 0 & 1 & 0 \\ 4 & -4 & 5 & 0\end{matrix}\right] \Rightarrow \left[\begin{matrix}1 & -1 & \frac{5}{4} & 0 \\ 0 & 1 & -\frac{3}{4} & 0 \\ 0 & 0 & 1 & 0\end{matrix}\right]$$
I took the resulting matrix to mean all of my solutions are equal to zero, but some reason I feel like I'm doing something wrong.
If you were given the ODE $y'-y=0$, for example, what happens when you set $y'=0$? You arrive at $y=0$, but that's not the solution to the ODE (well, it's one of them, but it's not the solution).
Written in matrix form, the system of ODEs looks like this: $$\begin{bmatrix}x'\\y'\\z'\end{bmatrix}=\begin{bmatrix}1&2&-1\\1&0&1\\4&-4&5\end{bmatrix}\begin{bmatrix}x\\y\\z\end{bmatrix}$$ The problem with your attempt is that you can't just set the derivatives equal to zero. You need to compute the eigenvalues/-vectors of the coefficient matrix to find solutions for $x$, $y$, and $z$.
Given this coefficient matrix $A$, you find the eigenvalues $\lambda_i$ by computing the determinant of the matrix and setting equal to zero. In other words, solving $\det(A-\lambda I)=0$. In this case, you have $$\begin{vmatrix}1-\lambda&2&-1\\1&-\lambda&1\\4&-4&5-\lambda\end{vmatrix}=0\implies\begin{cases}\lambda_1=1\\\lambda_2=2\\\lambda_3=3\end{cases}$$ and so on.
Added:
I just noticed that you specifically asked about solving by elimination, but while we're on the subject, I'm going to continue with the eigenvalue method. If/when I have time later I'll include a solution via elimination.
Let's compute that determinant. For no reason in particular, I'll use a cofactor expansion along the first column. We have
$$\begin{align*}\begin{vmatrix}\color{red}{1-\lambda}&2&-1\\\color{red}1&-\lambda&1\\\color{red}4&-4&5-\lambda\end{vmatrix}&=(\color{red}{1-\lambda})\begin{vmatrix}-\lambda&1\\-4&5-\lambda\end{vmatrix}-\color{red}1\begin{vmatrix}2&-1\\-4&5-\lambda\end{vmatrix}+\color{red}4\begin{vmatrix}2&-1\\-\lambda&1\end{vmatrix}\\[1ex] &=(1-\lambda)(\lambda^2-5\lambda+4)-(10-2\lambda-4)+4(2-\lambda)\\[1ex] &=-(\lambda-1)(\lambda-2)(\lambda-3) \end{align*}$$
and we arrive at the eigenvalues given above.
From here, we need to find the associated eigenvectors. For each $\lambda_i$, we want to find a vector $\vec{\eta}_i=\begin{bmatrix}\eta_1\\\eta_2\\\eta_3\end{bmatrix}$ such that $$A\vec{\eta}_i=\lambda_i\vec{\eta}_i \iff (A-\lambda_iI)\vec{\eta}_i=\vec{0}$$
Let's consider the first eigenvalue $\lambda_1=1$: $$(A-I)\vec{\eta}_1=\begin{bmatrix}0&2&-1\\1&-1&1\\4&-4&4\end{bmatrix}\begin{bmatrix}\eta_1\\\eta_2\\\eta_3\end{bmatrix}=\begin{bmatrix}0\\0\\0\end{bmatrix}$$ RREF'ing the matrix yields $$\begin{bmatrix}1&0&\frac{1}{2}\\0&1&-\frac{1}{2}\\0&0&0\end{bmatrix}\begin{bmatrix}\eta_1\\\eta_2\\\eta_3\end{bmatrix}=\begin{bmatrix}0\\0\\0\end{bmatrix}\implies\vec{\eta}_1=\begin{bmatrix}-\frac{1}{2}\eta_3\\\frac{1}{2}\eta_2\\\eta_3\end{bmatrix}$$ $\eta_3$ is a free variable. I would choose $\eta_3=2$, since this gives the final eigenvector in the convenient form of $\vec{\eta}_1=\begin{bmatrix}-1\\1\\2\end{bmatrix}$. I'll leave solving for the other two eigenvectors to you.
Once you get all three eigenvectors, the solution to the system will take the form $$\begin{bmatrix}x\\y\\z\end{bmatrix}=C_1\vec{\eta}_1e^{\lambda_1t}+C_2\vec{\eta}_2e^{\lambda_2t}+C_3\vec{\eta}_3e^{\lambda_3t}$$