When I teach linear algebra (it has been some years now), I always tell my students to never ever compute the inverse of a matrix, at least not if the matrix is much bigger than $3\times3$. If you need to solve $Ax=b$ for just one single $b$, do Gaussian elimination. If you need to do it for several $b$ values but a single $A$, compute the $LU$ decomposition of $A$ and use that to compute the required solutions.
You went wrong in that you solved the system and considered that the answer to the question. Both of these systems have infinite solutions but that is not what the question is asking. It is asking are the systems equivalent. One way to do that is solve them both and see that they give the same solution set. Another way to to transform one set of equations into the other using algebra.
$\begin{align}
-x_1+x_2+4x_3 &= 0 \\
+x_1+3x_2+8x_3 &= 0 \\
\hline
4x_2+12x_3 &= 0
\end{align}
$
Divide the last equation by $4$ and you get $x_2+3x_3=0$, which can also be viewed as $x_3=-3x_3$. Substituting this into one of the equations (say the second), we get $x_1+3x_2+8x_3=0 \iff x_1+3(-3x_3)+8x_3=0 \iff x_1-x_3=0$.
Sine we can transform the one system into the other, they are equivalent. This is often done using the augmented matrix version of an equation and Gaussian elimination (also called Gauss-Jordan elimination).
In response to the comment, label the first system's equations $a,b,c$, in order, and the second system's equations $d,e$, in order. The work above shows that $e=\frac{1}{4}a+\frac{1}{4}b$. Since we used $a$ and $b$ to get $e$, we have to pick a different pair to try and get $d$. It turns out that $d = \frac{2}{3}c-\frac{2}{3}a$ (though this is not the only possible answer).
The other direction is much simpler. Since $d$ has $x_1$ but not $x_2$, any $x_1$ terms in $a,b,c$ have to have come from $d$. The same is true for $e$ and $x_2$. Since $a$ has $-x_1$, we need a $-d$. Since $a$ has $x_2$, we need $e$. It turns out that this is enough to get the answer as the $x_3$ takes care of itself. So $a=e-d$. Similarly, $b=d+3e$ and $c=\frac{1}{2}d+e$.
Best Answer
Making the substitution $\tilde{y}=\log_2 y$ and $\tilde{z}=z(z+1)$ leads to the equivalent linear system $$ \begin{eqnarray} x &+& &\tilde{y} &+& &\tilde{z} &= 0,\\ 3x &+& 2&\tilde{y} &+& \frac{3}{2}&\tilde{z} &= 0,\\ 2x&-&&\tilde{y}&-&\frac{1}{2}&\tilde{z} &= 4 \end{eqnarray} $$ solved by $x=1$, $\tilde{y}=-3$, $\tilde{z}=2$. Hence $y=2^{\tilde{y}}=2^{-3}=1/8$ and $z$ is obtained by solving $z(z+1)=2$, that is, $z=1$ or $z=-2$.