For what value of $x$ is this matrix invertible? I've been given the following matrix $X$:
$$X=
\begin{bmatrix}
1 & 4 & 8 & 1 \\     
0 & 30 &1  &  0 \\    
0 &2&  0&  0 \\     
1 &2 & 9 & x \\
\end{bmatrix}
$$
and I have to determine for which value of $x$ the matrix is invertible.
Any hints/suggestions would be greatly appreciated.
So far I have substituted different values for $x$, but there seems to be an inverse for all values of $x$ above $1$.

$$X=
\begin{pmatrix}
1 & 4 & 8 & 1 \\     
0 & 30 &1  &  0 \\    
0 &2&  0&  0 \\     
1 &2 & 9 & x \\
\end{pmatrix}\stackrel{ R_4-R_1}\longrightarrow\;\; 
\begin{pmatrix}
1 & 4 & 8 & 1 \\     
0 & 30 &1  &  0 \\    
0 &2&  0&  0 \\     
0 &\!\!-2 & 1 & x-1 \\
\end{pmatrix}\stackrel{R_3-\frac1{15}R_2}{\stackrel{R_4+\frac1{15}R_2\;,\;R_2}\longrightarrow}
$$
$$\begin{pmatrix}
1 & 4 & 8 & 1 \\     
0 & 30 &1  &  0 \\    
0 &0& \!\! -\frac1{15}&  0 \\     
0 &0 & \frac{16}{15} & x-1 \\
\end{pmatrix}\stackrel{R_4+16R_3}\longrightarrow\begin{pmatrix}
1 & 4 & 8 & 1 \\     
0 & 30 &1  &  0 \\    
0 &0& \!\! -\frac1{15}&  0 \\     
0 &0 & 0 & x-1 \\
\end{pmatrix}$$
and the above matrix (and thus the original one as well) is singular iff the last row is all zeros, so...
