For which values of $x$ the matrix is invertible? The following matrix has coefficients in $\Bbb Z_{11}$:
$\left(\begin{matrix}
1 & 0 & 3 & 0 & 5 \\
0 & 3 & 0 & 5 & 0 \\
3 & 0 & x & 0 & 7 \\
0 & 5 & 0 & 7 & 0 \\
5 & 0 & 7 & 0 & 9
\end{matrix}\right)$
To determine for which values of $x$ it is invertible, I tried to find the correspondent triangular matrix so I can easily calculate the determinant and then understand for which values $x$ is $0$.   I have come to this point:
$\left(\begin{matrix}
1 & 0 & 3 & 0 & 5 \\
0 & 1 & 0 & 7 & 0 \\
0 & 0 & 2x & 0 & 3 \\
0 & 0 & 0 & 5 & 0 \\
0 & 0 & 3 & 0 & 6
\end{matrix}\right)$
I don't know how to remove the $3$ to make the matrix triangular. Any help?

The original matrix $A$ will not be invertible if and only if there is a nonzero vector $v=(v_1,\ldots,v_5)^T$ such that $Av=0$.
By the pattern of zeros of $A$ we see that the equations from $Av=0$ for $v_2,v_4$ are independent of those for $v_1,v_3,v_5$. Moreover we have $3v_2+5v_4=0=5v_2+7v_4$, which are independent of each other in $\mathbb{Z}_{11}$, so $v_2=0=v_4$.
Now we have to impose that the matrix for $v_1,v_3,v_5$ is not invertible. That matrix is equivalent modulo $11$ to
$$\begin{pmatrix}1 & 3 & 5 \\ 3 & x & -4\\ 5 & -4 & -2\end{pmatrix}.$$
Its determinant is equivalent modulo $11$ to $6x+3$, so $\det(A)\equiv 0\pmod{11}$ iff $x\equiv (-3)6^{-1}\equiv (-3)2\equiv 5\pmod{11}$.
