For which values of $x$ the matrix is invertible?

Question

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?

Answer 1

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 $x\equiv (-3)6^{-1}\equiv (-3)2\equiv 5\pmod{11}$.

Answer 2

The matrix is singular (and not invertible) if any row can be described as a linear combination of the other rows (other than the trivial combination).

Due to the placement of the zeros, it should be obvious that second and fourth rows must be independent from the third row.

But if we take the first plus the fifth minus twice the third we get

$(0,0,10-2x,0,0)$,

and we have a singular matrix if

$10-2x \equiv 0\pmod {11}$.