For which $n \in N$ is the following matrix invertible?

Question

For which $n \in N$ is the following matrix invertible? $$\left[\begin{array}{[c c c]} 10^{30}+5 & 10^{20}+4 & 10^{20}+6 \\ 10^{4}+2 & 10^{8}+7 & 10^{10}+2n \\ 10^{4}+8 & 10^{6}+4 & 10^{15}+9 \\ \end{array}\right]$$

My attempt: For the matrix to be invertible, it must be non-singular. To compute the determinant, I split the large determinant into smaller determinants using the column addition property but it got too tedious to compute (and too lengthy to type out here :P)

The answer:

Replacing even numbers by zero and odd numbers by one, we have $$|A| = \left| \begin{array}{c c c} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{array}\right|$$ which is an odd number and hence $|A|$ can not be zero. Hence A is invertible for all $n \in N$.

I did not understand how all the odd numbers were simply replaced with 1 and the even numbers with 0. I would appreciate it if someone could provide a different answer or explain the given answer.

Answer 1

Just expand the determinant using the first column. You see quickly that you get a sum of $3$ odd integers. Hence the determinant is an odd integer.

For example one term would be $[(10)^{30}+5] [((10)^{8}+7)((10)^{15}+9)-((10)^{10}+2n)((10)^{6}+4)]$ which is odd.

Answer 2

To put it algebraically: you have a map $q$ from $\Bbb Z$ to the field $\Bbb F_2 = \{0,1\}$ sending even numbers to $0$, odd numbers to $1$ (in fact the standard quotient map $\Bbb Z \to \Bbb Z{/}{2\Bbb Z} \simeq \{0,1\}$, so it's a ring homomorphism). This extends to a map from all integer matrices (of dimension $3 \times 3$) to one that maps to all $0,1$-matrices (of that same dimension). This will also be a ring homomorphism, and it's clear that map preserves the determinant function $\det$ in the sense that $$\det(q(A)) = q(\det(A))$$ for all such matrices $A$ (it's just a sum of products of coefficients, or their negation). So if $q(A)$ (the $0,1$-matrix from reducing modulo 2) has non-zero determinant (so $\det(q(A))=1$), $\det(A)$ was non-zero. That's the essence of this argument.