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.
Best Answer
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.