2

I was looking at an old post and want to learn about why the answer is true. I have taken linear algebra but never heard about this. Where can I learn this?

Impropio
  • 351
  • 1
    Phrased differently, any even number will cause any multiplication (of integers) that is is a part of to be even, and the sum of an even number to another number will not change its parity. The answer there explains that the only odd numbers worth looking at in the entire calculation of the determinant of that matrix will be along the diagonal and there is only one odd "pattern" in calculating the determinant, implying the determinant is in fact odd. Since zero is even, we know the determinant cannot be zero. – JMoravitz Jan 20 '18 at 03:02
  • 2
    Had we learned the determinant was even, this would not have been enough information to say whether the matrix was invertible or not and we would have needed to examine further or use a different technique entirely. – JMoravitz Jan 20 '18 at 03:04
  • The same trick is applied here. If the elements of a square matrix $M$ belong to $\mathbb{Z}$, so does $\det M$. A sufficient condition for $\det M\neq 0$ is that $\det M$ is odd, and the parity of $\det M$ can be found by computing the determinant of the original matrix reduced $!!\pmod{2}$. – Jack D'Aurizio Jan 20 '18 at 16:58

1 Answers1

1

Reduce all the entries modulo $2$. That is, odd numbers become $1$, even numbers become $0$. Then work out the determinant. Since that just involves multiplication, addition, and subtraction, the result you get is congruent modulo $2$ to the true answer. In this case, the result was 1, so you know the actual determinant is odd, hence nonzero.

saulspatz
  • 53,131
  • Why is the result is congruent modulo 2 to the determinant? What others operations can you do and the result will still be congruent to it? – Impropio Jan 20 '18 at 14:39
  • There's a pretty good article on modular arithmetic on Wikipedia. Generally, you can do addition subtraction and multiplication. For example if $x = q_1m + r_1, y = q_2m + r_2,$ then $xy= q_1q_2m^2 + (r_1q_2+r_2q_1)m+r_1r_2,$ so that the remainder of the product on division by $m$ is just the product of the remainders. When the modulus is prime, you can also handle division, but it's more complicated. – saulspatz Jan 20 '18 at 14:49