Why is $\det M$ equal to zero?

Question

Let $M$ be a $d \times d$ anti-symmetric matrix, i.e., $M^T = −M$.

Since $\det M = \det(−M^T) = \det (−M) = (−1)^d \det M$, it follows that $\det M = 0$ if $d$ is odd.

I am not able to understand why is $-\det(M) = 0$.

Or why $\det M = -\det M$ means that $\det M = 0$, meaning $M$ is not invertible.

Answer 1

Over the real or complex numbers, $2x=0$ implies $x=0,$ which is what you need to apply here, with $x=\det M.$

This would only be fail in fields of characteristic $2$, where $1+1=0.$

Answer 2

If you know $\mathrm{det}(M)$, then consider the matrix $-M$: every row has been multiplied by $-1$. Since each time you multiply a row by $-1$ the determinant also gets multiplied by $-1$, this means $$\mathrm{det}(-M) = -1^{d}\mathrm{det}(M).$$ But you also have that $$\mathrm{det}(M) = \mathrm{det}(M^{T})=\mathrm{det}(-M),$$ so this tells you that either $-1^d = 1$ or $\mathrm{det}(M) = 0$.