Determinant/Transposed matrix/Universal property/Fact/Proof

If ${\displaystyle {}M}$ is not invertible, then, due to fact, the determinant is ${\displaystyle {}0}$ and the rank is smaller than ${\displaystyle {}n}$. This does also hold for the transposed matrix, so that its determinant is again ${\displaystyle {}0}$. So suppose that ${\displaystyle {}M}$ is invertible. We reduce the statement in this case to the corresponding statement for the elementary matrices, which can be verified directly, see exercise. Because of fact, there exist elementary matrices ${\displaystyle {}E_{1},\ldots ,E_{s}}$ such that

${\displaystyle {}D=E_{s}\cdots E_{1}M\,}$

is a diagonal matrix. Due to exercise, we have

${\displaystyle {}{D^{\text{tr}}}={M^{\text{tr}}}{E_{1}^{\text{tr}}}\cdots {E_{s}^{\text{tr}}}\,}$

and

${\displaystyle {}{M^{\text{tr}}}={D^{\text{tr}}}({E_{s}^{\text{tr}}})^{-1}\cdots ({E_{1}^{\text{tr}}})^{-1}\,.}$

The diagonal matrix ${\displaystyle {}D}$ is not changed under transposing it. Since the determinants of the elementary matrices are also not changed under transposition, we get, using fact,

${\displaystyle {}{\begin{aligned}\det {M^{\text{tr}}}&=\det {\left({D^{\text{tr}}}({E_{s}^{\text{tr}}})^{-1}\cdots ({E_{1}^{\text{tr}}})^{-1}\right)}\\&=\det {D^{\text{tr}}}\cdot \det({E_{s}^{\text{tr}}})^{-1}\cdots \det({E_{1}^{\text{tr}}})^{-1}\\&=\det D\cdot \det {\left(E_{s}^{-1}\right)}\cdots \det {\left(E_{1}^{-1}\right)}\\&=\det {\left(E_{1}^{-1}\right)}\cdots \det {\left(E_{s}^{-1}\right)}\cdot \det D\cdot \\&=\det {\left(E_{1}^{-1}\cdots E_{s}^{-1}\cdot D\right)}\\&=\det M.\end{aligned}}}$