Determinant of a triangular matrix
Task number: 4433
What is the formula for the determinant of triangular matrices of the following type?
\[
\begin{pmatrix}
a_{11} & \dots & \dots & a_{2,n-1} & a_{1n} \\
a_{21} & \dots & a_{2,n-2} & a_{2,n-1} & 0 \\
\vdots & & & & \vdots \\
a_{n-1{,}1} & a_{n-1{,}2} & 0 & \dots& 0 \\
a_{n1} & 0 & \dots & \dots & 0
\end{pmatrix}
\]
Solution
The sign of the permutation corresponding to the antidiagonal depends on \(n\). It is a permutation \((n,n-1,\dots,1)\). It has \(\frac{n(n-1)}2\) inversions, and its sign is \((-1)^{\frac{n(n-1)}2}\).
\[\begin{vmatrix} a_{11} & \dots & \dots & a_{2,n-1} & a_{1n} \\ a_{21} & \dots & a_{2,n-2} & a_{2,n-1} & 0 \\ \vdots & & & & \vdots \\ a_{n-1{,}1} & a_{n-1{,}2} & 0 & \dots& 0 \\ a_{n1} & 0 & \dots & \dots & 0 \end{vmatrix} = (-1)^{\frac{n(n-1)}2} \prod\limits_{i=1}^n a_{i,n+1-i} \]The sign changes in two steps, that is \(+,-,-,+,+,-,-,\cdots\).
