Determinant with Unit Element in Otherwise Zero Column
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Theorem
Let $D$ be the determinant:
- $D = \begin{vmatrix}
1 & b_{12} & \cdots & b_{1n} \\ 0 & b_{22} & \cdots & b_{2n} \\
\vdots & \vdots & \ddots & \vdots \\
0 & b_{n2} & \cdots & b_{nn}
\end{vmatrix}$
Then:
- $D = \begin{vmatrix}
b_{22} & \cdots & b_{2n} \\
\vdots & \ddots & \vdots \\
b_{n2} & \cdots & b_{nn}
\end{vmatrix}$
Proof
We note that:
- $D = \begin{vmatrix}
1 & b_{12} & \cdots & b_{1n} \\ 0 & b_{22} & \cdots & b_{2n} \\
\vdots & \vdots & \ddots & \vdots \\
0 & b_{n2} & \cdots & b_{nn}
\end{vmatrix}$
is the transpose of:
- $D^\intercal = \begin{vmatrix}
1 & 0 & \cdots & 0 \\ b_{12} & b_{22} & \cdots & b_{n2} \\
\vdots & \vdots & \ddots & \vdots \\
b_{1n} & b_{2n} & \cdots & b_{nn}
\end{vmatrix}$
From Determinant with Unit Element in Otherwise Zero Row:
- $D^\intercal = \begin{vmatrix}
b_{22} & \cdots & b_{n2} \\
\vdots & \ddots & \vdots \\
b_{2n} & \cdots & b_{nn}
\end{vmatrix}$
The result follows by Determinant of Transpose.
$\blacksquare$