Determinant with Unit Element in Otherwise Zero Column

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.