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.

$\blacksquare$


Also see