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.

$\blacksquare$

Also see

• Determinant with Unit Element in Otherwise Zero Row