Determinant with Unit Element in Otherwise Zero Row

Theorem
Let $D$ be the determinant:


 * $D = \begin{vmatrix}

1 & 0 & \cdots & 0 \\ b_{21} & b_{22} & \cdots & b_{2n} \\ \vdots & \vdots & \ddots & \vdots \\ b_{n1} & 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 refer to the elements of:
 * $\begin{vmatrix}

1 & 0 & \cdots & 0 \\ b_{21} & b_{22} & \cdots & b_{2n} \\ \vdots & \vdots & \ddots & \vdots \\ b_{n1} & b_{n2} & \cdots & b_{nn} \end{vmatrix}$

as $\begin{vmatrix}b_{ij}\end{vmatrix}$.

Thus $b_{11} = 1, b_{12} = 0, \ldots, b_{1n} = 0$.

Then from the definition of determinant:

Now we note:

So only those permutations on $\N^*_n$ such that $\lambda \left({1}\right) = 1$ contribute towards the final summation.

Thus we have:
 * $D = \sum_{\mu} \operatorname{sgn} \left({\mu}\right) b_{2 \mu \left({2}\right)} \cdots b_{n \mu \left({n}\right)}$

where $\mu$ is the collection of all permutations on $\N^*_n$ which fix $1$.

Hence the result.