Determinant with Unit Element in Otherwise Zero Row

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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:

\(\displaystyle D\) \(=\) \(\displaystyle \sum_{\lambda} \left({\operatorname{sgn} \left({\lambda}\right) \prod_{k \mathop = 1}^n b_{k \lambda \left({k}\right)} }\right)\)
\(\displaystyle \) \(=\) \(\displaystyle \sum_{\lambda} \operatorname{sgn} \left({\lambda}\right) b_{1 \lambda \left({1}\right)} b_{2 \lambda \left({2}\right)} \cdots b_{n \lambda \left({n}\right)}\)


Now we note:

\(\displaystyle \lambda \left({1}\right) = 1\) \(\implies\) \(\displaystyle b_{1 \lambda \left({1}\right)} b_{2 \lambda \left({2}\right)} \cdots b_{n \lambda \left({n}\right)} = 1\)
\(\displaystyle \lambda \left({1}\right) \ne 1\) \(\implies\) \(\displaystyle b_{1 \lambda \left({1}\right)} b_{2 \lambda \left({2}\right)} \cdots b_{n \lambda \left({n}\right)} = 0\)


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.

$\blacksquare$


Sources