Determinant with Unit Element in Otherwise Zero Row
From ProofWiki
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:
| \(\displaystyle \) | \(\displaystyle \) | \(\displaystyle \) | \(\displaystyle D\) | \(=\) | \(\displaystyle \sum_{\lambda} \left({\operatorname{sgn} \left({\lambda}\right) \prod_{k=1}^n b_{k \lambda \left({k}\right)} }\right)\) | \(\displaystyle \) | \(\displaystyle \) | \(\displaystyle \) | |||
| \(\displaystyle \) | \(\displaystyle \) | \(\displaystyle \) | \(\displaystyle \) | \(=\) | \(\displaystyle \sum_{\lambda} \operatorname{sgn} \left({\lambda}\right) b_{1 \lambda \left({1}\right)} b_{2 \lambda \left({2}\right)} \cdots b_{n \lambda \left({n}\right)}\) | \(\displaystyle \) | \(\displaystyle \) | \(\displaystyle \) |
Now we note:
| \(\displaystyle \) | \(\displaystyle \) | \(\displaystyle \) | \(\displaystyle \lambda \left({1}\right) = 1\) | \(\implies\) | \(\displaystyle b_{1 \lambda \left({1}\right)} b_{2 \lambda \left({2}\right)} \cdots b_{n \lambda \left({n}\right)} = 1\) | \(\displaystyle \) | \(\displaystyle \) | \(\displaystyle \) | |||
| \(\displaystyle \) | \(\displaystyle \) | \(\displaystyle \) | \(\displaystyle \lambda \left({1}\right) \ne 1\) | \(\implies\) | \(\displaystyle b_{1 \lambda \left({1}\right)} b_{2 \lambda \left({2}\right)} \cdots b_{n \lambda \left({n}\right)} = 0\) | \(\displaystyle \) | \(\displaystyle \) | \(\displaystyle \) |
So only those permutations on $\N^*_n$ such that $\lambda \left({1}\right) = 1$ contribute towards the final sum.
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.
$\blacksquare$
