# Determinant with Unit Element in Otherwise Zero Row

## Theorem

Let $D$ be the determinant:

$D = \begin {vmatrix} 1 & 0 & \cdots & 0 \\ b_{2 1} & b_{2 2} & \cdots & b_{2 n} \\ \vdots & \vdots & \ddots & \vdots \\ b_{n 1} & b_{n 2} & \cdots & b_{n n} \end {vmatrix}$

Then:

$D = \begin {vmatrix} b_{2 2} & \cdots & b_{2 n} \\ \vdots & \ddots & \vdots \\ b_{n 2} & \cdots & b_{n n} \end {vmatrix}$

## Proof

We refer to the elements of:

$\begin {vmatrix} 1 & 0 & \cdots & 0 \\ b_{2 1} & b_{2 2} & \cdots & b_{2 n} \\ \vdots & \vdots & \ddots & \vdots \\ b_{n 1} & b_{n 2} & \cdots & b_{n n} \end {vmatrix}$

as $\begin {vmatrix} b_{i j} \end {vmatrix}$.

Thus $b_{1 1} = 1, b_{1 2} = 0, \ldots, b_{1 n} = 0$.

Then from the definition of determinant:

 $\ds D$ $=$ $\ds \sum_\lambda \paren {\map \sgn \lambda \prod_{k \mathop = 1}^n b_{k \map \lambda k} }$ $\ds$ $=$ $\ds \sum_{\lambda} \map \sgn \lambda b_{1 \map \lambda 1} b_{2 \map \lambda 2} \cdots b_{n \map \lambda n}$

Now we note:

 $\ds \map \lambda 1 = 1$ $\implies$ $\ds b_{1 \map \lambda 1} b_{2 \map \lambda 2} \cdots b_{n \map \lambda n} = 1$ $\ds \map \lambda 1 \ne 1$ $\implies$ $\ds b_{1 \map \lambda 1} b_{2 \map \lambda 2} \cdots b_{n \map \lambda n} = 0$

So only those permutations on $\N^*_n$ such that $\map \lambda 1 = 1$ contribute towards the final summation.

Thus we have:

$\displaystyle D = \sum_\mu \map \sgn \mu b_{2 \map \mu 2} \cdots b_{n \map \mu n}$

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

Hence the result.

$\blacksquare$