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:

Now we note:

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

Thus we have:
 * $\ds D = \sum_\mu \map \sgn \mu b_{2 \map \mu 2} \cdots b_{n \map \mu n}$

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

Hence the result.

Also see

 * Determinant with Unit Element in Otherwise Zero Column