Determinant with Unit Element in Otherwise Zero Row

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Theorem

Let $D$ be the determinant:

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


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.

$\blacksquare$


Also see


Sources