sum of two determinants

Question

If

\begin{vmatrix}a&c\\b&d\end{vmatrix} and \begin{vmatrix}a&e\\b&f\end{vmatrix}, then sum of these determinant can be written as in terms of another determinant given by \begin{vmatrix}a&c+e\\b&d+f\end{vmatrix}

is it right?

Answer 1

Yes. Three proofs, depending on three common definitions:

• Direct, by the formula $ad-bc$: expanding out, $$ \begin{vmatrix} a & c+e \\ b & d+f \end{vmatrix} = a(d+f)-b(c+e) = (ad-bc) + (af-be) = \begin{vmatrix} a & c \\ b & d \end{vmatrix} + \begin{vmatrix} a & e \\ b & f \end{vmatrix} $$.

• By the formula $\det{A} = \sum_{\sigma \in S_n} \epsilon(\sigma) \prod_{i} A_{i\sigma(i)} $, if $A_{ij} = B_{ij} = C_{ij}$ for $i \neq k$, $A_{kj}=B_{kj}+C_{kj}$, then \begin{align} \det{A} &= \sum_{\sigma \in S_n} \epsilon(\sigma) \prod_{i} A_{i\sigma(i)} = \sum_{\sigma \in S_n} \epsilon(\sigma) \left(\prod_{i\neq k} A_{i\sigma(i)} \right) A_{k\sigma(k)} \\ &= \sum_{\sigma \in S_n} \epsilon(\sigma) \left(\prod_{i\neq k} A_{i\sigma(i)} \right) (B_{k\sigma(k)}+C_{k\sigma(k)}) \\ &= \sum_{\sigma \in S_n} \epsilon(\sigma) \left(\prod_{i\neq k} A_{i\sigma(i)} \right) B_{k\sigma(k)} + \sum_{\sigma \in S_n} \epsilon(\sigma) \left(\prod_{i\neq k} A_{i\sigma(i)} \right) C_{k\sigma(k)} \\ &= \sum_{\sigma \in S_n} \epsilon(\sigma) \left(\prod_{i\neq k} B_{i\sigma(i)} \right) B_{k\sigma(k)}+\sum_{\sigma \in S_n} \epsilon(\sigma) \left(\prod_{i\neq k} C_{i\sigma(i)} \right) C_{k\sigma(k)}) \\ &= \det{B}+\det{C}.\end{align}

• If the determinant is defined as an alternating multilinear map on columns, this result is part of the definition (multilinear meaning that it is linear in each column).

Answer 2

The determinant is a multilinear function of its columns and so $$ \begin{vmatrix}a&c+e\\b&d+f\end{vmatrix} =\begin{vmatrix}a&c\\b&d\end{vmatrix} + \begin{vmatrix}a&e\\b&f\end{vmatrix} $$