# 行列式の和の性質の証明

$|\begin{array}{cccc}{a}_{11}& {a}_{12}& \cdots & {a}_{1n}\\ ⋮& ⋮& \ddots & ⋮\\ {a}_{t1}+{b}_{t1}& {a}_{t2}+{b}_{t2}& \cdots & {a}_{tn}+{b}_{tn}\\ ⋮& ⋮& \ddots & ⋮\\ {a}_{n1}& {a}_{n2}& \cdots & {a}_{nn}\end{array}|$

$=\sum _{n}\mathrm{sgn}\left(\begin{array}{ccccc}1& \cdots & t& \cdots & n\\ {i}_{1}& \cdots & {i}_{t}& \cdots & {i}_{n}\end{array}\right){a}_{11}·\cdots ·\left({a}_{t{i}_{t}}+{b}_{t{i}_{t}}\right)·\cdots ·{a}_{n{i}_{n}}$

$=\sum _{n}\mathrm{sgn}\left(\begin{array}{ccccc}1& \cdots & t& \cdots & n\\ {i}_{1}& \cdots & {i}_{t}& \cdots & {i}_{n}\end{array}\right)\left({a}_{11}·\cdots ·{a}_{t{i}_{t}}·\cdots ·{a}_{n{i}_{n}}+{a}_{11}·\cdots ·{b}_{t{i}_{t}}·\cdots ·{a}_{n{i}_{n}}\right)$

$=\sum _{n}\mathrm{sgn}\left(\begin{array}{ccccc}1& \cdots & t& \cdots & n\\ {i}_{1}& \cdots & {i}_{t}& \cdots & {i}_{n}\end{array}\right){a}_{11}·\cdots ·{a}_{t{i}_{t}}·\cdots ·{a}_{n{i}_{n}}$$+\sum _{n}\mathrm{sgn}\left(\begin{array}{ccccc}1& \cdots & t& \cdots & n\\ {i}_{1}& \cdots & {i}_{t}& \cdots & {i}_{n}\end{array}\right){a}_{11}·\cdots ·{b}_{t{i}_{t}}·\cdots ·{a}_{n{i}_{n}}$

$=|\begin{array}{cccc}{a}_{11}& {a}_{12}& \cdots & {a}_{1n}\\ ⋮& ⋮& \ddots & ⋮\\ {a}_{t1}& {a}_{t2}& \cdots & {a}_{tn}\\ ⋮& ⋮& \ddots & ⋮\\ {a}_{n1}& {a}_{n2}& \cdots & {a}_{nn}\end{array}|$$+|\begin{array}{cccc}{a}_{11}& {a}_{12}& \cdots & {a}_{1n}\\ ⋮& ⋮& \ddots & ⋮\\ {b}_{t1}& {b}_{t2}& \cdots & {b}_{tn}\\ ⋮& ⋮& \ddots & ⋮\\ {a}_{n1}& {a}_{n2}& \cdots & {a}_{nn}\end{array}|$

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