定数倍の性質の証明

$A=\left(\begin{array}{cccc}{a}_{11}& {a}_{12}& \cdots & {a}_{1n}\\ ⋮& ⋮& \ddots & ⋮\\ c{a}_{t1}& c{a}_{t2}& \cdots & c{a}_{tn}\\ ⋮& ⋮& \ddots & ⋮\\ {a}_{n1}& {a}_{n2}& \cdots & {a}_{nn}\end{array}\right)$

$B=\left(\begin{array}{cccc}{b}_{11}& {b}_{12}& \cdots & {b}_{1n}\\ ⋮& ⋮& \ddots & ⋮\\ {b}_{t1}& {b}_{t2}& \cdots & {b}_{tn}\\ ⋮& ⋮& \ddots & ⋮\\ {b}_{n1}& {b}_{n2}& \cdots & {b}_{nn}\end{array}\right)$$=\left(\begin{array}{cccc}{a}_{11}& {a}_{12}& \cdots & {a}_{1n}\\ ⋮& ⋮& \ddots & ⋮\\ c{a}_{t1}& c{a}_{t2}& \cdots & c{a}_{tn}\\ ⋮& ⋮& \ddots & ⋮\\ {a}_{n1}& {a}_{n2}& \cdots & {a}_{nn}\end{array}\right)$

とおく．

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

となる．これを以下のように式変形をする．

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

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

$=c|A|$

$|B|=c|A|$

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