# ベクトルの1次結合の計算

${x}_{1}\left(\begin{array}{c}{a}_{11}\\ {a}_{21}\\ ⋮\\ {a}_{n1}\end{array}\right)+{x}_{2}\left(\begin{array}{c}{a}_{12}\\ {a}_{22}\\ ⋮\\ {a}_{n2}\end{array}\right)+\cdots +{x}_{m}\left(\begin{array}{c}{a}_{1m}\\ {a}_{2m}\\ ⋮\\ {a}_{nm}\end{array}\right)$

(∵ 行列のスカラー倍列ベクトルは行列に含まれる)

$=\left(\begin{array}{c}{a}_{11}{x}_{1}\\ {a}_{21}{x}_{1}\\ ⋮\\ {a}_{n1}{x}_{1}\end{array}\right)+\left(\begin{array}{c}{a}_{12}{x}_{2}\\ {a}_{22}{x}_{2}\\ ⋮\\ {a}_{n2}{x}_{2}\end{array}\right)+\cdots +\left(\begin{array}{c}{a}_{1m}{x}_{m}\\ {a}_{2m}{x}_{m}\\ ⋮\\ {a}_{nm}{x}_{m}\end{array}\right)$

(∵ 行列の和)

$=\left(\begin{array}{c}{a}_{11}{x}_{1}+{a}_{12}{x}_{2}+\cdots +{a}_{1m}{x}_{m}\\ {a}_{21}{x}_{1}+{a}_{22}{x}_{2}+\cdots +{a}_{2m}{x}_{m}\\ ⋮\\ {a}_{n1}{x}_{1}+{a}_{n2}{x}_{2}+\cdots +{a}_{nm}{x}_{m}\end{array}\right)$

$=\left(\begin{array}{cccc}{a}_{11}& {a}_{12}& \cdots & {a}_{1m}\\ {a}_{21}& {a}_{22}& \cdots & {a}_{2m}\\ ⋮& ⋮& \ddots & ⋮\\ {a}_{n1}& {a}_{n2}& \cdots & {a}_{nm}\end{array}\right)\left(\begin{array}{c}{x}_{1}\\ {x}_{2}\\ ⋮\\ {x}_{m}\end{array}\right)$

$=\left({a}_{1},{a}_{2},\cdots ,{a}_{m}\right)\left(\begin{array}{c}{x}_{1}\\ {x}_{2}\\ ⋮\\ {x}_{m}\end{array}\right)$

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