数学试题讲解

证明若方程组

(I) {\begin{matrix} a_{11} y_{1} + a_{12} y_{2} + \dots + a_{1 n} y_{n} = b_{1} \\ a_{21} y_{1} + a_{22} y_{2} + \dots + a_{2 n} y_{n} = b_{2} \\ ⋮ \\ a_{m 1} y_{1} + a_{m 2} y_{2} + \dots + a_{m n} y_{n} = b_{m} \end{matrix}

有解．则方程组

(II) {\begin{array}{r} a_{11} x_{1} + a_{21} x_{2} + \dots + a_{m 1} x_{m} = 0 \\ a_{21} x_{1} + a_{22} x_{2} + \dots + a_{m 2} x_{m} = 0 \\ ⋮ \\ a_{1 n} x_{1} + a_{2 n} x_{2} + \dots + a_{m n} x_{m} = 0 \end{array}

的任意一组基

{(x_{1}, x_{2}, \dots, x_{m})}^{T}

必满足
（III）

b_{1} x_{1} + b_{2} x_{2} + \dots + b_{m} x_{m} = 0

证本题从方程组的不同形式出发加以证明．