a是Ax=b的解，b不为0，a1，a2，……，an-r是Ax=0的一组基，证（1）a，a1，a2，……，an-r线性无关

1)证明线性无关一般用定义就可以
假设
k0*a + k1*a1 + ...+ kn*(an-r) = 0
两边左乘A
A*k0*a + A*k1*a1+...+A*kn*(an-r) = 0
k0,k1都是数，乘法可以交换，有
k0*Aa + k1*Aa1 +...+ kn*A(an-r) = 0
由已知，Aa = b,Aki = 0(i = 1,2,...n)
上式变为
k0*b = 0，且b不为0，则只有k0 = 0
代入最开始第一个式子
k1*a1 + ...+ kn*(an-r) = 0
因为a1,a2,...,an-r 是一组基，一定是线性无关的，所以必有
k1 = k2 =...= kn = 0
综上，k0 = k1 = k2 =... =kn =0
a，a1，a2，……，an-r线性无关

第二问方法类似，就不详细写了
3）由于a1，a2，……，an-r是Ax=0的一组基，且a是Ax=b的解，根据非线性方程组解的结构由齐次方程的解加上特解组成，因此Ax=b的解为
c1*a1 + c2*a2 + ...+ cn*(an-r) + a(c1,c2,..,cn是任意常数)

Dear Wang Fei，
I'd like to tell you something about.

Dear Wang Fei，
I'd like to tell you something about.
I took part in the English Corner held by my school last month. The attdance included some teachers and students.
There were various kinds of activies, such as reading English newspaper, practing the oral English, learning English songs.
In m