线性代数,Let A be a 4 x 5 matrix and let U be the reduced row ec

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线性代数,Let A be a 4 x 5 matrix and let U be the reduced row echelon form of A.
Let A be a 4 x 5 matrix and let U be the reduced row echelon form of A.If
a1=[2; 1; -3; -2] and a2=[-1; 2; 3; 1],U=[1 0 2 0 -1; 0 1 3 0 -2; 0 0 0 1 5; 0 0 0 0 0],
where each semicolon represents a new line
(a) find a basis for N(A)
(b) given that x0 is a solution of Ax=b,where
b=[0; 5; 3; 4] and x0=[3; 2; 0; 2; 0]
(i) find all solutions to the system
(ii) determine the remaining column vectors of A

(a) 因为A的行最简形 U=
1 0 2 0 -1
0 1 3 0 -2
0 0 0 1 5
0 0 0 0 0
所以 N(A) 的一组基为 [-2;-3;1;0;0],[1;2;0;-5;1]
(b) (i)
Ax=b 的通解为:[3; 2; 0; 2; 0]+c1[-2;-3;1;0;0]+c2[1;2;0;-5;1]
(ii)
由于x0是Ax=b的解,所以有 3a1+2a2+2a4=b
所以 a4=(1/2)(b-3a1-2a2)=[-2;-1;3;4]
由(a),[-2;-3;1;0;0],[1;2;0;-5;1] 是Ax=0的解
所以 -2a1-3a2+a3=0,a1+2a2-5a4+a5=0
所以 a3=2a1+3a2=[1;8;3;-1],a5=-a1-2a2+5a4=[-10;-10;12;20]
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