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.Ifa1=[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

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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.Ifa1=[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
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

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.Ifa1=[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
(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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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.Ifa1=[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 线性代数特征值和特征向量证明题Let u be an eigenvector of A corresponding to an eigenvalue p ,and let H be the line in R^n through u and the origin .Explain why H is variant under A in the sense that Ax is in H whenever x is in H .Let K let`s all be a Let us all be a family 线性代数的可对角化证明题~Let A be a 4*4 matrix , prove that if A has 4 linearly independent eigenvectors, so does A^T证明：A是可对角化的, 存在 P·α·P^-1 A P=D 然后可逆然后就不知道了~ 线性代数可对角化的证明题~Let A be a 4*4 matrix ,prove that if A has 4 linearly independent eigenvectors,so does A^T证明：A是可对角化的,存在 P·α·P^-1 A P=D然后可逆P·α 是哪儿来的~ 一道线性代数题（英语）Let A be an n by n matrix with eigenvalues (including multiplicities) -1,-1,4,4,4.1.What is 2.The dimension of the row space of A is:3.The eigenvalues of the matrix A2 of A are:4.Is A invertible?5.The dimension of the 线性代数计算n阶行列式x a ...aa x ...a.........a a ...x 线性代数矩阵A~ 线性代数中矩阵A, |a|线性代数中是什么意思线性代数中rank(A, 线性代数R(a) 线性代数 R(A) 线性代数：若r(A) let A be a 4*4 matrix with the characteristic equation(1-λ^4).determine if A is diagonalizable. .Let A and B be vector subspaces of a vector space V .The intersection of A and B,A ∩ B,is the.Let A and B be vector subspaces of a vector space V .The intersection of A and B,A ∩ B,is the set {x ∈ V | x ∈ A and x ∈ B}.The union of A and B, There is a kind of love be called to let