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Let A be the s×m matrix, B be the m×n matrix, and X be the vector of n-dimensional unknown columns. Let A be the s×m matrix, B be the m×n matrix, and X be the vector of n-dimensional unknown columns. AB has the same rank as B, i.e. r (AB)=r (B) Let A be the s×m matrix, B be the m×n matrix, and X be the vector of n-dimensional unknown columns. Let A be the s×m matrix, B be the m×n matrix, and X be the vector of n-dimensional unknown columns. AB has the same rank as B, i.e. r (AB)= r (B)
According to the meaning, n column vectors e1, e2,... , En is the solution of Ax=0, and Aei is the ith column vector of A, so A=0 you can use the inverse method
By meaning, n column vectors e1, e2 of n-order identity matrix,... , En is the solution of Ax=0, and Aei is the ith column vector of A, so A=0 you can use the inverse method
By meaning, n column vectors e1, e2 of the n-th-order identity matrix,... , En is the solution of Ax=0, and Aei is the ith column vector of A, so A=0 you can use the inverse method
Suppose that the sum of the elements of each row of the matrix A of order n is zero, and the rank of A is n-1, then the general solution of the linear equation group AX=0 is ______. Suppose that the sum of the elements of row of the matrix A of order n is zero and the rank of A is n-1, then the general solution of the linear equation group AX=0 is ______.
The sum of the elements of each row of the matrix A of order n is zero,
Let (1,1,...,1) T (n column vectors of 1) be a solution of Ax=0,
Since the rank of A is: n-1,
Therefore, the dimension of the basic solution system is: n-r (A),
Therefore, the dimension of the basic solution system of A is 1,
Since (1,1,...,1) T is a solution of the equation and is not 0,
Therefore, the general solution of Ax=0 is: k (1,1,...,1) T.
Suppose A is a matrix of order n, b is a nonzero vector of dimension n, r1, r2 is the solution of the nonhomogeneous linear equation group AX=b, m is the solution of the homogeneous linear equation AX=0. If r1, r2, are not equal, prove that r1, r2, are linearly independent If the rank of A is n-1, prove the linear correlation of m, r1, r2 Suppose A is a matrix of order n, b is a nonzero vector of dimension n, r1, r2 is the solution of the nonhomogeneous linear equation group AX=b, m is the solution of the homogeneous linear equation AX=0. If r1, r2, are not equal, prove that r1, r2, are linearly independent If the rank of A is n-1, it is proved that m, r1, r2 are linearly related
If r1, r2 are linearly related, then r1, r2 are multiple,
If r1=kr2 and r1-r2 is the solution of homogeneous equation, r1-r2=(1-k) r2
So A (1-k) r2=(1-k) Ar2=0 is contradictory to Ar2=b! So it doesn't matter.
If the rank of A is n-1, then e is the basic solution system, so the general solution (c1 and c2 can be represented by e and r1, r2) has
X=c1e+r1, x=c2e+r2 are all valid, and there is (c1-c2) e+r1-r2=0, so the correlation
Let A be a 4*5 matrix with rank (A)=4, then for any 5-dimensional column vector b, the linear equation group AX=b A, there are infinite solutions, B, there is a unique solution, C has no solution, D, I can't be sure, my question is how b can be a 5-dimensional vector, it should be a 4-dimensional vector, because A is 4*5, X is 5*1, so b should be 4*1.
You're right. The question is wrong. B is only a 4-dimensional vector. The answer to this question should be A. The economic math team will help you solve it. Please accept it in time.
Let A be a matrix of 3*5, B be a 3-dimensional column vector, R (A)=3, then is there a solution to the equation group AX=B
Because r (A)= r (A|B), where A|B denotes the augmented matrix of A, the equation must have a solution.
Furthermore, because r (A)<5(the number of unknowns), the equation AX=B has infinite solutions.
Because r (A)= r (A|B), where A|B denotes the augmented matrix of A, the equation must have a solution.
Also, because r (A)<5(the number of unknowns), the equation AX=B has infinite solutions.
It is proved that α1,α2,...,αn are linearly independent if and only if any n-dimensional vector can be linearly represented by them. Let α1,α2,...αn be a set of n-dimensional vectors,
Necessity:α1,α2,...αn is linearly independent, for any n-dimensional vector X, let X=t1 1+t2 2,...+tn n, then the coefficient determinant of the system of equations composed of them is not 0, then through the theory of the system of equations, you can know that the system of equations has a solution, and the solution is unique. Sufficiency: any n-dimensional vector can be...
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