if one of the rows or columns of the matrix is a linear combination of the other rows or columns then the determinate of the matrix is 0

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Posted on 2004-04-12

hi, all,

Does anybody know how to prove that for the nxn matrix, if rank(A)<n, then det(A)=0;

any help will be highly appreciated.

Chris

Does anybody know how to prove that for the nxn matrix, if rank(A)<n, then det(A)=0;

any help will be highly appreciated.

Chris

2 Comments

Comment Utility

If rank(A) < n then one of the rows or columns of the matrix is a linear combination of the other rows or columns

if one of the rows or columns of the matrix is a linear combination of the other rows or columns then the determinate of the matrix is 0

if one of the rows or columns of the matrix is a linear combination of the other rows or columns then the determinate of the matrix is 0

Comment Utility

ozo,

thanks for the answer, I think I got the answer already.

Since the determinant of a matrix has such properties as follows.

1. any row or column of a matrix added to one row or column, the determinant will not change.

2. if matrix B is one row or column of a matrix A multiply a constant k, then the det(B)=k*det(A);

3. if any row or column of a matrix A is 0, then det(A)=0;

And if rank(A)<n, then we must have at least two rows or columns are linearly dependant, it means k*Ai+Aj=0(Ai and Aj are the vectors of matrix A).

So if we do this operation on matrix A to get a matrix B in which one vector is 0, then we can have det(A)=k*det(B)=0. Then we got it proved.

thanks for the answer, I think I got the answer already.

Since the determinant of a matrix has such properties as follows.

1. any row or column of a matrix added to one row or column, the determinant will not change.

2. if matrix B is one row or column of a matrix A multiply a constant k, then the det(B)=k*det(A);

3. if any row or column of a matrix A is 0, then det(A)=0;

And if rank(A)<n, then we must have at least two rows or columns are linearly dependant, it means k*Ai+Aj=0(Ai and Aj are the vectors of matrix A).

So if we do this operation on matrix A to get a matrix B in which one vector is 0, then we can have det(A)=k*det(B)=0. Then we got it proved.

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