Can a matrix be linearly independent if it is inconsistent?

The answer is yes. Ax = b can be inconsistent in that situation. If the columns are independent, then either the matrix is square, or there are fewer columns than the number of rows. Where there are more columns than the number of rows, the columns cannot be linearly independent.

How do you know if Ax B is inconsistent?

(1) Ax = b is inconsistent iff rank(A ) = rank[A |b ] iff [A |b ] contains a row in which the only nonzero entry lies in the last column, the b column. (2) Ax = b is consistent iff [A |b ] contains no row in which the only nonzero entry lies in the last column.

What does it mean when Ax B is inconsistent?

Then the equation is consistent (see Question 1). Question 4. If the equation Ax = b is inconsistent, then b is not in the set spanned by the columns of A.

How do you check if the columns of a matrix are linearly independent?

Given a set of vectors, you can determine if they are linearly independent by writing the vectors as the columns of the matrix A, and solving Ax = 0. If there are any non-zero solutions, then the vectors are linearly dependent. If the only solution is x = 0, then they are linearly independent.

Can a matrix have linearly independent columns but linearly dependent rows?

Therefore the columns of the row reduced echelon form matrix are linearly dependent. That’s so because all have zero in the same entry,thus have only n-1 “free” variables ,non-zero entries ,thus are n-1 dimensional,and n vectors in only n-1 dimensional space- can’t be linearly independent.

What makes columns linearly independent?

The columns of A are linearly independent if and only if A has a pivot in each column. The columns of A are linearly independent if and only if A is one-to-one. The rows of A are linearly dependent if and only if A has a non-pivot row.

Is ax b consistent for all B?

A consistent system involving free variables will have infinitely many solutions. A has only 6 linearly independent column vectors, so column vectors will span R6. Since the columns vectors span R6, by (2), the system Ax = b will be consistent for any choice of b.

How many solutions will the linear system Ax B have if b is in the column space of A?

one solution
Since the columns of A are linearly independent in the reduced row echelon form of A every column will have a pivot. Therefore the system Ax = b does not have free unknowns, hence it has exactly one solution.

Does Ax B have a solution for every b explain?

The equation Ax = b is solvable for every b. There are n − r = n − m free variables, so there are n − m special solutions to Ax = 0.

Are columns linearly independent?

How do you know if linearly independent?

We have now found a test for determining whether a given set of vectors is linearly independent: A set of n vectors of length n is linearly independent if the matrix with these vectors as columns has a non-zero determinant. The set is of course dependent if the determinant is zero.