Find a linearly independent set of vectors that spans the same subspace of r3 as that spanned by the vectors ⎡⎣⎢−1−23⎤⎦⎥, ⎡⎣⎢−13−3⎤⎦⎥, ⎡⎣⎢2−10⎤⎦⎥. a linearly independent spanning set for the subspace is: let a = describe all solutions of ax = 0. find a 3 times 3 matrix a such that

First, we need to know what is the dimension of the set of vectors

For that, we find the echelon form of the matrix we columns the vectors of the set.

Now, we use row operations

1. To row two of matrix A we subtract row 1 twice and to row three we add row 1 3 times. And we obtain the matrix:

2. To the third row of the matrix that we obtained in the previous step we added 6/5 times row two and we obtain the matrix

The echelon form has one free variable then the dimension of the set is one.

And observe that using backward substitution we have that

1.

2.

and the set the solutions of the system is

and therefore a linear independent set of vectors that spans the same subspace as that spanned by B is

answer is 12345678900987654321234567890987654321

step-by-step explanation:

answer: the answer is fre sha voca do

step-by-step explanation: