Question

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

Answer 1

Answer:

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

$B= \{\left[\begin{array}{c}-1\\-2\\3\end{array}\right], \left[\begin{array}{c}-1\\3\\-3\end{array}\right], \left[\begin{array}{c}2\\-1\\0\end{array}\right] \}$

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

$A=\left[\begin{array}{ccc}-1&-1&2\\-2&3&-1\\3&-3&0\end{array}\right]$

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:

$\left[\begin{array}{ccc}-1&-1&2\\0&5&-5\\0&-6&6\end{array}\right]$

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

$\left[\begin{array}{ccc}-1&-1&2\\0&5&-5\\0&0&0\end{array}\right]$

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.

$5y-5z=0\\y=z$

2.

$-x-y+2z=0\\-x-z+2z=0\\x=z$

and the set the solutions of the system $Ax=0$ is

$\{(x,y,z)=(t,t,t)=t(1,1,1): t\in\mathbb{R}\}$

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

$\{(1,1,1)\}$