# Column space

## Homework Statement

Does b = [ 2 15 ]T lie in the column of the matrix A

[1 -3]
[2 5]

## Homework Equations

a basis of CS(U) forms a basis for the corresponding columns for CS(A)

## The Attempt at a Solution

I reduced the given matrix into row echelon form and determined its column space. But did not figure out if [2 15]T lies in the column space of A.

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Your matrix A reduces to the identity matrix in reduced row echelon form; so then the column space is made up of all the columns of the original matrix;

$$\text{Col}(A)=\left\{ \begin{pmatrix} 1 \\ -3 \end{pmatrix} , \begin{pmatrix} 2 \\ 5 \end{pmatrix} \right\}$$

So does the vector they're asking lie in that space? In other words is it a linear combination of those vectors in the space?

Is the

$$\text{Col}(A)=\left\{ \begin{pmatrix} 1 \\ 2 \end{pmatrix} , \begin{pmatrix} -3 \\ 5 \end{pmatrix} \right\}$$

or

$$\text{Col}(A)=\left\{ \begin{pmatrix} 1 \\ -3 \end{pmatrix} , \begin{pmatrix} 2 \\ 5 \end{pmatrix} \right\}$$
?
The given matrix is

$$\begin{pmatrix} 1; -3\\ 2; 5 \end{pmatrix}$$

thanks, i resolved it!