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so I'm trying to prove by setting up a augmented matrix with vector b [3 2] is in col(A). I am trying to find if vector b spans col(A).

$$A =\begin{bmatrix} 1 & 0 & -1 \\1 & 1& 1 \end{bmatrix}$$

so the augmented matrix is:

$$\left[\begin{array}{rrr|r} 1 & 0 & -1 & 3\\ 1 & 1 & 1 & 2 \end{array}\right]$$

so what I start by is row reducing and get:

$$\left[\begin{array}{rrr|r} 1 & 0 & -1 & 3\\ 0 & 1 & 2 & -1 \end{array}\right]$$

but from here it looks like I can't get linear independence, but my book says this is in col(A). Can someone show me where I am going wrong?

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  • $\begingroup$ What is A and respectively col(A) ? $\endgroup$ – brick Nov 5 '14 at 23:25
  • $\begingroup$ @brick A is the matrix left of the 2 " | " to see if the matrix is a basis I must see if vector b is in the columns of A. $\endgroup$ – Joshhw Nov 5 '14 at 23:26
  • $\begingroup$ If you trying to prove that $(3, 2)$ is a linear combination of the columns of $А$, then it's obvious since the first and the second column are linearly independent and therefore form a basis of $\mathbb{R}^2$. $\endgroup$ – brick Nov 5 '14 at 23:34
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I'm not sure what linear independence has to do with anything here. You've shown that every row has a pivot, so $\text{Col}(A)$ spans all of $\mathbb R^2$, including $b$. Indeed, we can explicitly express $b$ as a linear combination of the original columns of $A$ by inspection: $$ \begin{bmatrix} 3 \\ 2 \end{bmatrix} = 3\begin{bmatrix} 1 \\ 1 \end{bmatrix} + (-1) \begin{bmatrix} 0 \\ 1 \end{bmatrix} + 0 \begin{bmatrix} -1 \\ 1 \end{bmatrix} $$

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  • $\begingroup$ I'm trying to find out if vector v is in the basis for the matrix A $\endgroup$ – Joshhw Nov 5 '14 at 23:31

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