Mathematics OpenStudy (anonymous):

determine if the vector b is in teh span of the columns of the matrix A A= 1 2 3 4 5 6 7 8 9 b= 10 11 12 OpenStudy (shadowfiend):

I *believe* the way to do this is to see if there is a multiple of each of the column vectors of $$\bf{A}$$ that will produce $$\vec{b}$$. OpenStudy (shadowfiend):

But don't quote me on that! OpenStudy (anonymous):

do you row reduce it? OpenStudy (shadowfiend):

Hm, no a multiple of the sums. So I believe you have to break down the columns into three vectors: \begin{align} \bf{A} = \left[\begin{matrix}1 & 2 & 3\\4 & 5 & 6\\7 & 8 & 9\end{matrix}\right] \end{align} You turn this into three vectors: \begin{align} \vec{a_1} = \left[\begin{matrix}1\\4\\7\end{matrix}\right] & \vec{a_2} = \left[\begin{matrix}2\\5\\8\end{matrix}\right] & \vec{a_3} = \left[\begin{matrix}3\\6\\9\end{matrix}\right] \end{align} OpenStudy (shadowfiend):

Then, if you can find an $$x, y, z$$ such that: $x\vec{a_1} + y\vec{a_2} + z\vec{a_3} = \vec{b}$ Then I believe $$\vec{b}$$ is in the span of the columns of $$\bf{A}$$. OpenStudy (shadowfiend):

You end up with a system of 3 equations, one for each of the rows multiplied by x/y/z, and you can solve to see if there is a solution for x/y/z. OpenStudy (anonymous):

then do u do the agumented of row reduced? OpenStudy (shadowfiend):

I have no idea! Hehe. Like I said, I don't remember *too* much about this, and I've posted more or less everything I do remember. Sorry :/ OpenStudy (anonymous):

okay thank you :)

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