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Let $\mathbf{u}=\left[\begin{array}{r}{2} \\ {-3} \\ {2}\end{array}\right]$ and $A=\left[\begin{array}{rrr}{5} & {8} & {7} \\ {0} & {1} & {-1} \\ {1} & {3} & {0}\end{array}\right] .$ Is u in the subset of $\mathbb{R}^{3}$ spanned by the columns of $A ?$ Why or why not?

$u$ is not a subset of $\mathbb{R^{3}}$ spanned by the column of $A$

Algebra

Chapter 1

Linear Equations in Linear Algebra

Section 4

The Matrix Equation Ax D b

Introduction to Matrices

Campbell University

Harvey Mudd College

Lectures

01:32

In mathematics, the absolu…

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now to determine if something is in the span of the columns of some matrix. Um, that's just a fancy way of saying, Is there some solution or some vector X? When we multiply this, we end up, um, getting you out of it. And one way we can do this is by signing up in Augmented Matrix and Row reducing it. So let's go ahead and do that. 58701 negative 1130 And then over here on the right, this will be to negative 32 Now, if you can use a calculator to help you out here, um and you can just plug this in and do the road reduced echelon form of it. You actually end up with this matrix. So 1030 01 negative 100001 And from this notice that we end up with this saying zero is equal toe one, which is definitely not true. So but this implies is we have and inconsistent system, which means there is no solution. So there is no vector X where if we multiply this by our matrix A, we end up getting you And so this also implies further is that you is not in the span of the column of a. So this is going to be what our conclusion ends up big. Um, but if for whatever reason you're not allowed to just use a calculator on, you actually have to do the road reduction by hand. We'll do that as well. Now, normally, we would try to get it. Kind of like what they have here. They have, like, this upper triangular form. Um, you could also put into the lower triangular form without any trouble. Um, and the reason why I'm going to actually do this in the upper I mean, the lower versus the upper. Because notice that we have a one here and we have a one here so we could just clear everything above it. And then it gives us that nice lower triangular matrix. You could do it Either way. It doesn't really matter. You'll get the same answer regardless. I also just don't want to divide that first row by five to get fractions because I try to stay as far away from fractions, if I can. Yeah, so let's go ahead and clear this five out here, so we would need to do, uh, negative five row three plus Rogue one. And so we're only changing broad one. So five become zero negative. 15 plus eight is going to be, um, negative. Seven zero times sank to five. That's just zero. So that's seven. And then two times negative five is negative. 10. Add that to to that would be negative eight. And then the rest of the matrix stays the same. So 01 negative one negative. Three, 30 030101032 And now notice that if we were to multiply this second row, we can cancel this seven. Um, if we multiply the second row by seven, I think didn't say what I was doing s 07 row two plus row one. So this is going to become zero that 10 in the negative one time. Seven that zero. And when we add that to seven and then we get negative 21 plus negative eight, which is going to be negative 29 and then the rest of the matrix stays the same. So you'll see we get something slightly different from the reduced form. But we still end up with our contradiction here because this says zero is equal to negative 29 which again, inconsistent system implies you is not in the span of the columns of a because remember, Like I was saying before, this is just a fancy way of saying, Is there a solution to this matrix equation here? And if you get a inconsistent system, that means there is no solution.

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