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Let $\mathbf{u}=\left[\begin{array}{r}{2} \\ {-1}\end{array}\right]$ and $\mathbf{v}=\left[\begin{array}{l}{2} \\ {1}\end{array}\right] .$ Show that $\left[\begin{array}{l}{h} \\ {k}\end{array}\right]$ is inSpan $\{\mathbf{u}, \mathbf{v}\}$ for all $h$ and $k$

For nill $h$ and $k , \left[ \begin{array} { l } { h } \\ { k } \end{array} \right]$ is in $\operatorname { Span } \{ \mathbf { u } , v \}$

Algebra

Chapter 1

Linear Equations in Linear Algebra

Section 3

Vector Equations

Introduction to Matrices

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in this exercise we want to determine if this vector HK were H and K are any real numbers is in the span of you envy provide here? Let's start off by saying that HK is in span You ve if and only if h and K as a vector is a linear combination of you envy Algebraic Lee That would be if and only if x one times you plus x two times V is equal to HK So if his victory system is consistent then we'll say yes. HK is in the span so to determine if this victory questions consistent, we can go toe undocumented matrix where the first column is going to be you to native one calling to his V which is 21 and then we augment with the right hand side, which is HK Now. When we roll reduce we only have to go to echelon form But let me first take the falling row operation where we inter chains rose ones and two So my new row one is negative 11 k and Row two is now to to h now This way we won't have to divide by two making this entry of one and forcing a fraction over here. So next let's take this is our pivot entry and eliminate the entry down below. We could multiply row one by to add it to Row two, and if we copy Row one, the result would be zero four and hpe plus two K. Now let's analyze the pivots. We have a pivot and calm one, a pivot and column too. And so it follows that this is not a pivot column and the values of H and K don't not make that determination because of the non zero entry here. So that means that this system is consistent next one ew plus X two v equals H K is consistent for all h and K in our and so that means you envy is a linear combination. Very Excuse me, H and K is a linear combination of you envy. And so that means that HK is in the span of the vectors you envy and this holds for all values of H and K. Since our vector HK here was left in arbitrary form

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