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adoni [48]
3 years ago
7

If u, v, and w are nonzero vectors in r 2 , is w a linear combination of u and v?

Mathematics
1 answer:
Tju [1.3M]3 years ago
8 0
Not necessarily. \mathbf u and \mathbf v may be linearly dependent, so that their span forms a subspace of \mathbb R^2 that does not contain every vector in \mathbb R^2.

For example, we could have \mathbf u=(0,1) and \mathbf v=(0,-1). Any vector \mathbf w of the form (r,0), where r\neq0, is impossible to obtain as a linear combination of these \mathbf u and \mathbf v, since

c_1\mathbf u+c_2\mathbf v=(0,c_1)+(0,-c_2)=(0,c_1-c_2)\neq(r,0)

unless r=0 and c_1=c_2.
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3a+2x-3y=15<br>Solve for a.<br>Then Solve put the value for a =<br>a+7÷10​
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\huge\bf Question:–

\sf \longmapsto \: 3a+2x−3y=15

\bf \huge \: To  \: Find:–

\boxed{\bf \: Value\: of  \: A}

\huge\bf Solution:–

\sf \longmapsto \: 3a+2x−3y=15

\boxed{ \bf \: Add -2x  \: to  \: both  \: sides}

\sf \longmapsto \: 3a+2x−3y+−2x=15+−2x

\sf \longmapsto \: 3a−3y=−2x+15

\boxed{ \bf \: Add  \: 3y  \: to \:  both  \: sides}

\sf \longmapsto \: 3a−3y+3y=−2x+15+3y

\sf \longmapsto \: 3a=−2x+3y+15

\boxed{\bf \:  \: Divide  \: both  \: sides \:  by \:  3}

\sf \longmapsto \:  \dfrac{3a}{3} =  \dfrac{−2x+3y+15}{3}

\boxed{\bf \:  Cross \: Multiply}

\boxed{\sf \longmapsto \: a =  \dfrac{ - 2}{3} x + y + 5}

______________________________________

\bf \: Put\:The\: Value

\sf \longmapsto \: \dfrac{−2x+3y+15}{3}  +7÷10

\boxed{\bf \: Distribute}

\sf \longmapsto \: \dfrac{ - 2}{3} x+y+5+ \dfrac{7}{10}

\boxed{\bf \: Combine \:  Like \:  terms}

\sf \longmapsto \: \bigg( \dfrac{ - 2}{3} x\bigg) + y +\bigg( 5 +  \dfrac{7}{10} \bigg)

\sf \longmapsto \:  \dfrac{ - 2}{3} x + y +  \dfrac{57}{10}

______________________________________

\boxed{\bf The Answer\: is:–}

\boxed{{\underline{\bf\dfrac{ - 2}{3} x + y +  \dfrac{57}{10}} }}

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