Linear Combination of Vectors

  • Thread starter shan
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  • #1
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The question asks: Write down the vector b as the linear combination of vectors, v1, v2, v3.

To which I got:

b = [tex]\left(\begin{array}{cc}1\\3\\2\end{array}\right)[/tex] = [tex]\left(\begin{array}{cc}1\\-2\\2\end{array}\right)\frac{7}{2} + \left(\begin{array}{cc}0\\1\\-3\end{array}\right)0 + \left(\begin{array}{cc}-1\\4\\-2\end{array}\right)\frac{5}{2}[/tex]

(as A was the matrix made up of those three vectors, and the scalars were the answers from the system Ax=b)

Then the question asks: Determine whether the vector v1 is the linear combination of vectors, v2, v3, b.

Could someone tell me how to find the scalar for vector b?

At the moment I have:

[tex]\left(\begin{array}{cc}0\\1\\-3\end{array}\right)0+ \left(\begin{array}{cc}-1\\4\\-2\end{array}\right)\frac{5}{2}+ \left(\begin{array}{cc}\frac{7}{2}\\0\\\frac{5}{2}\end{array}\right)[/tex]

And I don't know what number to multiply the last vector by...
 

Answers and Replies

  • #2
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Shouldnt the linear combinations look like this?

[tex]
\left(\begin{array}{cc}1\\3\\2\end{array}\right)=
1\left(\begin{array}{cc}1\\0\\0\end{array}\right)+
3\left(\begin{array}{cc}0\\1\\0\end{array}\right)+
2\left(\begin{array}{cc}0\\0\\1\end{array}\right)
[/tex]
 
  • #3
arildno
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Erienion said:
Shouldnt the linear combinations look like this?

[tex]
\left(\begin{array}{cc}1\\3\\2\end{array}\right)=
1\left(\begin{array}{cc}1\\0\\0\end{array}\right)+
3\left(\begin{array}{cc}0\\1\\0\end{array}\right)+
2\left(\begin{array}{cc}0\\0\\1\end{array}\right)
[/tex]
No, he were to express the vector b as a linear combination in terms of the three originally given vectors.
 
  • #4
arildno
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shan:
Just multiply your original (vector) equation with [tex]\frac{2}{7}[/tex] and rearrange your terms.
 
  • #5
dextercioby
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Erenion,u used 3 vectors that could form a basis in [itex]\mathbb{C}^{3} [/itex]...I think this property will be lost,once u try to express one of the basis vectors as a lin.combo.of the other basis vectors & the original vector...

Since the OP's problem dooesn't mention linear independence & completitude,i guess what he's done is correct and he can solve point "b",if he's more careful with his algebra...

Daniel.
 
  • #6
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arildno said:
shan:
Just multiply your original (vector) equation with [tex]\frac{2}{7}[/tex] and rearrange your terms.
hmm but I don't understand why? As I need to determine whether v1 is a linear combination of vectors v2, v3 and b, if I multiplied the original vector equation with [tex]\frac{2}{7}[/tex], wouldn't I change the length of v1?
 
  • #7
arildno
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shan said:
hmm but I don't understand why? As I need to determine whether v1 is a linear combination of vectors v2, v3 and b, if I multiplied the original vector equation with [tex]\frac{2}{7}[/tex], wouldn't I change the length of v1?
What do you mean?
v1 is the vector (1,-2,2), right?
 
  • #8
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arildno said:
What do you mean?
v1 is the vector (1,-2,2), right?
Yeah... ok maybe I'm misunderstanding. By original vector equation, you mean this one?

[tex]\left(\begin{array}{cc}1\\-2\\2\end{array}\right)\frac{7}{2} + \left(\begin{array}{cc}0\\1\\-3\end{array}\right)0 + \left(\begin{array}{cc}-1\\4\\-2\end{array}\right)\frac{5}{2}[/tex]

So you're saying to multiply the above equation by \frac{2}{7} and rearrange to make v1 (1, -2, 2) the subject?
 
  • #9
arildno
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shan said:
Yeah... ok maybe I'm misunderstanding. By original vector equation, you mean this one?

[tex]\left(\begin{array}{cc}1\\-2\\2\end{array}\right)\frac{7}{2} + \left(\begin{array}{cc}0\\1\\-3\end{array}\right)0 + \left(\begin{array}{cc}-1\\4\\-2\end{array}\right)\frac{5}{2}[/tex]

So you're saying to multiply the above equation by \frac{2}{7} and rearrange to make v1 (1, -2, 2) the subject?
I take it you meant:
b=[tex]\left(\begin{array}{cc}1\\-2\\2\end{array}\right)\frac{7}{2} + \left(\begin{array}{cc}0\\1\\-3\end{array}\right)0 + \left(\begin{array}{cc}-1\\4\\-2\end{array}\right)\frac{5}{2}[/tex]

Yes, that's the equation I was referring to..
 
  • #10
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Ah ok, I got it. Thanks for the help :)
 

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