Finding the Scalar for a Vector in a Linear Combination

[shan]

The question asks: Write down the vector b as the linear combination of vectors, v1, v2, v3.

To which I got:

b = \left(\begin{array}{cc}1\\3\\2\end{array}\right) = \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}

(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:

\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)

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

 

[Erienion]

Shouldnt the linear combinations look like this?

<br /> \left(\begin{array}{cc}1\\3\\2\end{array}\right)=<br /> 1\left(\begin{array}{cc}1\\0\\0\end{array}\right)+<br /> 3\left(\begin{array}{cc}0\\1\\0\end{array}\right)+<br /> 2\left(\begin{array}{cc}0\\0\\1\end{array}\right)<br />

 

[arildno]

Shouldnt the linear combinations look like this?

<br /> \left(\begin{array}{cc}1\\3\\2\end{array}\right)=<br /> 1\left(\begin{array}{cc}1\\0\\0\end{array}\right)+<br /> 3\left(\begin{array}{cc}0\\1\\0\end{array}\right)+<br /> 2\left(\begin{array}{cc}0\\0\\1\end{array}\right)<br />

No, he were to express the vector b as a linear combination in terms of the three originally given vectors.

 

[arildno]

shan:
Just multiply your original (vector) equation with \frac{2}{7} and rearrange your terms.

 

[dextercioby]

Erenion,u used 3 vectors that could form a basis in \mathbb{C}^{3}...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.

 

[shan]

shan:
Just multiply your original (vector) equation with \frac{2}{7} 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 \frac{2}{7}, wouldn't I change the length of v1?

 

[arildno]

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 \frac{2}{7}, wouldn't I change the length of v1?

What do you mean?
v1 is the vector (1,-2,2), right?

 

[shan]

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?

\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}

So you're saying to multiply the above equation by \frac{2}{7} and rearrange to make v1 (1, -2, 2) the subject?

 

[arildno]

Yeah... ok maybe I'm misunderstanding. By original vector equation, you mean this one?

\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}

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=\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}

Yes, that's the equation I was referring to..

 

[shan]

Ah ok, I got it. Thanks for the help :)