Linear Algebra Vector Question

**qzno** · September 16th 2011, 02:56 PM

1. In the following case, determine whether or not u is in the plane spanned by v and w.

u = $\left(\begin{array}{cc}13\\7\end{array}\right)$ , v = $\left(\begin{array}{cc}-5\\6\end{array}\right)$ , w = $\left(\begin{array}{cc}4\\9\end{array}\right)$

What I was thinking I had to do was do the following:
$\left(\begin{array}{cc}x\\y\end{array}\right)$ = a $\left(\begin{array}{cc}-5\\6\end{array}\right)$ + b $\left(\begin{array}{cc}4\\9\end{array}\right)$

and then from here I would solve for a and b but I have no idea if it's right

Thanks in advance to whoever helps!

**Plato** · September 16th 2011, 03:44 PM

Originally Posted by qzno

1. In the following case, determine whether or not u is in the plane spanned by v and w.

u = $\left(\begin{array}{cc}13\\7\end{array}\right)$ , v = $\left(\begin{array}{cc}-5\\6\end{array}\right)$ , w = $\left(\begin{array}{cc}4\\9\end{array}\right)$

What I was thinking I had to do was do the following:
$\left(\begin{array}{cc}x\\y\end{array}\right)$ = a $\left(\begin{array}{cc}-5\\6\end{array}\right)$ + b $\left(\begin{array}{cc}4\\9\end{array}\right)$

and then from here I would solve for a and b but I have no idea if it's right

Thanks in advance to whoever helps!

One question: Is it true $\left| {\begin{array}{rl} { - 5} & 4 \\ 6 & 9 \\ \end{array} } \right| \ne 0\;?$

AND why is that question significant?

**qzno** · September 16th 2011, 04:06 PM

i dont really understand what you are asking Plato

**Plato** · September 16th 2011, 04:45 PM

Originally Posted by qzno

i dont really understand what you are asking Plato

That may well why you have trouble with the question in the first place.
The question I asked you has everything to do with if this system has a unique system.
$\left\{ \begin{gathered} - 5a + 4b = x \hfill \\ 6a + 9b = y \hfill \\ \end{gathered} \right.$

Does that system have a unique solution?

**qzno** · September 16th 2011, 04:55 PM

i solved for a and b and got really weird numbers so im guessing it doesnt have one.?

**Plato** · September 16th 2011, 05:09 PM

Originally Posted by qzno

i solved for a and b and got really weird numbers so im guessing it doesnt have one.?

Are you really studying linear algebra?
Does $\left[ {\begin{array}{*{20}c} { - 5} & 4 \\ 6 & 9 \\ \end{array} } \right]^{ - 1}$ exist?

If so, then there is a unique solution.

You may need to seek live help from your instructor.

**HallsofIvy** · September 17th 2011, 04:36 AM

[QUOTE=qzno;680875]1. In the following case, determine whether or not u is in the plane spanned by v and w.

u = $\left(\begin{array}{cc}13\\7\end{array}\right)$ , v = $\left(\begin{array}{cc}-5\\6\end{array}\right)$ , w = $\left(\begin{array}{cc}4\\9\end{array}\right)$

What I was thinking I had to do was do the following:
$\left(\begin{array}{cc}x\\y\end{array}\right)$ = a $\left(\begin{array}{cc}-5\\6\end{array}\right)$ + b $\left(\begin{array}{cc}4\\9\end{array}\right)$ [quote]
I see no reason to use "x" and "y" when you are given specific components for u.
Why not $\begin{pmatrix}13 \\ 7\end{pmatrix}= a\begin{pmatrix}-5 \\ 6 \end{pmatrix}+ b\begin{pmatrix} 4 \\ 9 \end{pmatrix}$
which gives you two equations for a and b.

and then from here I would solve for a and b but I have no idea if it's right

Thanks in advance to whoever helps!

By the way, using "\begin{pmatrix}... \end{pmatrix} is a little easier than "\(\begin{array}{cc}... \end{array}\right)"! The "p" is for "parentheses" \begin{bmatrix} gives brackets:

\begin{pmatrix} 4 \\ 9\end{pmatrix}:
$\begin{pmatrix} 4 \\ 9 \end{pmatrix}$

\begin{matrix} 4 \\ 9 \end{pmatrix}:
$\begin{bmatrix} 4 \\ 9 \end{bmatrix}$

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