# Linear algebra plane span

• Sep 15th 2008, 04:41 PM
Linnus
Linear algebra plane span
Let $\displaystyle a_1= \begin{bmatrix} 1 \\ 4 \\ -2 \end{bmatrix}, a_2= \begin{bmatrix} -2 \\ -3 \\ 7 \end{bmatrix}$ and $\displaystyle b= \begin{bmatrix} 4 \\ 1 \\ h \end{bmatrix}$. For what value(s) of h is b in the plane spanned by $\displaystyle a_1$and $\displaystyle a_2$?
• Sep 15th 2008, 05:00 PM
ThePerfectHacker
Quote:

Originally Posted by Linnus
Let $\displaystyle a_1= \begin{bmatrix} 1 \\ 4 \\ -2 \end{bmatrix}, a_2= \begin{bmatrix} -2 \\ -3 \\ 7 \end{bmatrix}$ and $\displaystyle b= \begin{bmatrix} 4 \\ 1 \\ h \end{bmatrix}$. For what value(s) of h is b in the plane spanned by $\displaystyle a_1$and $\displaystyle a_2$?

You are asking for what $\displaystyle h$ does there exists real numbers $\displaystyle a,b$ so that $\displaystyle a\bold{a_1}+b\bold{a_2} = \bold{b}$?

This leads to the equations,
$\displaystyle a-2b = 4$
$\displaystyle 4a -3b = 1$
$\displaystyle -2a+7b = h$

We want this system to be consistent i.e. it has solutions.
Convert this equation into a matrix,
$\displaystyle \begin{bmatrix}1 & -2 & 4 \\ 4 & - 3 & 1 \\ -2 & 7 & h \end{bmatrix}$

Now use Gauss-Jordan elimination to reduce it you should get on the bottom row something like: $\displaystyle [ 0 ~ \ ~ 0 ~ \ ~ \text{ expression of h} ]$.
Then in order for it to be consider that expression of $\displaystyle h$ needs to be equal to zero.