Simple Linear Algebra matrix problem

[anniecvc]

Homework Statement

Let a₁ = [1 4 -2] supposed to be a column
Let a₂ = [-2 -3 7] supposed to be a column
Let b = [4 1 h] supposed to be a column

For what value(s) of h is b in the plane spanned by a₁ and a₂?

The Attempt at a Solution

Well, the first thing I did as in 99% of all Linear Alg problems was to put it into an augmented matrix:

[1 -2 l 4]
[4 -3 l 1]
[-2 7 l h]

which became from -4R1 +R2 -> R2:

[1 -2 l 4]
[0 5 l -15]
[-2 7 l h]

and I'm stuck. I could do 2R1+ R3 -> R3, but this doesn't seem to render a correct answer of h= 7/2.

 

[Muphrid]

I think the simpler solution is to construct the whole 3x3 matrix and take the determinant. If b lies in the plane of a1, a2, then the three of them put together no longer span all of 3D space, and the determinant of the matrix will be zero.

 

[I like Serena]

Hi anniecvc!

No need to be stuck.
You can divide row 2 by 5: R2/5 -> R2

Then you can use row 2 to simplify row 1.

From rows 1 and 2 you get the solution for the linear combination.

Substitute those in the equation represented by row 3 and you get h.

 

[HallsofIvy]

Personally, I prefer to do problems like this right from the definitions.
b is in the space spanned by a_1 and a_2, then it is a linear combination of them. That is, there must exist numbers x and y such that
$$x\begin{bmatrix}1 \\ 4\\ -2\end{bmatrix}+ y\begin{bmatrix}-2 \\ -3 \\ 7\end{bmatrix}= \begin{bmatrix}4 \\ 1\\ h\end{bmatrix}$$
which is the same as
$$\begin{bmatrix}x- 2y \\ 4x- 3y \\ -2x+ 7y\end{bmatrix}= \begin{bmatrix}4 \\ 1 \\ h\end{bmatrix}$$

which, in turn, is equivalent to the three equations x- 2y= 4, 4x- 3y= 1, -2x+7y= h.

Solve the first two equations for x and y and put those values into the third to find h.

 

[ougoah]

If the required answer is 7/2, it is wrong, as can be checked easily. What you described initially does work.

 

[anniecvc]

Ougoah, you're right, the answer is h=-17, which using Serena's help comes out nicely. Thank you! Ivy, that does work too but I think defeats the purpose of being in a linear algebra class rather than an Alg 2 class. Still, thanks for the insight!

 

[I like Serena]

You're welcome!