# Matrices - Linear equations

• Jan 11th 2011, 05:43 PM
TN17
Matrices - Linear equations
http://i1097.photobucket.com/albums/...s_/Problem.jpg

I don't quite understand how to use that augmented matrix to prove anything about a, b and c.
Any ideas on where to start?
• Jan 11th 2011, 07:40 PM
TN17
The above image seems a bit hard to read.
Here is another question similar in format, though, and it's clearer:
http://i1097.photobucket.com/albums/...sics_/help.jpg

It's part of our matrices unit but I'm not sure how to incorporate that into this question.
• Jan 11th 2011, 08:09 PM
Prove It
25. You know four points that lie on the curve, $\displaystyle (0, 10), (1, 7), (3, -11), (4, -14)$.

So that means they all satisfy the equation $\displaystyle y = a\,x^3 + b\,x^2 + c\,x + d$.

So
$\displaystyle 10 = a\cdot 0^3 + b\cdot 0^2 + c\cdot 0 + d$

$\displaystyle 7 = a\cdot 1^3 + b\cdot 1^2 + c\cdot 1 + d$

$\displaystyle -11 = a\cdot 3^3 + b\cdot 3^2 + c\cdot 3 + d$

$\displaystyle -14 = a\cdot 4^3 + b\cdot 4^2 + c\cdot 4 + d$.

So now you can set up your matrix equation...

$\displaystyle \left[\begin{matrix}0 & 0 & 0 & 1 \\ 1 & 1 & 1 & 1\\ 27 & 9 & 3 & 1 \\ 64 & 16 & 4 & 1\end{matrix}\right]\left[\begin{matrix}a \\ b\\ c\\d \end{matrix}\right] = \left[\begin{matrix}\phantom{-}10 \\ \phantom{-}7 \\ -11 \\ -14\end{matrix}\right]$
• Jan 11th 2011, 08:25 PM
TN17
.I usually write the constants in the same brackets as the large matrix here (to the furthest right side of the matrix, since that's what we were taught). So:

0 0 0 1 10
1 1 1 1 7
27 9 3 1 -11
64 16 4 1 -14
Is that right?

Would I use Gauss-Jordan to reduce that into reduced row-echelon?
I'm still confused about how to find the variables from there.
• Jan 11th 2011, 08:38 PM
TN17
.
• Jan 12th 2011, 07:44 AM
HallsofIvy
[QUOTE=TN17;604092].I usually write the constants in the same brackets as the large matrix here (to the furthest right side of the matrix, since that's what we were taught). So:

0 0 0 1 10
1 1 1 1 7
27 9 3 1 -11
64 16 4 1 -14
Is that right?/[quote]
Yes, that is what is meant by the "augmented" matrix.

Quote:

Would I use Gauss-Jordan to reduce that into reduced row-echelon?
I'm still confused about how to find the variables from there.
Yes, you certainly could use Gauss-Jordan. I would be inclined to swap the first row to the last initially to get
$\begin{bmatrix}1 & 1 & 1 & 1 & 7 \\ 27 & 9 & 3 & 1 & -11\\ 64 & 16 & 4 & 1 & -14 \\ 0 & 0 & 0 & 1 & -14\end{bmatrix}$
Then subtract 27 times the first row from the second, 64 times the first row from the third, etc.
• Jan 12th 2011, 07:31 PM
Prove It
If you have a CAS, it's quite simple if you know that the solution to the matrix equation

$\displaystyle \mathbf{A}\mathbf{x} = \mathbf{b}$

is

$\displaystyle \mathbf{x} = \mathbf{A}^{-1}\mathbf{b}$...
• Jan 13th 2011, 09:22 AM
HallsofIvy
Quote:

Originally Posted by Prove It
If you have a CAS, it's quite simple if you know that the solution to the matrix equation

$\displaystyle \mathbf{A}\mathbf{x} = \mathbf{b}$

is

$\displaystyle \mathbf{x} = \mathbf{A}^{-1}\mathbf{b}$...

On the other hand, if you want to actually learn some linear algebra there is no better way than doing it by hand!
• Jan 13th 2011, 02:28 PM
Prove It
Quote:

Originally Posted by HallsofIvy
On the other hand, if you want to actually learn some linear algebra there is no better way than doing it by hand!

I disagree, using matrix algebra to solve matrix equations and then programming a computer/CAS to perform the tedious calculations that you are almost certain to make a mistake with is far more beneficial.
• Jan 13th 2011, 02:40 PM
pickslides
I'll meet you guys halfway. To maximise understanding the OP should know how to solve a system like the one above by hand. Once confident with such methods then use technology to save time.