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?

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- Jan 11th 2011, 05:43 PMTN17Matrices - 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 PMTN17
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 ourbut I'm not sure how to incorporate that into this question.__matrices unit__ - Jan 11th 2011, 08:09 PMProve It
25. You know four points that lie on the curve, $\displaystyle \displaystyle (0, 10), (1, 7), (3, -11), (4, -14)$.

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

So

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

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

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

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

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

$\displaystyle \displaystyle \left[\begin{matrix}0 & 0 & 0 & 1 \\ 1 & 1 & 1 & 1\\ 27 & 9 & 3 & 1 \\ 64 & 16 & 4 & 1\end{matrix}\right]\left[\begin{matrix}a \\ b\\ c\\d \end{matrix}\right] = \left[\begin{matrix}\phantom{-}10 \\ \phantom{-}7 \\ -11 \\ -14\end{matrix}\right]$ - Jan 11th 2011, 08:25 PMTN17
.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 PMTN17
.

- Jan 12th 2011, 07:44 AMHallsofIvy
[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.

$\displaystyle \begin{bmatrix}1 & 1 & 1 & 1 & 7 \\ 27 & 9 & 3 & 1 & -11\\ 64 & 16 & 4 & 1 & -14 \\ 0 & 0 & 0 & 1 & -14\end{bmatrix}$

Then subtract 27 times the first row from the second, 64 times the first row from the third, etc. - Jan 12th 2011, 07:31 PMProve It
If you have a CAS, it's quite simple if you know that the solution to the matrix equation

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

is

$\displaystyle \displaystyle \mathbf{x} = \mathbf{A}^{-1}\mathbf{b}$... - Jan 13th 2011, 09:22 AMHallsofIvy
- Jan 13th 2011, 02:28 PMProve It
- Jan 13th 2011, 02:40 PMpickslides
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.