# Thread: Matrices - Linear equations

1. ## Matrices - Linear equations

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?

2. The above image seems a bit hard to read.
Here is another question similar in format, though, and it's clearer:

It's part of our matrices unit but I'm not sure how to incorporate that into this question.

3. 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...

$\displaystyle \left[\begin{matrix}0 & 0 & 0 & 1 \\ 1 & 1 & 1 & 1\\ 27 & 9 & 3 & 1 \\ 64 & 16 & 4 & 1\end{matrix}\right]\left[\begin{matrix}a \\ b\\ c\\d \end{matrix}\right] = \left[\begin{matrix}\phantom{-}10 \\ \phantom{-}7 \\ -11 \\ -14\end{matrix}\right]$

4. .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.

5. .

6. [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.

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.

7. 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

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

8. 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!

9. 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.

10. 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.