# Matrix Vector Transpose

1. Mar 15, 2012

### Ruminator

To find the least squares polynomial of degree 2 to
approximate points (X,Y) given in the table

X_____________Y

1_____________36
1.9_____________-49
2.9_____________23
4_____________-38
5.1_____________-14

the normal equations to be solved have the form
A c = v, where A is the matrix (NOT NECESSARILY DISPLAYED TO THE
PRECISION USED FOR THE CALCULATIONS)
_ _
| |
| _____ 1______ _____ 2______ _____ 3______ |
| _____ 4______ _____ 5______ _____ 6______ |
| _____ 7______ _____ 8______ _____ 9______ |
|_ _|

and c is the coefficient vector for the polynomial;
and v is the transpose of the vector:
_ _
| |
| _____10______ _____11______ _____12______ |
|_ _|

The solution vector c (FOUND USING DOUBLE PRECISION
CALCULATIONS with the ORIGINAL DATA above) is
the transpose of vector:
_ _
| |
| _____13______ _____14______ _____15______ |
|_ _|

2. Relevant equations

Ac=v

3. The attempt at a solution

I believe I've got the first part; i.e. solving the matrix:

5.1________14.9_________55.03
14.9_______55.03________228.899
55.03______228.899______1017.2803

Now, I'm clueless as to how to proceed from here on.