SOLUTION: Use the matrix method to write the particular equation of a function in the form y=ax^2+bx+c Containing (0,5) (2,13) and (3,26)

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Question 1125911: Use the matrix method to write the particular equation of a function in the form y=ax^2+bx+c
Containing (0,5) (2,13) and (3,26)

Found 2 solutions by MathLover1, greenestamps:
Answer by MathLover1(20850) About Me  (Show Source):
You can put this solution on YOUR website!
Use the matrix method to write the particular equation of a function in the form y=ax%5E2%2Bbx%2Bc
Containing (0,5) (2,13) and (3,26)

---------------
Quadratic Form:
y+=+ax%5E2+%2B+bx+%2B+c
Substitute each pair of x,y values then solve for a,b,c
y+=+ax%5E2+%2B+bx+%2B+c...........(0,5)
0+=+a%2A5%5E2+%2B+b%2A5+%2B+c
0+=+25a%2B+5b+%2B+c.............eq.1

y+=+ax%5E2+%2B+bx+%2B+c..........(2,13)
13+=+a%2A2%5E2+%2B+b%2A2+%2B+c
13+=+4a%2B+2b+%2B+c..........eq.2

y+=+ax%5E2+%2B+bx+%2B+c..........(3,26)
26+=+a%2A3%5E2+%2B+b%2A3+%2B+c
26+=+9a%2B+3b+%2B+c...........eq.3

solve the system:
25a%2B+5b+%2B+c=0.............eq.1
4a%2B+2b+%2B+c=13...........eq.2
+9a%2B+3b+%2B+c=26...........eq.3
-------------------------------------
Your matrix:
matrix%283%2C4%2C25%2C5%2C1%2C0%2C%0D%0A4%2C2%2C1%2C13%2C%0D%0A9%2C3%2C1%2C26%29

Make the pivot in the 1st column by dividing the 1st row by 25



Eliminate the 1st column



Make the pivot in the 2nd column by dividing the 2nd row by 6/5



Eliminate the 2nd column



Make the pivot in the 3rd column by dividing the 3rd row by -1/5



Eliminate the 3rd column




Solution set:
a+=+-26%2F3
a+=+169%2F3
c+=+-65

y+=+-%2826%2F3%29x%5E2+%2B+%28169%2F3%29x+-65


Answer by greenestamps(13216) About Me  (Show Source):
You can put this solution on YOUR website!


Unfortunately, the other tutor wrote her first equation incorrectly (using the point (5,0) instead of (0,5); so all her work was wasted....

Given any system of equations, there are endless different paths to get to the solution using the matrix method (by which I assume you mean Gauss-Jordan elimination). Here is the path I followed.

Initial matrix:

matrix%283%2C4%2C0%2C0%2C1%2C5%2C4%2C2%2C1%2C13%2C9%2C3%2C1%2C26%29

The first thing I'm going to do is switch rows 1 and 3, because the current row 1 is in exactly the form I want for row 3:

matrix%283%2C4%2C9%2C3%2C1%2C26%2C4%2C2%2C1%2C13%2C0%2C0%2C1%2C5%29

First objective: make (1,1) entry equal to 1. I choose to avoid fractions where possible, so, instead of dividing row 1 by 9, I'm going to replace row 1 with row 1 minus 2 times row 2 (9-4*2 = 1):

matrix%283%2C4%2C1%2C-1%2C-1%2C0%2C4%2C2%2C1%2C13%2C0%2C0%2C1%2C5%29

Next objective: get (2,1) and (3,1) entries equal to 0. (3,1) entry is already 0; to get (2,1) entry equal to 0, replace row2 with row 2 minus 4 times row 1:

matrix%283%2C4%2C1%2C-1%2C-1%2C0%2C0%2C6%2C5%2C13%2C0%2C0%2C1%2C5%29

Next objective: get (2,2) entry equal to 1. The only choice is to divide row 2 by 6:

matrix%283%2C4%2C1%2C-1%2C-1%2C0%2C0%2C1%2C5%2F6%2C13%2F6%2C0%2C0%2C1%2C5%29

Next objective: get (1,2) entry equal to 0 using the 1 in (2,2):

matrix%283%2C4%2C1%2C0%2C-1%2F6%2C13%2F6%2C0%2C1%2C5%2F6%2C13%2F6%2C0%2C0%2C1%2C5%29

Next objective: get (1,3) and (2,3) equal to 0 using the 1 in (3,3):

matrix%283%2C4%2C1%2C0%2C0%2C3%2C0%2C1%2C0%2C-2%2C0%2C0%2C1%2C5%29

The solution to the system is a=3, b=-2, c=5; the polynomial is

y+=+3x%5E2-2x%2B5