SOLUTION: hello im a student in high school sturggling with matrices x= 0 1 2 3 4..... y= 1 2 4 7 11 ........... using the matrices , how would i find the equation using the MATRICE

Algebra ->  Matrices-and-determiminant -> SOLUTION: hello im a student in high school sturggling with matrices x= 0 1 2 3 4..... y= 1 2 4 7 11 ........... using the matrices , how would i find the equation using the MATRICE      Log On


   



Question 145163: hello
im a student in high school sturggling with matrices
x= 0 1 2 3 4.....
y= 1 2 4 7 11 ...........
using the matrices , how would i find
the equation using the MATRICES! ?>?>>
please help me
i will appreciate it so much
i have a test tomorrow :(
please help

Answer by Edwin McCravy(20065) About Me  (Show Source):
You can put this solution on YOUR website!
hello
im a student in high school sturggling with matrices
x= 0 1 2 3 4.....
y= 1 2 4 7 11 ...........
using the matrices , how would i find
the equation using the MATRICES! ?>?>>
please help me
i will appreciate it so much
i have a test tomorrow :(
please help

First we try a first degree polynomial with the first two
points, amd see if it works with the other numbers:

Try  y+=+Ax+%2B+B

Substitute the first two points (x,y) = (0,1) and (x,y) = (1,2)

     1+=+A%280%29+%2B+B
     2+=+A%281%29+%2B+B

That system is

B+=+1
A%2BB=2

So B = 1 and substituting in A%2BB=2

A%2B1=2 and A=1

So y+=+Ax+%2B+B becomes

   y+=+1x+%2B+1

   y+=+x+%2B+1  

Now we check by substituting the other three given points:

Substituting (x,y)=(2,4)

   y+=+x+%2B+1
   4+=+2+%2B+1
     4+=+3

Oh, oh.  It doesn't check, so there is no use to 
check the other two.  We must try the next higher 
degree polynomial, so we try

     y+=+Ax%5E2+%2B+Bx+%2B+C

Substitute the first three points (x,y) = (0,1), (x,y) = (1,2),
and (2,4)

     1+=+A%280%29%5E2+%2B+B%280%29%2BC
     2+=+A%281%29%5E2+%2B+B%281%29%2BC
     4+=+A%282%29%5E2+%2B+B%282%29%2BC

That system is

1=C
2=A%2BB%2BC
4=4A%2B2B%2BC

Write the system as:

0A%2B0B%2B1C=1
1A%2B1B%2B1C=2
4A%2B2B%2B1C=4

Write as a matrix:



Swap row 1 and row 2 to get a 1 in the upper
left:




Need a 0 where the 4 on the left bottom is.
So multiply row 1 by -4 and add to row 3



Swap rows 2 and 3 to get a 0 where the -2 is



This is in triangular form, 3 0's at the bottom
left.  Convert back to a system of equations:

1A%2B1B%2B1C=2
0A-2B-3C=-4
0A%2B0B%2BC=1

or

A%2BB%2BC=2
  -2B-3C=-4
    C=1

Substitute C=1 into -2B-3C=-4 

    -2B-3C=-4
  -2B-3%281%29=-4
     -2B-3=-4
       -2B=-1
         B=1%2F2  
 
Substitute B=1%2F2 and C=1 into A%2BB%2BC=2

  A%2BB%2BC=2
A%2B1%2F2%2B1=2

Multiply through by 2 to clear of fractions:

2A%2B1%2B2=4
2A%2B3=4
2A=1
A=1%2F2

So y+=+Ax%5E2+%2B+Bx+%2B+C becomes:

   y+=+%281%2F2%29x%5E2+%2B+%281%2F2%29x+%2B+1   

Now we check by substituting the other three given points:

Substituting (x,y)=(2,4)

   y+=+%281%2F2%29x%5E2+%2B+%281%2F2%29x+%2B+1 
   4+=+%281%2F2%292%5E2+%2B+%281%2F2%292+%2B+1
   4+=+%281%2F2%294+%2B+1+%2B+1  
   4+=+2%2B1%2B1
   4+=+4  

That checks.

Substituting (x,y)=(3,7)

   y+=+%281%2F2%29x%5E2+%2B+%281%2F2%29x+%2B+1 
   7+=+%281%2F2%293%5E2+%2B+%281%2F2%293+%2B+1
   7+=+%281%2F2%299+%2B+3%2F2+%2B+1  
   7+=+9%2F2%2B3%2F2%2B1
   7+=+12%2F2%2B1
   7+=+6%2B1
   7+=+7

That also checks.

Finally, substitute (x,y)=(4,11)

   y+=+%281%2F2%29x%5E2+%2B+%281%2F2%29x+%2B+1 
   11+=+%281%2F2%294%5E2+%2B+%281%2F2%294+%2B+1
   11+=+%281%2F2%2916+%2B+2+%2B+1  
   11+=+8%2B2%2B1
   11+=+11

So that also checks, so the answer is

   y+=+%281%2F2%29x%5E2+%2B+%281%2F2%29x+%2B+1 

Edwin