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 

<pre><font size = 4 color = "indigo"><b>
First we try a first degree polynomial with the first two
points, amd see if it works with the other numbers:

Try  {{{y = Ax + B}}}

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

     {{{1 = A(0) + B}}}
     {{{2 = A(1) + B}}}

That system is

{{{B = 1}}}
{{{A+B=2}}}

So B = 1 and substituting in {{{A+B=2}}}

{{{A+1=2}}} and {{{A=1}}}

So {{{y = Ax + B}}} becomes

   {{{y = 1x + 1}}}

   {{{y = x + 1}}}  

Now we check by substituting the other three given points:

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

   {{{y = x + 1}}}
   {{{4 = 2 + 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^2 + Bx + C}}}

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

     {{{1 = A(0)^2 + B(0)+C}}}
     {{{2 = A(1)^2 + B(1)+C}}}
     {{{4 = A(2)^2 + B(2)+C}}}

That system is

{{{1=C}}}
{{{2=A+B+C}}}
{{{4=4A+2B+C}}}

Write the system as:

{{{0A+0B+1C=1}}}
{{{1A+1B+1C=2}}}
{{{4A+2B+1C=4}}}

Write as a matrix:

{{{drawing(100,50,0,4,0,4,line(2.1,-1,2.1,5), 

(matrix(3,4,0,0,1,1,1,1,1,2,4,2,1,4))  )}}}

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

{{{drawing(100,50,0,4,0,4,line(2.1,-1,2.1,5), 
(matrix(3,4,1,1,1,2,0,0,1,1,4,2,1,4))  )}}}


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

{{{drawing(110,50,0,4,0,4,line(3,-1,3,5), 
(matrix(3,4,1,1,1,2,0,0,1,1,0,-2,-3,-4))  )}}}

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

{{{drawing(110,50,0,4,0,4,line(3,-1,3,5), 
(matrix(3,4,1,1,1,2,0,-2,-3,-4,0,0,1,1))  )}}}

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

{{{1A+1B+1C=2}}}
{{{0A-2B-3C=-4}}}
{{{0A+0B+C=1}}}

or

{{{A+B+C=2}}}
  {{{-2B-3C=-4}}}
    {{{C=1}}}

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

    {{{-2B-3C=-4}}}
  {{{-2B-3(1)=-4}}}
     {{{-2B-3=-4}}}
       {{{-2B=-1}}}
         {{{B=1/2}}}  
 
Substitute {{{B=1/2}}} and {{{C=1}}} into {{{A+B+C=2}}}

  {{{A+B+C=2}}}
{{{A+1/2+1=2}}}

Multiply through by 2 to clear of fractions:

{{{2A+1+2=4}}}
{{{2A+3=4}}}
{{{2A=1}}}
{{{A=1/2}}}

So {{{y = Ax^2 + Bx + C}}} becomes:

   {{{y = (1/2)x^2 + (1/2)x + 1}}}   

Now we check by substituting the other three given points:

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

   {{{y = (1/2)x^2 + (1/2)x + 1}}} 
   {{{4 = (1/2)2^2 + (1/2)2 + 1}}}
   {{{4 = (1/2)4 + 1 + 1}}}  
   {{{4 = 2+1+1}}}
   {{{4 = 4}}}  

That checks.

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

   {{{y = (1/2)x^2 + (1/2)x + 1}}} 
   {{{7 = (1/2)3^2 + (1/2)3 + 1}}}
   {{{7 = (1/2)9 + 3/2 + 1}}}  
   {{{7 = 9/2+3/2+1}}}
   {{{7 = 12/2+1}}}
   {{{7 = 6+1}}}
   {{{7 = 7}}}

That also checks.

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

   {{{y = (1/2)x^2 + (1/2)x + 1}}} 
   {{{11 = (1/2)4^2 + (1/2)4 + 1}}}
   {{{11 = (1/2)16 + 2 + 1}}}  
   {{{11 = 8+2+1}}}
   {{{11 = 11}}}

So that also checks, so the answer is

   {{{y = (1/2)x^2 + (1/2)x + 1}}} 

Edwin</pre>