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}}}) 


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Quadratic Form:
{{{y = ax^2 + bx + c}}}

Substitute each pair of {{{x}}},{{{y}}} values then solve for {{{a}}},{{{b}}},{{{c}}}

{{{y = ax^2 + bx + c}}}...........({{{0}}},{{{5}}})
{{{0 = a*5^2 + b*5 + c}}}
{{{0 = 25a+ 5b + c}}}.............eq.1


{{{y = ax^2 + bx + c}}}..........({{{2}}},{{{13}}})
{{{13 = a*2^2 + b*2 + c}}}
{{{13 = 4a+ 2b + c}}}..........eq.2


{{{y = ax^2 + bx + c}}}..........({{{3}}},{{{26}}})
{{{26 = a*3^2 + b*3 + c}}}
{{{26 = 9a+ 3b + c}}}...........eq.3


solve the system:

 {{{25a+ 5b + c=0}}}.............eq.1
 {{{4a+ 2b + c=13}}}...........eq.2
{{{ 9a+ 3b + c=26}}}...........eq.3
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Your matrix:

{{{matrix(3,4,25,5,1,0,
4,2,1,13,
9,3,1,26)}}}


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

{{{matrix(3,4, 1,	1/5,	1/25,	0,
4,	2,	1,	13,
9,	3,	1,	26)}}}



Eliminate the 1st column

{{{matrix(3,4, 
1,	1/5,	1/25,	0,
0,	6/5,	21/25,	13,
0,	6/5,	16/25,	26)}}}



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

{{{matrix(3,4,1,	1/5,	1/25,	0,
0,	1,	7/10,	65/6,
0,	6/5,	16/25,	26)}}}



Eliminate the 2nd column

{{{matrix(3,4,1,	0,	-1/10,	-13/6,
0,	1,	7/10,	65/6,
0,	0,	-1/5,	13)}}}



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

{{{matrix(3,4,1,	0,	-1/10,	-13/6,
0,	1,	7/10,	65/6,
0,	0,	1,	-65)}}}



Eliminate the 3rd column


{{{matrix(3,4,1,	0,	0,	-26/3,
0,	1,	0,	169/3,
0,	0,	1,	-65)}}}



Solution set:

{{{a = -26/3}}}
{{{a = 169/3}}}
{{{c = -65}}}


{{{y = -(26/3)x^2 + (169/3)x -65}}}