find values of a, b and c when $y=ax^2 +bx + c$ is a curve? find values of $a, b$ and $c$ when $y=ax^2 +bx + c$ is a curve and the passes through the points $(x_1, y_1), (x_2, y_2), (x_3, y_3)$. The question is needed to be solved using matrices that involves row operations. Please help can't get a proper eq or could understand how that matrix will be made
 A: According to the problem statement, we know: $$y_1=a(x_1)^2+bx_1+c$$ $$y_2=a(x_2)^2+bx_2+c$$ $$y_3=a(x_3)^2+bx_3+c.$$
The $(x_i)^2$'s etc. are constants in these equations.
Thus we have the system of equations
$$\begin{bmatrix}
(x_1)^2 & x_1 & 1 \\
(x_2)^2 & x_2 & 1 \\
(x_3)^2 & x_3 & 1 \\
\end{bmatrix}
\begin{bmatrix}
a \\ b \\ c \\
\end{bmatrix}
=
\begin{bmatrix}
y_1 \\ y_2 \\ y_3 \\
\end{bmatrix}
.$$
Now we solve this system of equations to find $a,b,c$

To get you started with the row operations, we take the augmented matrix
$$\left[\begin{array}{rrr|r}
(x_1)^2 & x_1 & 1 & y_1 \\
(x_2)^2 & x_2 & 1 & y_2 \\
(x_3)^2 & x_3 & 1 & y_3 \\
\end{array}\right]$$
and perform Gauss-Jordan Elimination.
We perform the row operations $$R_2 \gets R_2-\frac{(x_2)^2}{(x_1)^2} R_1$$ and $$R_3 \gets R_3-\frac{(x_3)^2}{(x_1)^2} R_3.$$  (Here we have assumed $x_1 \neq 0$.  If it happens that $x_1=0$, we should swap rows $1$ and $3$, then continue.   We wouldn't have both $x_1=0$ and $x_3=0$, otherwise we only know two points on the curve.)
This gives
$$\left[\begin{array}{rrr|r}
(x_1)^2 & x_1 & 1 & y_1 \\
0 & x_2-\frac{(x_2)^2}{(x_1)} & 1-\frac{(x_2)^2}{(x_1)^2} & y_2-\frac{(x_2)^2}{(x_1)^2}y_1 \\
0 & x_3-\frac{(x_3)^2}{(x_1)} & 1-\frac{(x_3)^2}{(x_1)^2} & y_3-\frac{(x_3)^2}{(x_1)^2}y_1 \\
\end{array}\right].$$
These row operations were selected specifically to create the $0$'s in the first column.
It's going to get messy, but aside from that, there's no real obstacle preventing us from doing it.
Now take $$R_3 \gets R_3-\frac{x_3-\frac{(x_3)^2}{(x_1)}}{x_2-\frac{(x_2)^2}{(x_1)}}R_2.$$  (Here we have assumed $x_2-\frac{(x_2)^2}{(x_1)} \neq 0$, but if it were, then $x_2=x_1$, and again we only have two points on the curve.)
I'll let you finish the rest.
A: You can use Lagrange Interpolation.
Notice that $L_1(x)=\cfrac {(x-x_2)(x-x_3)}{(x_1-x_2)(x_1-x_3)}$ 
has the property that $L_1(x_1)=1$ and $L_1(x_2)=L_1(x_3)=0$
If you define $L_2(x)$ and $L_3(x)$ analogously, you will find that $$p(x)=y_1L_1(x)+y_2L_2(x)+y_3L_3(x)$$ is a polynomial which takes the requisite values.
