Unique Cubic Polynomial Hermite Interpolation. A practice problem from my textbook (not homework/assignment).
Show that there is a unique polynomial $P_{3}(x)$ where
$p_{3}(x_{0}) = f(x_{0}),\space   
p_{3}(x_{2}) = f(x_{2}),
p'_{3}(x_{1}) = f'(x_{1}), p^{''}_{3}(x_{1}) = f^{''}(x_{1})$ where $f(x)$ is a given function.
I'm looking towards using Hermite Interpolation but I"m unsure of how to start the problem. If anyone could just provide a hint to push me in the right direction, it would be appreciated.
 A: Let $p_3(x) = ax^3 + bx^2 + cx + d$ and without restricting generality suppose that $x_0 < x_1 < x_2$. Then in order for $p_3(x)$ to satisfy the given equations, its coefficients must satisfy the linear system
$$
\begin{bmatrix}
1 & x_0 & x_0^2 & x_0^3\\
1 & x_2 & x_2^2 & x_2^3\\
0 & 1 & 2x_1 & 3x_1^2\\
0 & 0 & 2 & 6x_1
\end{bmatrix}
\begin{bmatrix}
d\\
c\\
b\\
a
\end{bmatrix}
=
\begin{bmatrix}
f(x_0)\\
f(x_2)\\
f'(x_1)\\
f''(x_1)
\end{bmatrix},
$$
where each row corresponds to one of the equations.
Row reducing the coefficient matrix to a partial RREF we obtain
$$
\begin{bmatrix}
1 & x_0 & x_0^2 & x_0^3\\
0 & 1 & x_2+x_0 & x_2^2 + x_0x_2 + x_0^2\\
0 & 0 & 2x_1-x_2-x_0 & 3x_1^2 - (x_2^2 + x_0x_2 + x_0^2)\\
0 & 0 & 2 & 6x_1
\end{bmatrix}
$$
From here it is easy to compute the determinant, which is
$$
6x_1^2 + 2x_2^2 + 2x_0^2 + 2x_0x_2 -6x_1x_2-6x_0x_1 = 3(x_1-x_0)^2 + 3(x_2 - x_1)^2 - (x_2-x_0)^2
$$
Now if you let $d_1 = x_1 - x_0 > 0$ and $d_2 = x_2 - x_1 > 0$, then $x_2 - x_0 = d_1 + d_2$. Thus the determinant may be written as
$$
3d_1^2 + 3d_2^2 - (d_1 + d_2)^2 = 2d_1^2 - 2d_1d_2 + 2d_2^2 = (d_1 - d_2)^2 + d_1^2 + d_2^2 > 0
$$
Since the determinant is nonzero, the linear system must have a unique solution. Hence, there exists a unique polynomial $p_3(x)$ that satisfies the given equations.
