Prove that $\det(A)=p_1p_2-ba={bf(a)-af(b)\over b-a}$ Let $f(x)=(p_1-x)\cdots (p_n-x)$ $p_1,\ldots, p_n\in \mathbb R$ and let $a,b\in \mathbb R$ such that $a\neq b$
Prove that $\det A={bf(a)-af(b)\over b-a}$ where $A$ is the matrix:
$$\begin{pmatrix}p_1 & a & a & \cdots & a \\ b & p_2 & a & \cdots & a \\ \vdots & \vdots & \vdots & \ddots & \vdots \\ b & b & b & \cdots & p_n \end{pmatrix}$$
that is the entries $k_{ij}=a$ if $i<j$, $k_{ij}=p_i$ if $i=j$ and $k_{ij}=b$ if $i>j$
I tried to do it by induction over $n$. The base case for $n=2$ is easy $\det(A)=p_1p_2-ba={bf(a)-af(b)\over b-a}$
The induction step is where I don´t know what to do. I tried to solve the dterminant by brute force(applying my induction hypothesis for n and prove it for n+1) but I don´t know how to reduce it. It gets horrible.
I would really appreciate if you can help me with this problem. Any comments, suggestions or hints would be highly appreciated
 A: Hint. To emphasise the dependence of the $A$ and $f$ on $n$ as well as the $p_i$s, rewrite them as $A(p_1,\ldots,p_n)$ and $f(x; p_1,\ldots,p_n)$.
Now, in the induction step, subtract the last column by the last but one column. Then apply Laplace expansion. The $(n,n-1)$-th minor is $\det A(p_1,\ldots,p_{n-2},b)$, while the $(n,n)$-th minor is $\det A(p_1,\ldots,p_{n-1})$. Hence
\begin{align}
\det A
&=
(p_{n-1}-a)\det A(p_1,\ldots,p_{n-2},b)\\
&\phantom{=}+
(p_n-b)\det A(p_1,\ldots,p_{n-1})\\
&=
(p_{n-1}-a)\frac{bf(a; p_1,\ldots,p_{n-2},b)-af(b; p_1,\ldots,p_{n-2},b)}{b-a}\\
&\phantom{=}+
(p_n-b)\frac{bf(a; p_1,\ldots,p_{n-1})-af(b; p_1,\ldots,p_{n-1})}{b-a}.
\end{align}
The rest is straightforward if you write each $f$ as a product of linear factors according to the definition.
A: Here is a possible proof without induction. The idea is to consider $\det A$ as a function of $p_n$.
We define the function $F: \Bbb R \to \Bbb R$ as
$$
 F(p) = \begin{vmatrix}
p_1 &a &\ldots &a &a \\
b &p_2 &\ldots &a &a \\
\vdots &\vdots &\ddots &\vdots &\vdots \\
b &b &\ldots &p_{n-1} &a\\
b &b &\ldots &b &p
\end{vmatrix}
$$
$F$ is a linear function of $p$ and therefore completely determined by its values at two different arguments.
$F(a)$ and $F(b)$ can be computed easily:
 By subtracting the last row from all previous rows we get
$$
 F(a) = \begin{vmatrix}
p_1 &a &\ldots &a &a \\
b &p_2 &\ldots &a &a \\
\vdots &\vdots &\ddots &\vdots &\vdots \\
b &b &\ldots &p_{n-1} &a\\
b &b &\ldots &b &a
\end{vmatrix}
=
 \begin{vmatrix}
p_1-b &a-b &\ldots &a-b &0\\
0 &p_2-b &\ldots &a-b &0 \\
\vdots &\vdots &\ddots &\vdots &\vdots\\
0 &0 &\ldots &p_{n-1}-b &0 \\
b &b &\ldots &b &a
\end{vmatrix} \\
$$
i.e.
$$
 F(a) = a(p_1-b)\cdots (p_{n-1}-b) \, .
$$
In the same way (or by using the symmetry in $a$ and $b$) we get
$$
F(b) = b (p_1-a)\cdots (p_{n-1}-a) \, .
$$
Now we can compute $\det A = F(p_n)$ with linear interpolation:
$$
 \det A =  \frac{b-p_n}{b-a} F(a) + \frac{p_n-a}{b-a} F(b) \\
 = \frac{- a(p_1-b)\cdots (p_{n-1}-b)(p_n-b) + b(p_1-a)\cdots (p_{n-1}-a)(p_n-a) }{b-a} \\
 = \frac{-af(b) + bf(a)}{b-a} \, .
$$
