Cfrtjryk

$A =begin{vmatrix}a^2 & b^2 & c^2 & d^2\ (a-1)^2 & (b-1)^2 & (c-1)^2 & (d-1)^2 \ (a-2)^2 & (b-2)^2 & (c-2)^2 & (d-2)^2\ (a-3)^2 & (b-3)^2 & (c-3)^2 & (d-3)^2end{vmatrix}$

I have been trying to solve this determinant for a while now and haven't gotten very far.

I have been using row reduction to try and create a lower triangle but keep getting caught up with the amount of ugly multiplication involved and cant seem to find an easier way.

Any help with this question would be much appreciated.

Use row transformations in the following manner:

$R_1to R_1-R_2\R_2to R_2-R_3\R_3to R_3-R_4$

$=begin{vmatrix}2a-1 & 2b-1 & 2c-1 & 2d-1\ 2a-3 & 2b-3 & 2c-3 & 2d-3 \ 2a-5 & 2b-5 & 2c-5 & 2d-5\ (a-3)^2 & (b-3)^2 & (c-3)^2 & (d-3)^2end{vmatrix}$

$R_1to R_1-R_2\R_2to R_2-R_3$

$=begin{vmatrix}2 & 2 & 2 & 2\ 2 & 2 & 2 & 2 \ 2a-5 & 2b-5 & 2c-5 & 2d-5\ (a-3)^2 & (b-3)^2 & (c-3)^2 & (d-3)^2end{vmatrix}$

Determinant having identical rows evaluates to $0$. You could go an extra mile and perform $R_1to R_1-R_2$ to see why this is true.

@Shubham Johri : There is a simple proof that this determinant, or more precisely its transpose, is zero.

Let us consider the following matrix-vector product :

$$begin{bmatrix}
a^2 & (a-1)^2 & (a-2)^2 & (a-3)^2\
b^2 & (b-1)^2 & (b-2)^2 & (b-3)^2\
c^2 & (c-1)^2 & (c-2)^2 & (c-3)^2\
d^2 & (d-1)^2 & (d-2)^2 & (d-3)^2
end{bmatrix}
begin{bmatrix}
-1\ 3\-3\ 1
end{bmatrix}=begin{bmatrix}
0\0\0\0
end{bmatrix}.$$
Thus this matrix has a non-zero kernel ; therefore its determinant is equal to $0$.

Remark : it wasn't necessary to explicitate the vector with coordinates $-1,3,-3,1$ (have you noticed that it uses binomial coefficients ?) ; we could simply have said that such a vector with $p,q,r,s$ coordinates exist because in the 3-dimensional vector space $mathbb{R}[X]_{deg leq 2}$, any set of more than 3 vectors is necessarily linked by a zero linear combination :

$$p X^2 + q (X-1)^2 + r (X-2)^2 + s (X-3)^2 = 0.$$