How should I figure out the following determinants?
It is required to use $|AB| = |A||B|$ to figure them out.
(1) $D_1 = \begin{vmatrix}
1+x_1y_1 & 1+x_1y_2 & \dots & 1+x_1y_n \\
1+x_2y_1 & 1+x_2y_2 & \dots & 1+x_2y_n \\
\vdots & \vdots & \ddots & \vdots \\
1+x_ny_1 & 1+x_ny_2 & \dots & 1+x_ny_n
\end{vmatrix}$
(2) $D_2 = \begin{vmatrix}
1 & \cos(a_1-a_2) & \cos(a_1-a_3) & \dots & \cos(a_1-a_n) \\
\cos(a_1-a_2) & 1 & \cos(a_2-a_3) & \dots & \cos(a_2-a_n) \\
\cos(a_1-a_3) & \cos(a_2-a_3) & 1 & \dots & \cos(a_3-a_n) \\
\vdots & \vdots & \vdots & \ddots & \vdots \\
\cos(a_1-a_n) & \cos(a_2-a_n) & \cos(a_3-a_n) & \dots & 1
\end{vmatrix}$
(3) $D_3 = \begin{vmatrix}
a & a & a & a \\
a & a & -a & -a \\
a & -a & a & -a \\
a & -a & -a & a
\end{vmatrix}$
(4) Let $s_k=a_1^k+a_2^k+a_3^k+a_4^k \quad (k=1,2,3,4,5,6)$,
$$D_4 = \begin{vmatrix}
4 & s_1 & s_2 & s_3 \\
s_1 & s_2 & s_3 & s_4 \\
s_2 & s_3 & s_4 & s_5 \\
s_3 & s_4 & s_5 & s_6
\end{vmatrix}$$
My Attempt:
(1) I noticed for any $i,j$,
$1+x_iy_j = \begin{bmatrix} 1 & x_i \end{bmatrix}
\begin{bmatrix} 1 \\ y_j \end{bmatrix}$.
So, the matrix corresponds $D_1$ equals $\begin{bmatrix}
1 & x_1 \\ 1 & x_2 \\ \vdots & \vdots \\ 1 & x_n
\end{bmatrix}_{n \times 2}
\begin{bmatrix}
1 & 1 & \dots & 1 \\
y_1 & y_2 & \dots & y_n
\end{bmatrix}_{2 \times n}$.
But it's not helpful at all. 🙁
(2) It's obvious that
$a_{ij} = \cos(a_i – a_j) = \cos a_i \cos a_j + \sin a_i \sin a_j$.
(3) I've no idea about this problem at all. All I came up with, is
$$D_3 = a^4 \begin{vmatrix}
1 & 1 & 1 & 1 \\
1 & 1 & -1 & -1 \\
1 & -1 & 1 & -1 \\
1 & -1 & -1 & 1
\end{vmatrix}$$
(4) I noticed that $a_{ij} = s_{i+j-2}$. But still, not helpful.
Plz give me some hints. Thx in advance.
Best Answer
For the fourth determinant start with $$ V=\pmatrix{1 & 1 & 1 & 1\\a_1 & a_2 & a_3 & a_4\\a_1^2 & a_2^2 & a_3^2 & a_4^2\\a_1^3 & a_2^3 & a_3^3 & a_4^3\\} $$ and notice that $$ \begin{pmatrix} 4 & s_1 & s_2 & s_3 \\ s_1 & s_2 & s_3 & s_4 \\ s_2 & s_3 & s_4 & s_5 \\ s_3 & s_4 & s_5 & s_6 \end{pmatrix}=VV^{T}. $$
The determinant of $V$ is well-known to be $\prod_{i<j}(a_i-a_j)$, so $D_4=(\prod_{i<j}(a_i-a_j))^2$.