Solve determinants with $|AB| = |A||B|$ 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.
 A: 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 $D_4=(\prod_{i<j}(a_i-a_j))^2$.
A: You can use matrix rank inequality:
$$r(AB) \le \min\{r(A),r(B)\}$$
For $D_1$ you wrote your matrix as $AB$, and both ranks of $A$ and $B$ are $\le 2$ so its rank is $\le 2$ by the above formula. Therefore if $n \ge 3$ then $D_1 = 0$. For $n=1, 2$ you can calculate manually. You get
$$|1+x_1y_1| = 1+x_1y_1, \quad \begin{vmatrix} 1+x_1y_1 & 1+x_1y_2 \\ 1 + x_2y_1 & 1+x_2y_2\end{vmatrix} = (x_1-y_1)(x_2-y_2).$$
Similarly for $D_2$ you can write your matrix as
$$\begin{bmatrix} \cos a_1 & \sin a_1 \\ \vdots & \vdots \\ \cos a_n & \sin a_1\end{bmatrix}\begin{bmatrix} \cos a_1 & \cdots & \cos a_n \\ \sin a_1 & \cdots & \sin a_n\end{bmatrix}$$
so again for $n \ge 3$ we get $D_2 = 0$, and for $n=1,2$ calculate manually.
For $D_3$ it's not hard to guess eigenvalues and eigenvectors. If
$$A = \begin{bmatrix} 1 & 1 & 1 & 1 \\ 1 & 1 & -1 & -1 \\ 1 & -1 & 1 & -1 \\ 1 & -1 & -1 & 1 \end{bmatrix}$$
then $$A\begin{bmatrix} -1 \\ 1 \\ 1 \\ 1\end{bmatrix} = -2\begin{bmatrix} -1 \\ 1 \\ 1 \\ 1\end{bmatrix}, \quad  A\begin{bmatrix} 1 \\ 0 \\ 0 \\ 1\end{bmatrix}=2\begin{bmatrix} 1 \\ 0 \\ 0 \\ 1\end{bmatrix}, \quad A\begin{bmatrix} 1 \\ 0 \\ 1 \\ 0\end{bmatrix}=2\begin{bmatrix} 1 \\ 0 \\ 1 \\ 0\end{bmatrix}, \quad A\begin{bmatrix} 1 \\ 1 \\ 0 \\ 0\end{bmatrix}=2\begin{bmatrix} 1 \\ 1 \\ 0 \\ 0\end{bmatrix}$$
so $\det A = -2 \cdot 2 \cdot 2 \cdot 2 = -16$ and hence $D_3 = -16a^4$.
A: for D_3 case;I think LU decomposition can help to rewrite as $AB$ and use $|AB|=|A||B|$
$$\left(\begin{matrix}
1 & 0 & 0 & 0 \\
0 & 0 & 1 & 0 \\
0 & 1 & 0 & 0 \\
0 & 0 & 0 & 1
\end{matrix}\right).\left(\begin{matrix}
a & a & a & a \\
a & a & -a & -a \\
a & -a & a & -a \\
a & -a & -a & a
\end{matrix}\right)=\\
\left(\begin{matrix}
1 & 0 & 0 & 0 \\
1 & 1 & 0 & 0 \\
1 & 0 & 1 & 0 \\
1 & 1 & 1 & 1
\end{matrix}\right).\left(\begin{matrix}
a & a & a & a \\
0 & -2*a & 0 & -2*a \\
0 & 0 & -2*a & -2*a \\
0 & 0 & 0 & 4*a
\end{matrix}\right)$$ so det=$$(1.1.1.1).(a.(-2a).(-2a).(4a))=16a^4$$so det$D_3 = \begin{vmatrix}
a & a & a & a \\
a & a & -a & -a \\
a & -a & a & -a \\
a & -a & -a & a
\end{vmatrix}=-16a^4$
A: for the case $\left(\begin{matrix}
x_1*y_1+1 & x_1*y_2+1 & x_1*y_3+1 \\
x_2*y_1+1 & x_2*y_2+1 & x_2*y_3+1 \\
x_3*y_1+1 & x_3*y_2+1 & x_3*y_3+1
\end{matrix}\right)$
also LU decopmostion
$$\left(\begin{matrix}
x_1*y_1+1 & x_1*y_2+1 & x_1*y_3+1 \\
x_2*y_1+1 & x_2*y_2+1 & x_2*y_3+1 \\
x_3*y_1+1 & x_3*y_2+1 & x_3*y_3+1
\end{matrix}\right)=\\
\left(\begin{matrix}
1 & 0 & 0 \\
\frac{x_2*y_1+1}{x_1*y_1+1} & 1 & 0 \\
\frac{x_3*y_1+1}{x_1*y_1+1} & \frac{x_1-x_3}{x_1-x_2} & 1
\end{matrix}\right)\left(\begin{matrix}
x_1*y_1+1 & x_1*y_2+1 & x_1*y_3+1 \\
0 & \frac{x_1*y_1-x_2*y_1-x_1*y_2+x_2*y_2}{x_1*y_1+1} & \frac{x_1*y_1-x_2*y_1-x_1*y_3+x_2*y_3}{x_1*y_1+1} \\
0 & 0 & 0
\end{matrix}\right)$$so $$\left|\begin{matrix}
1+x_1*y_1 & 1+x_1*y_2 & 1+x_1*y_3 \\
1+x_2*y_1 & 1+x_2*y_2 & 1+x_2*y_3 \\
1+x_3*y_1 & 1+x_3*y_2 & 1+x_3*y_3
\end{matrix}\right|=0$$
