Solve determinants with $|AB| = |A||B|$

Question

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 $s_k=a_1^k+a_2^k+a_3^k+a_4^k \quad (k=1,2,3,4,5,6)$, $$D_4 = \begin{vmatrix} 4 & s_1 & s_2 & s_3 \\ s_1 & s_2 & s_3 & s_4 \\ s_2 & s_3 & s_4 & s_5 \\ s_3 & s_4 & s_5 & s_6 \end{vmatrix}$$

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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.

Answer 1

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 $D_4=(\prod_{i<j}(a_i-a_j))^2$.

Answer 2

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 $D_3 = -16a^4$.

Answer 3

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$D_3 = \begin{vmatrix} a & a & a & a \\ a & a & -a & -a \\ a & -a & a & -a \\ a & -a & -a & a \end{vmatrix}=-16a^4$

Answer 4

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 $$\left|\begin{matrix} 1+x_1*y_1 & 1+x_1*y_2 & 1+x_1*y_3 \\ 1+x_2*y_1 & 1+x_2*y_2 & 1+x_2*y_3 \\ 1+x_3*y_1 & 1+x_3*y_2 & 1+x_3*y_3 \end{matrix}\right|=0$$