# 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.

• Some hints: for 1), your observation tells us something about $D_1$'s rank, which tells us something about $D_1$'s determinant for sufficiently large $n$. Similarly, for 2), your observation allows you to write $D_2$ as a sum of two rank $1$ matrices, which tells us something about $D_2$'s determinant for sufficiently large $n$. Feb 27, 2021 at 6:53

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 $$\prod_{i, so $$D_4=(\prod_{i.

• :Very nice Idea. How it started in your brain? +1 Feb 27, 2021 at 8:11
• I learned it from my teachers ;-). It is how one calculates the discriminant of a quartic, $D_4$. Feb 27, 2021 at 10:29

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 $$D_3 = -16a^4$$.

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

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