九州大学 システム情報科学府 情報理工学専攻・電気電子工学専攻 2024年9月実施 線形代数

标签：

• Kyushu-University
• Matrix-Determinant

Author

祭音Myyura (co-authored with ChatGPT 5.5 Thinking)

Description

(1) Find the determinants of the following matrices $A$ and $B$, respectively.

$$A=\begin{bmatrix}2 & 0 & 2 \\4 & 0 & 9 \\1 & 1 & 2\end{bmatrix},\qquad B=\begin{bmatrix}1 & 2 & 3 & 4 \\4 & 6 & 8 & 10 \\2 & 4 & 6 & 8 \\3 & 5 & 7 & 9\end{bmatrix}.$$

(2) For any upper triangular matrix $A=(a_{ij})$ of order $n$, show

$$\det(A)=\prod_{i=1}^{n} a_{ii},$$

where $\det(A)$ denotes the determinant of $A$.

题目描述

1. 分别求矩阵

   $$A=\begin{bmatrix}2&0&2\\4&0&9\\1&1&2\end{bmatrix}, \qquad B=\begin{bmatrix} 1&2&3&4\\ 4&6&8&10\\ 2&4&6&8\\ 3&5&7&9 \end{bmatrix}$$

   的行列式。

2. 对任意 $n$ 阶上三角矩阵 $A=(a_{ij})$，证明

   $$\det A=\prod_{i=1}^na_{ii}.$$

Kai

(1)

First compute $\det(A)$.

$$A=\begin{bmatrix}2 & 0 & 2 \\4 & 0 & 9 \\1 & 1 & 2\end{bmatrix}.$$

Expanding along the first row, we get

$$\det(A)=2\begin{vmatrix}0 & 9 \\1 & 2\end{vmatrix}+2\begin{vmatrix}4 & 0 \\1 & 1\end{vmatrix}.$$

Thus,

$$\det(A)=2(0\cdot 2-9\cdot 1)+2(4\cdot 1-0\cdot 1).$$

Therefore,

$$\det(A)=-18+8=-10.$$

Hence,

$$\boxed{\det(A)=-10}.$$

Next compute $\det(B)$.

$$B=\begin{bmatrix}1 & 2 & 3 & 4 \\4 & 6 & 8 & 10 \\2 & 4 & 6 & 8 \\3 & 5 & 7 & 9\end{bmatrix}.$$

Notice that the third row is twice the first row:

$$\begin{bmatrix}2 & 4 & 6 & 8\end{bmatrix}=2\begin{bmatrix}1 & 2 & 3 & 4\end{bmatrix}.$$

Therefore, the rows of $B$ are linearly dependent, so

$$\det(B)=0.$$

Hence,

$$\boxed{\det(B)=0}.$$

(2)

Let $A=(a_{ij})$ be an upper triangular matrix of order $n$. By the Leibniz formula for determinants,

$$\det(A)=\sum_{\sigma\in S_n}\operatorname{sgn}(\sigma)\prod_{i=1}^{n} a_{i,\sigma(i)}.$$

Since $A$ is upper triangular, we have

$$a_{ij}=0\qquad\text{if } i>j.$$

Therefore, for the product

$$\prod_{i=1}^{n} a_{i,\sigma(i)}$$

to be nonzero, it is necessary that

$$i\leq \sigma(i)\qquad\text{for all } i=1,2,\dots,n.$$

However, since $\sigma$ is a permutation of $\{1,2,\dots,n\}$, we have

$$\sum_{i=1}^{n} \sigma(i)=\sum_{i=1}^{n} i.$$

Thus, if $\sigma(i)\geq i$ for all $i$, then we must have

$$\sigma(i)=i\qquad\text{for all } i.$$

Hence, the only term in the Leibniz formula that can be nonzero comes from the identity permutation. Therefore,

$$\det(A)=\prod_{i=1}^{n} a_{ii}.$$

Hence,

$$\boxed{\det(A)=\prod_{i=1}^{n} a_{ii}}.$$