早稲田大学 創造理工学研究科 経営システム工学専攻 2014年8月実施 线性代数

标签：

• Waseda-University
• Matrix-Rank

Author

思齐塾, 祭音Myyura

Description

行の基本操作(elementary row operations)を行って、以下の行列 $A$ の階数(rank)を求めよ.

$$A = \begin{bmatrix} 1 & 4 & 7 & 10 \\ 2 & 5 & 8 & 11 \\ 3 & 6 & 9 & 12 \end{bmatrix}$$

题目描述

使用初等行变换求矩阵

$$A= \begin{bmatrix} 1&4&7&10\\ 2&5&8&11\\ 3&6&9&12 \end{bmatrix}$$

的秩。

Kai

Apply elementary row operations to find the rank of matrix A.

$$A = \begin{bmatrix} 1 & 4 & 7 & 10 \\ 2 & 5 & 8 & 11 \\ 3 & 6 & 9 & 12 \end{bmatrix}$$

Subtract 2 times the first row from the second row, and 3 times the first row from the third row:

$$\begin{bmatrix} 1 & 4 & 7 & 10 \\ 0 & -3 & -6 & -9 \\ 0 & -6 & -12 & -18 \end{bmatrix}$$

Multiply the second row by -1/3:

$$\begin{bmatrix} 1 & 4 & 7 & 10 \\ 0 & 1 & 2 & 3 \\ 0 & -6 & -12 & -18 \end{bmatrix}$$

Add 6 times the second row to the third row:

$$\begin{bmatrix} 1 & 4 & 7 & 10 \\ 0 & 1 & 2 & 3 \\ 0 & 0 & 0 & 0 \end{bmatrix}$$

The matrix is now in row echelon form. There are two non-zero rows. Therefore, the rank of the matrix is 2.

$$rank(A) = 2$$