数学のブログ

行列 行列の分割 正方行列、4次、2次、積

手を動かしてまなぶ 線形代数 (藤岡 敦(著)、裳華房)の第1章(行列)、3(行列の分割)、基本問題の問3.3の解答を求めてみる。

問3.3

A = [ 0 - 1 1 0 ] B = [ - 1 0 0 1 ] C = [ 0 - 1 - 1 0 ]

とおく。

I 2 = [ A O O A ] [ A O O A ] = [ A 2 O O A 2 ] = [ - 1 0 0 0 0 - 1 0 0 0 0 - 1 0 0 0 0 - 1 ]
J 2 = [ O B - B O ] [ O B - B O ] = [ - B 2 O O - B 2 ] = [ - 1 0 0 0 0 - 1 0 0 0 0 - 1 0 0 0 0 - 1 ]
K 2 = [ O C - C O ] [ O C - C O ] = [ - C 2 O O - C 2 ] = [ - 1 0 0 0 0 - 1 0 0 0 0 - 1 0 0 0 0 - 1 ]
I J = [ A O O A ] [ O B - B O ] = [ O A B - A B O ] = [ 0 0 0 - 1 0 0 - 1 0 0 1 0 0 1 0 0 0 ]
J I = [ O B - B O ] [ A O O A ] = [ O B A - B A O ] = [ 0 0 0 1 0 0 1 0 0 - 1 0 0 - 1 0 0 0 ]
J K = [ O B - B O ] [ O C - C O ] = [ - B C O O - B C ] = [ 0 - 1 0 0 1 0 0 0 0 0 0 - 1 0 0 1 0 ]
K J = [ O C - C O ] [ O B - B O ] = [ - C B O O - C B ] = [ 0 1 0 0 - 1 0 0 0 0 0 0 1 0 0 - 1 0 ]
K I = [ O C - C O ] [ A O O A ] = [ O C A - C A O ] = [ 0 0 - 1 0 0 0 0 1 1 0 0 0 0 - 1 0 0 ]
I K = [ A O O A ] [ O C - C O ] = [ O A C - A C O ] = [ 0 0 1 0 0 0 0 - 1 - 1 0 0 0 0 1 0 0 ]

コード、入出力結果(Wolfram Language, Jupyter)

i = {{0, -1, 0, 0}, {1, 0, 0, 0}, {0, 0, 0, -1}, {0, 0, 1, 0}}

{{0, -1, 0, 0}, {1, 0, 0, 0}, {0, 0, 0, -1}, {0, 0, 1, 0}}
j = {{0, 0, -1, 0}, {0, 0, 0, 1}, {1, 0, 0, 0}, {0, -1, 0, 0}}

{{0, 0, -1, 0}, {0, 0, 0, 1}, {1, 0, 0, 0}, {0, -1, 0, 0}}
k = {{0, 0, 0, -1}, {0, 0, -1, 0}, {0, 1, 0, 0}, {1, 0, 0, 0}}

{{0, 0, 0, -1}, {0, 0, -1, 0}, {0, 1, 0, 0}, {1, 0, 0, 0}}
i // TraditionalForm
Output
j // TraditionalForm
Output
k // TraditionalForm
Output
Dot[i, i] // TraditionalForm
Output
Dot[j, j] // TraditionalForm
Output
Dot[j, j] // TraditionalForm
Output
Dot[i, j] // TraditionalForm
Output
Dot[j, i] // TraditionalForm
Output
Dot[j, k] // TraditionalForm
Output
Dot[k, j] // TraditionalForm
Output
Dot[k, i] // TraditionalForm
Output
Dot[i, k] // TraditionalForm
Output