设A=(1,1,1; 1,1,-1; 1,-1,1),设B=(1,2,3,; -1,-2,4; 0,5,1;) 求AtB
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![设A=(1,1,1; 1,1,-1; 1,-1,1),设B=(1,2,3,; -1,-2,4; 0,5,1;) 求AtB](/uploads/image/z/10210856-32-6.jpg?t=%E8%AE%BEA%3D%EF%BC%881%2C1%2C1%EF%BC%9B+1%2C1%2C-1%EF%BC%9B+1%2C-1%2C1%EF%BC%89%2C%E8%AE%BEB%3D%EF%BC%881%2C2%2C3%2C%EF%BC%9B+-1%2C-2%2C4%EF%BC%9B+0%2C5%2C1%EF%BC%9B%EF%BC%89+%E6%B1%82AtB)
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设A=(1,1,1; 1,1,-1; 1,-1,1),设B=(1,2,3,; -1,-2,4; 0,5,1;) 求AtB
设A=(1,1,1; 1,1,-1; 1,-1,1),设B=(1,2,3,; -1,-2,4; 0,5,1;) 求AtB
设A=(1,1,1; 1,1,-1; 1,-1,1),设B=(1,2,3,; -1,-2,4; 0,5,1;) 求AtB
B^T=
1 -1 8
2 3 2
0 4 1
A=
1 1 1
1 1 -1
1 -1 1
所以
B^TA
=
(1-1+8 1-1-8 1+1+8
2+3+2 2+3-2 2-3+2
0+4+1 0+4-1 0-4+1)
=
(8 -8 10
8 3 1
5 3 -3)