请问下图中的线性代数题如何解答?求详细解题过程,谢谢!
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4(3) (A, E) =
[0 0 1 2 1 0 0 0]
[1 0 2 0 0 1 0 0]
[0 1 0 2 0 0 1 0]
[2 1 0 0 0 0 0 1]
初等行变换为
[1 0 2 0 0 1 0 0]
[0 1 0 2 0 0 1 0]
[0 0 1 2 1 0 0 0]
[0 1 -4 0 0 -2 0 1]
初等行变换为
[1 0 2 0 0 1 0 0]
[0 1 0 2 0 0 1 0]
[0 0 1 2 1 0 0 0]
[0 0 -4 -2 0 -2 -1 1]
初等行变换为
[1 0 0 -4 -2 1 0 0]
[0 1 0 2 0 0 1 0]
[0 0 1 2 1 0 0 0]
[0 0 0 6 4 -2 -1 1]
初等行变换为
[1 0 0 0 2/3 -1/3 -2/3 2/3]
[0 1 0 0 -4/3 2/3 4/3 -1/3]
[0 0 1 0 -1/3 2/3 1/3 -1/3]
[0 0 0 1 2/3 -1/3 -1/6 1/6]
A ^(-1) = (1/6)*
[ 4 -2 -4 4]
[-8 4 8 -2]
[-2 4 2 -2]
[ 4 -2 -1 1]
5. AX = B
x = A^(-1) B
(A, B) =
[1 2 3 1 2]
[0 1 2 0 1]
[4 5 3 1 0]
初等行变换为
[1 2 3 1 2]
[0 1 2 0 1]
[0 -3 -9 -3 -8]
初等行变换为
[1 0 -1 1 0]
[0 1 2 0 1]
[0 0 -3 -3 -5]
初等行变换为
[1 0 0 2 5/3]
[0 1 0 -2 -7/3]
[0 0 1 1 5/3]
后两列即为 X。
[0 0 1 2 1 0 0 0]
[1 0 2 0 0 1 0 0]
[0 1 0 2 0 0 1 0]
[2 1 0 0 0 0 0 1]
初等行变换为
[1 0 2 0 0 1 0 0]
[0 1 0 2 0 0 1 0]
[0 0 1 2 1 0 0 0]
[0 1 -4 0 0 -2 0 1]
初等行变换为
[1 0 2 0 0 1 0 0]
[0 1 0 2 0 0 1 0]
[0 0 1 2 1 0 0 0]
[0 0 -4 -2 0 -2 -1 1]
初等行变换为
[1 0 0 -4 -2 1 0 0]
[0 1 0 2 0 0 1 0]
[0 0 1 2 1 0 0 0]
[0 0 0 6 4 -2 -1 1]
初等行变换为
[1 0 0 0 2/3 -1/3 -2/3 2/3]
[0 1 0 0 -4/3 2/3 4/3 -1/3]
[0 0 1 0 -1/3 2/3 1/3 -1/3]
[0 0 0 1 2/3 -1/3 -1/6 1/6]
A ^(-1) = (1/6)*
[ 4 -2 -4 4]
[-8 4 8 -2]
[-2 4 2 -2]
[ 4 -2 -1 1]
5. AX = B
x = A^(-1) B
(A, B) =
[1 2 3 1 2]
[0 1 2 0 1]
[4 5 3 1 0]
初等行变换为
[1 2 3 1 2]
[0 1 2 0 1]
[0 -3 -9 -3 -8]
初等行变换为
[1 0 -1 1 0]
[0 1 2 0 1]
[0 0 -3 -3 -5]
初等行变换为
[1 0 0 2 5/3]
[0 1 0 -2 -7/3]
[0 0 1 1 5/3]
后两列即为 X。
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