线代 计算行列式 2 1 0 0 0 1 2 1 0 0 0 1 2 1 0 0 0 1 2 1
2个回答
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记原行列式为D(5),则
D(5) = 2*D(4) - 1*det(1,1,0,0;0,2,1,0;0,1,2,1;0,0,1,2)
= 2*D(4) - 1*1*D(3)
= 2*D(4) - D(3)
= 2*[2*D(3) - D(2)] - D(3)
= 3*D(3) - 2*D(2)
= 3*[2*D(2) - D(1)] - 2*D(2)
= 4*D(2) - 3*D(1)
= 4*(2*2-1*1) - 3*2
= ……
D(5) = 2*D(4) - 1*det(1,1,0,0;0,2,1,0;0,1,2,1;0,0,1,2)
= 2*D(4) - 1*1*D(3)
= 2*D(4) - D(3)
= 2*[2*D(3) - D(2)] - D(3)
= 3*D(3) - 2*D(2)
= 3*[2*D(2) - D(1)] - 2*D(2)
= 4*D(2) - 3*D(1)
= 4*(2*2-1*1) - 3*2
= ……
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