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pure maths problem
show that if (B+C)^-1 = B^-1+C^-1,then detB = detC pls help thx
最佳解答:
Is using the following method? (B+C)^-1 = B^-1+C^-1 (B+C)(B+C)^-1 = (B+C)(B^-1+C^-1) I = I+BC^-1+CB^-1+I 0 = BC^-1+CB^-1+I (BC)(0) = (BC)(BC^-1+CB^-1+I) 0 = B^2+C^2+BC (B-C)(0) = (B-C)(B^2+C^2+BC) 0 = B^3-C^3 B^3 = C^3 detB^3 = detC^3 [detB]^3 = [detC^3]^3 detB = det C 2008-01-08 23:48:25 補充： correct for some mistake , the last two steps should be[detB]^3 = [detC]^3detB = det C
其他解答:
if (B+C)^-1= B^-1+C^-1 let be writ B^-1 = b, C^-1 = c we assumed B and C is invertible. both side times B+C 0 = 1+Bc+Cb times on right with C 0 = C + B + CbC .... (1) times on right with B 0 = B + BcB +C ...... (2) combine (1), (2) BcB=CbC det (AB)= det(A) det (B) det(B)^2/ det (C) = det(C)^2/ det (B) det (B)^3 = det (C)^3 detB = detC proved.
pure maths problem
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發問:show that if (B+C)^-1 = B^-1+C^-1,then detB = detC pls help thx
最佳解答:
Is using the following method? (B+C)^-1 = B^-1+C^-1 (B+C)(B+C)^-1 = (B+C)(B^-1+C^-1) I = I+BC^-1+CB^-1+I 0 = BC^-1+CB^-1+I (BC)(0) = (BC)(BC^-1+CB^-1+I) 0 = B^2+C^2+BC (B-C)(0) = (B-C)(B^2+C^2+BC) 0 = B^3-C^3 B^3 = C^3 detB^3 = detC^3 [detB]^3 = [detC^3]^3 detB = det C 2008-01-08 23:48:25 補充： correct for some mistake , the last two steps should be[detB]^3 = [detC]^3detB = det C
其他解答:
if (B+C)^-1= B^-1+C^-1 let be writ B^-1 = b, C^-1 = c we assumed B and C is invertible. both side times B+C 0 = 1+Bc+Cb times on right with C 0 = C + B + CbC .... (1) times on right with B 0 = B + BcB +C ...... (2) combine (1), (2) BcB=CbC det (AB)= det(A) det (B) det(B)^2/ det (C) = det(C)^2/ det (B) det (B)^3 = det (C)^3 detB = detC proved.
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