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標題:
發問:
A and B are two different 2 x 2 matrices. If A is not the inverse of B and B is not the inverse of A. Is it possible that AB = I? If yes, please give example, thanks.
最佳解答:
If A = (a11, a12 ... a21, a22), B = (b11, b12 ... b21, b22) then AB = (a11b11+a12b21, a11b12+a12b22 ... a21b11+a22b21, a21b12+a22b22) So, if AB = I, then a11b11+a12b21 = a21b12+a22b22 = 1 ?? (i) a11b12+a12b22 = a21b11+a22b21 = 0 ?? (ii) From (ii), we get b22 = -a11b12/a12, b21 = -a21b11/a22 Sub. Into (i), we get, a11b11 - a12a21b11/a22 = 1 and a21b12 - a22a11b12/a12 = 1 ==> b11 = a22/(a11a22 - a12a21) and b12 = a12/(a12a21 - a11a22) Let Δ = a11a22 - a12a21, then b11 = a22/Δ, b12 = -a12/Δ, b21 = -a21/Δ, b22 = a11/Δ So, B = (1/Δ)(a22, -a12 ... -a21, a11), which is the inverse of A (Δ is the det. of A). Conclusion : If AB = I, then B should be the inverse of A. 2014-07-08 20:40:05 補充： 感謝 自由自在 知識長的指導。 As det(AB) = det(I) = 1, so det(A)<>0, so A is invertible.
其他解答:
Thank you Andrew. 其實 inverse 已經是這樣 define，我在想是否有必要再作推論： http://en.wikipedia.org/wiki/Invertible_matrix|||||自由自在: Your idea can be further extended A is invertible, so A(A^{-1}) = I A(A^{-1}) - AB = 0 A(A^{-1} - B) = 0 A^{-1} - B = 0 B = A^{-1} So B must be inverse of A, and therefore A must be inverse of B. The proof applies to any dimension and any underlying field.|||||If we can also mention det(AB)=det(I)=1 So det(A)<>0 so A is invertible.|||||I think... If AB=I, then A^(-1) AB = A^(-1) I [ A^(-1) A ] B = A^(-1) I B = A^(-1) B=A^(-1) ∴B is the inverse of A. ∴No.
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Matrix發問:
A and B are two different 2 x 2 matrices. If A is not the inverse of B and B is not the inverse of A. Is it possible that AB = I? If yes, please give example, thanks.
最佳解答:
If A = (a11, a12 ... a21, a22), B = (b11, b12 ... b21, b22) then AB = (a11b11+a12b21, a11b12+a12b22 ... a21b11+a22b21, a21b12+a22b22) So, if AB = I, then a11b11+a12b21 = a21b12+a22b22 = 1 ?? (i) a11b12+a12b22 = a21b11+a22b21 = 0 ?? (ii) From (ii), we get b22 = -a11b12/a12, b21 = -a21b11/a22 Sub. Into (i), we get, a11b11 - a12a21b11/a22 = 1 and a21b12 - a22a11b12/a12 = 1 ==> b11 = a22/(a11a22 - a12a21) and b12 = a12/(a12a21 - a11a22) Let Δ = a11a22 - a12a21, then b11 = a22/Δ, b12 = -a12/Δ, b21 = -a21/Δ, b22 = a11/Δ So, B = (1/Δ)(a22, -a12 ... -a21, a11), which is the inverse of A (Δ is the det. of A). Conclusion : If AB = I, then B should be the inverse of A. 2014-07-08 20:40:05 補充： 感謝 自由自在 知識長的指導。 As det(AB) = det(I) = 1, so det(A)<>0, so A is invertible.
其他解答:
Thank you Andrew. 其實 inverse 已經是這樣 define，我在想是否有必要再作推論： http://en.wikipedia.org/wiki/Invertible_matrix|||||自由自在: Your idea can be further extended A is invertible, so A(A^{-1}) = I A(A^{-1}) - AB = 0 A(A^{-1} - B) = 0 A^{-1} - B = 0 B = A^{-1} So B must be inverse of A, and therefore A must be inverse of B. The proof applies to any dimension and any underlying field.|||||If we can also mention det(AB)=det(I)=1 So det(A)<>0 so A is invertible.|||||I think... If AB=I, then A^(-1) AB = A^(-1) I [ A^(-1) A ] B = A^(-1) I B = A^(-1) B=A^(-1) ∴B is the inverse of A. ∴No.
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