Matrix

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.

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.