Finding the inverse of a matrix using transformations?

[parshyaa]

We use A = I.A as equation and then by transforming only A of LHS and I of RHS we come to I = P.A and we say that P is the inverse of matrix A
My question is that why we only tranform A and I, why A of RHS is left as it is during the transformation, or why transformation do not take place in both the part of RHS

 

[FactChecker]

Each transformation step, T, is multiplication on the left: TA = TIA. It is not also multiplying the A of the RHS. TA ≠ TITA

 

[parshyaa]

TA ≠ TITA

Can you prove it
Or is it like i am transforming only one matrix on LHS therefore i have to transform only one from RHS

 

[FactChecker]

You must do the same things on both sides to keep them equal. Left multiply both sides to get TI = TIA. That keeps the two sides equal. Given a series of transformations T_n...T₂T₁, You have T_n...T₂T₁I = T_n...T₂T₁IA.

You should not "double up" on the right-side transformations like T_n...T₂T₁IT_n...T₂T₁A

 

[parshyaa]

You must do the same things on both sides to keep them equal. Left multiply both sides to get TI = TIA. That keeps the two sides equal. Given a series of transformations T₁T₂...T_n, You have T₁T₂...T_nI = T₁T₂...T_nIA.

You should not "double up" on the right-side transformations like T₁T₂...T_nIT₁T₂...T_nA

Is their a proof which says that if A =BC
Then the transformation of A is done simultaneously with transformation of B or C but not with both, i know that first one is right but is their a proof

 

[FactChecker]

Matrix multiplication is similar to multiplication of real numbers, except that it is not commutative and there is not a guaranteed multiplicative inverse. This should be proven early in the theory of matrices. So this question is very similar to a question about real numbers: Since 6 = 2×3, is 4×6 = 4×2×3 or should it be 4×6 = 4×2×4×3

PS. Studying the similarity and differences of linear transformations, matrices, real numbers, etc. is a good motivation for the study of abstract algebra. That is where these issues are addressed directly.

 

[parshyaa]

I got this proof from an old book but its really satisfying

 

[FactChecker]

Good proof. One way to quickly see that your alternative hypothesis ( TA = TITA ) was wrong is to realize that the real numbers are 1x1 matrices. So a counterexample in the reals is also a counterexample in matrices. That can sometimes allow you avoid trying to prove something that is wrong. Of course, there are things that are true for the reals that are not true for general matrices, so checking something in the real numbers won't always tell you if something is wrong for matrices.

 

[parshyaa]

Good proof. One way to quickly see that your alternative hypothesis ( TA = TITA ) was wrong is to realize that the real numbers are 1x1 matrices. So a counterexample in the reals is also a counterexample in matrices. That can sometimes allow you avoid trying to prove something that is wrong. Of course, there are things that are true for the reals that are not true for general matrices, so checking something in the real numbers won't always tell you if something is wrong for matrices.

Actually my focus was in poving the general case instid of checking the result