Undergrad Finding the inverse of a matrix using transformations?

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The discussion centers on the process of finding the inverse of a matrix through transformations, specifically questioning why only one side of the equation is transformed at a time. It emphasizes that to maintain equality, the same transformations must be applied to both sides of the equation. The participants explore the implications of matrix multiplication, noting its non-commutative nature and the lack of guaranteed inverses, drawing parallels to real number multiplication. A proof is mentioned that highlights the necessity of simultaneous transformations in certain cases, while also acknowledging that not all properties of real numbers apply to matrices. The conversation ultimately underscores the importance of understanding linear transformations and their theoretical foundations in abstract algebra.
parshyaa
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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
 
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Each transformation step, T, is multiplication on the left: TA = TIA. It is not also multiplying the A of the RHS. TA ≠ TITA
 
FactChecker said:
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
 
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 Tn...T2T1, You have Tn...T2T1I = Tn...T2T1IA.

You should not "double up" on the right-side transformations like Tn...T2T1ITn...T2T1A
 
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FactChecker said:
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...Tn, You have T1T2...TnI = T1T2...TnIA.

You should not "double up" on the right-side transformations like T1T2...TnIT1T2...TnA
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
 
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.
 
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IMG_20170627_212308393_HDR.jpg

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I got this proof from an old book but its really satisfying
 
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.
 
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FactChecker said:
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
 

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