linear algebra - Product of inverse matrices $ (AB)^ {-1 ...
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linear algebra - Intuitve explanation for $(AB)^{-1} = B^{-1}A^{-1 ...
2020年9月16日 · So I can do a mathematical proof of $(AB)^{-1} = B^{-1}A^{-1}$. But I would like a deeper, I guess more applicable, understanding.
abstract algebra - Check my proof that $(ab)^{-1} = b^{-1} a^{-1 ...
$\begingroup$ Note in my response to JavaMan's proof above-your proof has the major flaw in that it assumes the inverse is 2 sided and unique and the definition of a group doesn't require that.
proof verification - Proving $(ab)^{-1} = a^{-1}b^{-1}$, if $a,b\ne 0 ...
2019年2月14日 · Stack Exchange Network. Stack Exchange network consists of 183 Q&A communities including Stack Overflow, the largest, most trusted online community for developers to learn, share their knowledge, and build their careers.
linear algebra - Intrinsic proof for $(I + AB)^{-1}A = A(I + BA)^{-1 ...
2016年4月25日 · $\begingroup$ I think it's difficult to use the inverse formula, simply because if you do, then one side has a term that contains $(I-AB)^{-1}$ and the other side contains $(I-BA)^{-1}$, so you're basically back where you started. $\endgroup$
real analysis - Proof Verification: $ (ab)^{-1}=a^{-1}b^{-1 ...
Note that $(ab)^{-1}$ is the unique solution for $(ab)\cdot x=1$. Now, for us to show that $(ab)^{-1}=a^{-1}b^{-1}$ what we need to do is to verify that $(ab)(a^{-1}b^{-1})=1$. Using associativity, commutativity for multiplication and existence of multiplicative …
Give a counterexample to show that $(AB)^{-1} \\neq A^{-1}B^{-1}$
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If $AB = I$ then $BA = I$ - Mathematics Stack Exchange
2010年9月2日 · Dilawar says in 2. that he knows linear dependence! So I will give a proof, similar to that of TheMachineCharmer, which uses linear independence.
Boolean Algebra A + AB = A - Mathematics Stack Exchange
Hi I have a question about the following algebra rule A + AB = A My textbook explains this as follows A + AB = A This rule can be proved as such: Step 1: Dustributive Law: A + AB = A*1 = A(1+B) Huh...
abstract algebra - Is it true that $aH = bH$ iff $ab^{-1} \in H ...
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