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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.

$\begingroup$ 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 …

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 …

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 …

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 …

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 …

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.

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 = …

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 …