Question

Can it be shown that

\[{(\textbf{S~T})}^{-1}= \dots\]
\[\dots=(\textbf{T}^{-1}\textbf{S}^{-1})\]

Answer 1

my head is my only house unless it rains

Answer 2

*
\[{(\textbf{ST})}^{-1}= \dots\]
\[\dots=(\textbf{T}^{-1}\textbf{S}^{-1})\]

Answer 3

it is always true

Answer 4

\[(ab)^{-1}=b^{-1}a^{-1}\] because \(abb^{-1}a^{-1}=aa^{-1}=I\)

Answer 5

i dont think i really understand the concept of the inverse of a matrix/ies

Answer 6

true for functions, elements of a group, anywhere you have inverses. i thas nothing to do with matrices

Answer 7

why do i have to sow it that way it isnt a logical agrument

Answer 8

hold on

Answer 9

it is to do with matrices because the order is what is important

Answer 10

what is the meaning of \((ab)^{-1}\)? it is the inverse of \((ab)\) i.e. it is the element that when you multiply by \((ab)\) you should get \((ab)(ab)^{-1}=(ab)^{-1}(ab)=I\) the identity

Answer 11

so no matter what you have, be it numbers, matrices, functions, arbitrary elements of some group, premuations, anything at all
it is always true that \((ab)^{-1}=b^{-1}a^{-1}\) always
only reason we don't usually see it is because numbers commute whereas functions and matrices do not

Answer 12

i know it is true i just cant prove it to my self mathematically

Answer 13

so what is needed to show that \((MN)^{-1}=N^{-1}M^{-1}\) is that when you multiply by \(MN\) you get \(I\)

Answer 14

and it has to work, because
\[MNN^{-1}M^{-1}=MIM^{-1}=MM^{-1}=I\]

Answer 15

this is true but isn's the same as what i am asking

Answer 16

why is this the best evidence 'it has to work' because of this other expression is equal to I it just like a convoluted proof that isnt really self justifying

Answer 17

ok lets ask the question this way. what is the meaning of \((ab)^{-1}\)?

Answer 18

i am thinking you are looking for some kind of mechanics that will make it work maybe? i am not sure

Answer 19

suppose we have two functions with inverses say domain all real numbers \(f,g,g^{-1}, f^{-1}\) and composition is defined. then what is the inverse of \(f\circ g\)? i.e. what is \((f\circ g)^{-1}\)?

Answer 20

To prove similar equation { where the tilde ~ is denoting the transpose

\[\widetilde{(\textbf{ST})}=\widetilde{(\textbf{S}_{ik}\textbf{T}_{kj})}_{ij}\]\[= \sum\limits_{k=1}^n s_{jk}t_{ki}\]\[=\sum\limits_{k=1}^n \widetilde{t_{ik}} \widetilde{s_{kj}}\]\[=(\widetilde{ \textbf{T}_{ik}} ~\widetilde {\textbf{S}_{kj} })_{ij}\]\[=(\widetilde{ \textbf{T}}~\widetilde {\textbf{S} })\]

Answer 21

Well, we are relying on a couple of things for this proof. First, we rely on associativity, that is
(ab)(b'a') = a(bb')a'
second, we rely on the definition of inverses to get bb' = I, the identity element. That gets us to
a*I*a'
Then we rely on the definition of the identity element to get
a*a'
finally we rely on the definition of inverses again to get
a*a' = I
So we showed that (ab)(b'a') = I
Therefore, (b'a') = (ab)'

Answer 22

i dont know what is
\[(f \circ g)^{-1}=\]

Answer 23

it is \(g^{-1}f^{-1}\)

Answer 24

nice tilda btw!

Answer 25

also nice trout mask, i would have used it if i had thought of it first

Answer 26

Ha ha right right
Just dig it
That's right "The Mascara Snake"
Fast 'n bulbous
Tight also

Answer 27

\[ (\textbf{ST}) {(\textbf{ST})}^{-1} = \textbf {I}\]
\[\textbf{T}{(\textbf{ST})}^{-1} = \textbf {S}^{-1}\]
\[{(\textbf{ST})}^{-1} = \textbf{T}^{-1}\textbf {S}^{-1}\]