Question

(adj B)^-1 = adj(B^-1)

Answer 1

( adj B)^-1 =\[\frac{ 1 }{ \det B } B\]

Answer 2

??

Answer 3

are you allowed to use

\[A^{-1}=\frac{1}{\det(A)}\text{adj}(A)\]

Answer 4

yess

Answer 5

sorry OS went down ... use the above defn ... you will get \( |B| B = |B|B\)

Answer 6

\[\text{adj}(A)=\det(A)A^{-1}\]

\[(\text{adj}(A))^{-1}=(\det(A)A^{-1})^{-1}=(\det(A))^{-1}(A^{-1})^{-1}=\frac{1}{\det(A)}A\]

Answer 7

how about this?

Answer 8

Replace the A's above by B's

then use \[\text{adj}(B)=\det(B)B^{-1}\]


replace \(B\) with \(B^{-1}\)

Answer 9

imean by the other one....

Answer 10

what do you mean by "by the other one..."?

Answer 11

here

Answer 12

see the attachment please

Answer 13

yes...I told you what you could do to solve that

Answer 14

??

Answer 15

use \[\text{adj}(B)=\det(B)B^{-1}\]

replace \(B\) by \(B^{-1}\). What do you get

Answer 16

dont know

Answer 17

\[\text{adj}(B^{-1})=\det(B^{-1})(B^{-1})^{-1}\]

simplify this (the right hand side)

Answer 18

I??

Answer 19

no

Answer 20

then?

Answer 21

if you can't finish from here then you need to go and review the properties of invertible matrices

Answer 22

I've practically done the entire problem

Answer 23

what i get if adj.adj A

Answer 24

I don't even see what that means.
@Zarkon 's proof is almost complete and totally clear. If you don't understand any of it you really do need to review matrix properties as he said.