Question

find the matrix C if (AC)^-1 = B.

MY SOLUTION IS
C=A^-1B^-1
it is correct or wrong

Answer 1

Is this kind of like inverse of the product?

Answer 2

If \((AB)^{-1} = B^{-1}A^{-1}\) , you have ...\[(AC)^{-1} = B\]\[(AC)^{-1} = A^{-1}B^{-1}\]

Answer 3

Correction: \((AC)^{-1} = C^{-1}A^{-1}\)

Answer 4

Therefore: \(C^{-1}A^{-1} = B\).....


Along those lines?!

Answer 5

next step will be to post mutiply both sides by A

Answer 6

ohhh and \(AA^{-1} = I\)?

Answer 7

\(C^{-1}A^{-1} A = BA\)

and noting that ...yes, that :)

Answer 8

\(C^{-1} = BA\)

then finding C is easy now :)

Answer 9

But id \(AA^{-1} = I\), where does the \(I\) go?

Answer 10

IX or XI
both = X

I is the identity matrix, mutiplied with any matrix results in the same matrix

only catch is that both the matries X and I must have same order

Answer 11

so
\(C^{-1}I = C^{-1}\)

Answer 12

If the order is the same it's kind of like multiplying by 1...

Answer 13

right!

Answer 14

@hbr27 you can participate in the conversation, if you have something to contribute :)

Answer 15

\[(BA)^{-1} = A^{-1}B^{-1}\]

Answer 16

so @hbr27
your solution is indeed correct :)