Linear Algebra: Solve B^(-1) with A, C, D

[sam0617]

I'm looking for help to this problem. Here is my attempt:

I being the identity matrix and B^(-1) being B to the negative 1st power.

A (B B^(-1)) C D = B^(-1) I
so A I C D = B^(-1)
so A C D = B^(-1)

Thank you for any help.

 

[chiro]

I'm looking for help to this problem. Here is my attempt:

I being the identity matrix and B^(-1) being B to the negative 1st power.

A (B B^(-1)) C D = B^(-1) I
so A I C D = B^(-1)
so A C D = B^(-1)

Thank you for any help.

Hey sam0617 and welcome to the forums.

Assuming you start off with ABCD = I, that operation will not work. In terms of matrix operations with multiplication, you can only left multiply or right multiply, and you have to do the same thing for each side.

So if you want to pre-multiply by B^(-1), then you will get B^(-1) x ABCD = B^(-1) x I = B^(-1) which is not equal to ACD.

Given these hints, what is your next step?

 

[sam0617]

Hey sam0617 and welcome to the forums.

Assuming you start off with ABCD = I, that operation will not work. In terms of matrix operations with multiplication, you can only left multiply or right multiply, and you have to do the same thing for each side.

So if you want to pre-multiply by B^(-1), then you will get B^(-1) x ABCD = B^(-1) x I = B^(-1) which is not equal to ACD.

Given these hints, what is your next step?

I think I figured it out. I can't just do what I did on my attempt. (I wrote that on my test but owell)

A^(-1)ABCD = A^(-1)I
BCD = A^(-1)
BCDA=A^(-1) A
BCDA = I
B^(-1)BCDA = B^(-1)I
CDA = B^(-1)

Thank you for the help.