Solving Matrix Equations: Inverse of nxn (n=2)

[DiamondV]

Homework Statement

Homework Equations

Inverse of an (nxn) (n=2 only) square matrix:

The Attempt at a Solution

The answer provided in the solutions does the exact same thing except, where my ?? are. It does A = BCB^-1. Where as I do A = CBB^-1. When I was doing this question I was wondering the same thing. I know matrix multiplication isn't associative( AB isn't equal to BA), so how do I know which way to form the equations? I mean how am I meant to know whether its meant to be A=CBB^-1 or BCB^-1?[/B]

 

[Samy_A]

Homework Statement

Homework Equations

Inverse of an (nxn) (n=2 only) square matrix:

The Attempt at a Solution

The answer provided in the solutions does the exact same thing except, where my ?? are. It does A = BCB^-1. Where as I do A = CBB^-1. When I was doing this question I was wondering the same thing. I know matrix multiplication isn't associative( AB isn't equal to BA), so how do I know which way to form the equations? I mean how am I meant to know whether its meant to be A=CBB^-1 or BCB^-1?[/B]

Matrix multiplication is associative. You probably meant to write that matrix multiplication isn't commutative.

Since matrix multiplication isn't commutative, when you multiply one side of an equation to the left, you must multiply the other side of the equation to the left as well. Same for multiplying to the right.
So if $X=Y$, then $AX=AY$ and $XA=YA$, but not necessarily $XA=AY$.
(Here $X,\ Y, \ A$ are square matrices of the same dimension.)

Now in your case: apply this rule to $B^{-1}AB=C$ to get $A=BCB^{-1}$.

 

[DiamondV]

Matrix multiplication is associative. You probably meant to write that matrix multiplication isn't commutative.

Since matrix multiplication isn't commutative, when you multiply one side of an equation to the left, you must multiply the other side of the equation to the left as well. Same for multiplying to the right.
So if $X=Y$, then $AX=AY$ and $XA=YA$, but not necessarily $XA=AY$.
(Here $X,\ Y, \ A$ are square matrices of the same dimension.)

Now in your case: apply this rule to $B^{-1}AB=C$ to get $A=BCB^{-1}$.

Ah. so essentially whatever order they are in the at the left is the same order they get applied to on the right?

 

[Ray Vickson]

Ah. so essentially whatever order they are in the at the left is the same order they get applied to on the right?

No: they do not have the same order on the two sides.Look again, carefully.

 

[DiamondV]

No: they do not have the same order on the two sides.Look again, carefully.

Not understanding it completely. So say if I multiply the left side by some variable X and I put it to the left of whatever is already there, I have to do the same to the right? like say A=C, would be XA = XC?

 

[Samy_A]

Not understanding it completely. So say if I multiply the left side by some variable X and I put it to the left of whatever is already there, I have to do the same to the right? like say A=C, would be XA = XC?

Yes, this is correct. If A=C, then XA=XC.

Your previous post was ambiguous.

 

[DiamondV]

Yes, this is correct. If A=C, then XA=XC.

Your previous post was ambiguous.

Ah. Thanks a lot!

 

[HallsofIvy]

You are given that B^{-1}AB= C. Knowing that matrix multiplication is not commutative, get rid of the "B^{-1}" on the left by multiplying, by B, on both sides on the left: B(B^{1}AB)= BC. Because matrix multiplication is "associative" that is the same as (BB^{-1})AB= AB= BC. And to get rid of the "B" or the right, multiply on both sides by B^{-1} on the right to get (AB)B^{-1}=A(BB^{-1})= A= BCB^{-1}. It's just a matter of keeping track of which side you are on!