Algebraic manipulation of identity matrix

  • Thread starter KataKoniK
  • Start date
I was doing a few practice questions and was a bit confused about how the solution manual manipulated the I identity matrix. For example,

Just say you had the following

I - 2A = B

where A and B are 2x2 matrices. B is given, but we need to find A.

Therefore, shouldn't it be

-2A = B - I
A = -(1/2) * (B - I)?

Because the book did it like this

2A = I - B
A = (1/2) * (1 - B)

Can anyone tell me why? Because both statements do not seem to be equivalent. (unless I'm missing something)

Thanks in advance.
 
466
4
I think you may have had one of those brain lapses that I hate so much. The statements are equivalent, as the negative sign will distribute.

-(1/2)*(B-I) = (1/2)*-(B-I) = (1/2)*(I-B)
 
Argh, thanks a bunch. So it's just basic algebraic manipulations correct? No special things I have to do to bring I to the other side?
 
466
4
Nope, nothing special. Matrix addition and scalar multiplication satisfy all of the properties of regular addition and multiplication (commutativity, associativity, distributivity, etc.), so basic manipulations will suffice.
 
Thanks a lot!
 

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