Proving the Inverse Existence of C and its Value through Matrix Proof

[Rizzamabob]

"Show that if a square matrix C satisfies
C^3 + C^2 + C + I = 0
then the inverse C^-1 exists and
C^-1 = -(C^2 + C + I)

 

[mathphys]

hint

all you need to remember is that
c^(-1)c=I
and
c^2 = cc
c^3 = ccc
and the associative property of matrix multiplication
P.S. c^(-1) is taken here as the left inverse of c

 

[Rizzamabob]

Ok, i didn't realize i could write matrix's like that and deal with them just as variables...
OK so this is what i did...

CCC + CC + C + I = 0 (re arrange)
C = - (CCC + CC + I) (I = C^(1) C)
C = - (CCC + CC + (C^(-1)C)) (GET C^(-1)C on other side)
C^(-1)C = - (CCC + CC + C) (DIVIDE BY C)
C^(-1) = - (CC + C + 1) (1 = Identity matrix)
C^(-1) = - (CC + C + I)

DONE !? :!)

 

[mathphys]

you are okay up to your third equation

"Show that if a square matrix C satisfies
C^3 + C^2 + C + I = 0
then the inverse C^-1 exists and
C^-1 = -(C^2 + C + I)

Lets do it simpler...

Just multiple your left hand side (every term on it !) by C^(-1) and the right hand side by the same.

C^(-1) ( C^3 + C^2+ C +I) = C^(-1) 0

What do you get from here is the result that you're looking for

 

[mathphys]

Because well, you cannot exactly treat matrices as variables..

The Division (Divide by C) is not defined, just the multiplication.
and the multiplication of a matrix by its (left) inverse has the property that

C^(-1)C=I

And because the matrix multiplication is associative, if

C^2=CC then C^(-1)C^2= C^(-1)CC = (C^(-1) C) C = I C = C

something similar for C^3 or an arbitrary power of C