Homework Help: Matrix Proof help please

1. Nov 20, 2005

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) :surprised

2. Nov 20, 2005

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

3. Nov 20, 2005

Rizzamabob

Ok, i didn't realise 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 !? :!!) :uhh:

4. Nov 20, 2005

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

5. Nov 20, 2005

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