# Matrix Question!

1. Nov 8, 2007

### t_n_p

The problem statement, all variables and given/known data

Give the following matrix A = [2 1; 3 2] show that (A^2)-4A+I=0 where I is the 2x2 identity matrix. Hence use your result to deduce the inverse of A.

The attempt at a solution

I can do the show part no problem, but I'm stuck on how to find inverse of A. I multiply both sides of the equation by inverse A and get..

(A^-1)(A^2) - 4A(A^-1) + I(A^-1) = 0(A^-1)

Can you apply index laws on matrices? Any suggestions on what I should do next? Thanks!

2. Nov 8, 2007

### Dick

The equation you just wrote simplifies to A-4I+A^(-1)=0. Do you see how? Can you solve that for A^(-1)? You were practically done!

3. Nov 8, 2007

### t_n_p

yeah i see now, was just unsure whether I could apply index law for the first set of terms. just to refresh my memory!

So any matrix multiplied by its inverse gives the identity matrix of the same dimensions and

any matrix multiplied by the identity matrix gives the original matrix

4. Nov 8, 2007

### Dick

What index law? You are just using A*A^(-1)=I. You mean A^n*A^m=A^(n+m)? Sure you can.

5. Nov 8, 2007

### t_n_p

Yeah I just wanted to know if you could use A^n*A^m=A^(n+m).

Now everything is ok!

6. Nov 9, 2007

### HallsofIvy

Staff Emeritus
It's not really a matter of exponents (not "indices"), just the distributive law. You already know that (A^2)-4A+I=0 so obviously 4A- A^2= A(4I-A^2)= I. Now use the definition of "multiplicative inverse".

7. Nov 9, 2007

### ZioX

Note that A does not have an inverse iff the characteristic polynomial has zero as a root iff the constant term of the characteristic polynomial is zero.

8. Nov 10, 2007

### HallsofIvy

Staff Emeritus
Yes, but the fact that (A^2)-4A+I=0 shows that neither of those is true!