# Inverse and Power Matrices

• Oct 23rd 2008, 05:50 PM
aaronrj
Inverse and Power Matrices
I have a simple 2 x 2 matrix, a =

-1 2
1 3

Correct me if I'm wrong, but the inverse of this matrix is:

1/5 -2/5
-1/5 -3/5

How would I take a^3? Would it be:

-1^3 2^3
1^3 3^3

What would (A^-1)^3 be?
• Oct 23rd 2008, 06:46 PM
Shyam
First calculate \$\displaystyle A^2= A.A \$ then find \$\displaystyle A^3\$ after.

\$\displaystyle A^3\$ will be \$\displaystyle =A^2.A\$
• Oct 24th 2008, 04:02 AM
HallsofIvy
Quote:

Originally Posted by aaronrj
I have a simple 2 x 2 matrix, a =

-1 2
1 3

Correct me if I'm wrong, but the inverse of this matrix is:

1/5 -2/5
-1/5 -3/5

Yes, you are wrong. It should be fairly easy to see that, multiplying the first row of the first matrix by the first column of the second, (-1)(1/5)+ 2(-1/5)= -3/5 when it should be 1. It looks like you have just divided the numbers in the matrix by the determinant. That does NOT give the inverse matrix.

Quote:

How would I take a^3? Would it be:

-1^3 2^3
1^3 3^3

What would (A^-1)^3 be?
It looks like either you do not know how to multiply matrices or you simply have not thought about multiplying matrices in connection with either of these problems. What is A multiplied by itself?