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## Homework Statement

Use diagonalization to find A^9 where A = [0 8;2 0]

## The Attempt at a Solution

I perform (A-λI) to find characteristic equation and hence eigenvalues of 4 and -4. These yield eigenvectors of [2;1] and [2;-1] respectively.

Now I know [0 8;2 0][2 2;1 -1] = [2 2;1 1][4 0;0 -4]

Rearranging gives [0 8;2 0] = [2 2;1 -1][4 0;0 -4][2 2;1 -1]^-1

Now to raise to the power of 9 gives, [0 8;2 0]^9 = [2 2;1 -1][4 0;0 -4]^9[2 2;1 -1]^-1

Now I know that any matrix multiplied by its inverse will give the identity matrix, hence this leaves me with [0 8;2 0]^9 = [4 0;0 -4]^9

Therefore by my working, A^9 = [4^9 0; 0 (-4)^9]

The answer suggests [0 2*4^9; .5*4^9 0]

Can someone see where I went wrong?