Diagonalizing a 3x3 Matrix: Troubleshooting the P-1AP = D Expression

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I have to diagonalize this matrix and I've found the eigenvalues and vectors and they're linearly independent but I can't get the expression P-1AP = D to work.

It's a 3x3 matrix, (-11,-46,-3),(0,12,0),(-1,-2,-9) with eigenvalues of 12,-12 and [STRIKE]8[/STRIKE] -8.
The eigenvectors are (-2,1,0), (3,0,1) and (-1,0,1).

The (P) I made was using the vectors but D is never a diagonal matrix am I missing something?

Edited to correct typo.
 
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Did you make the columns of P (not the rows) out of the eigenvectors? What did you get for P-1?
 


vela said:
Did you make the columns of P (not the rows) out of the eigenvectors? What did you get for P-1?

My matrix P was (-2,3,-1), (1,0,0), (0,1,1)
 


P-1AP comes out diagonal here with that matrix.
 


IF you say it's diagonal than my matrix multiplication skills must be in need of help... I'll try again and see. Thanks for your help!
 


I get the third eigenvalue to be -8, not 8 as you showed. That wouldn't have made a difference in your calculation of P and P-1 though.
 
Prove $$\int\limits_0^{\sqrt2/4}\frac{1}{\sqrt{x-x^2}}\arcsin\sqrt{\frac{(x-1)\left(x-1+x\sqrt{9-16x}\right)}{1-2x}} \, \mathrm dx = \frac{\pi^2}{8}.$$ Let $$I = \int\limits_0^{\sqrt 2 / 4}\frac{1}{\sqrt{x-x^2}}\arcsin\sqrt{\frac{(x-1)\left(x-1+x\sqrt{9-16x}\right)}{1-2x}} \, \mathrm dx. \tag{1}$$ The representation integral of ##\arcsin## is $$\arcsin u = \int\limits_{0}^{1} \frac{\mathrm dt}{\sqrt{1-t^2}}, \qquad 0 \leqslant u \leqslant 1.$$ Plugging identity above into ##(1)## with ##u...

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