FM5002-HW2-2.1.12

The eigenvalues for m are 22 2 3 1 1 4 we let a and 0

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Unformatted text preview: t M A 2 22 t 0 A 0 0 0 2t 0 0 0 xt we conclude that y t zt 22 0 0 0 2t A0 0 0 2t 10 C 3 10 22 t 3 10 2t 10 0 2t 10 1 0A 1 22 t 9 tM 22 t 9 22 t A1 0 2 0 t. We therefore have that 00 0 0 3 22 t 3 3 10 22 t 3 10 2t 22 t 2t 3 10 10 22 t 9 10 22 t 0 1 2 3 2t 9 1 2 3 22 t 22 t . Finally, 20 1 2 1 2t 39 20 1 2t 39 20 2t 13 20 20 2t 13 1 20 2 A1 20 22 t 2 10 22 t A1t MA A 20 0 2t 10 10 10 0 2t 3 10 10 A A 1 t M AA 1 tM 2 1 2 9 22 t 20 3 22 t 20 1 2t 20 3 2t 20 ....
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