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Exercise 1.7.16

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Using determinants we see that : * $S$ : $c > 0$ , $c^{2} - 1 > 0$ , and $c^{3} - 3 C + 2 > 0$ , we conclude $c > 1$ * $T$ : $d - 4 > 0$ and $5 d + 24 + 24 - 9 d - 16 - 20 > 0$ , i.e. $12 - 4 d > 0$ , we conclude that no $d$ that can make $T$ positive definite.

niuers

2020-03-20 00:00

Exercise 1.7.16

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