Unformatted text preview: 1 2 3 2 d 4 3 4 5 are 1, d4 and4 d + 8. These are positive when, simultaneously, d > 4 and 2 > d , that is never. 15 ) We know that for any ~x 6 = ~ 0 that ~x T A~x > 0 and ~x T B~x > 0. So for any ~x 6 = ~ 0 we have ~x T ( A + B ) ~x = ~x T A~x + ~x T B~x > 0. 25 ) A = C T C = ± 3 0 1 2 ²± 3 1 0 2 ² = ± 9 3 3 5 ² . For ± 4 8 8 25 ² , the ﬁrst elimination step is to add2 times row 1 to row 2. This is multiplying by ± 1 02 1 ² on the left, so ± 4 8 8 25 ² = ± 1 0 2 1 ²± 4 8 0 9 ² = ± 1 0 2 1 ²± 4 0 0 9 ²± 1 2 0 1 ² so C = ± 2 0 0 3 ²± 1 2 0 1 ² = ± 2 4 0 3 ² ....
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 '05
 HUI
 Math, Linear Algebra, Algebra, Harshad number, 1920, xT Ax, xT Bx, upper left determinants

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