Exercise 1.7.14

Answers

Let S = [ade d b f efc ], then we have

xTSx = [x1x2x3 ] [ ade d b f efc ] [x1 x2 x3 ] = ax12 + bx 22 + cx 32 + 2dx 1x2 + 2ex1x3 + 2fx2x3 = 4(x1 x2 + 2x3)2 = 4x12 + 4x 22 + 16x 32 8x 1x2 + 16x1x3 16x2x3

Solve for a,b,c,d,e,f, we have S = [ 4 4 8 4 4 8 8 816 ]

This matrix has pivots of 4,0,0, it has rank of 1, it’s determinant is zero. The eigenvalues are: 0,0,24

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2020-03-20 00:00
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