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Exercise 3.4.8

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If $c_{1} (w_{2} + w_{3}) + c_{2} (w_{1} + w_{3}) + c_{3} (w_{1} + w_{2}) = 0$ then $(c_{2} + c_{3}) w_{1} + (c_{1} + c_{3}) w_{2} +$ $(c_{1} + c_{2}) w_{3} = 0 .$ Since the $w$ ’s are independent, $c_{2} + c_{3} = c_{1} + c_{3} = c_{1} + c_{2} = 0 .$ The only solution is $c_{1} = c_{2} = c_{3} = 0 .$ Only this combination of $v_{1}, v_{2}, v_{3}$ gives $0$ . (changing $- 1$ ’s to 1 ’s for the matrix $A$ in solution 7 above makes $A$ invertible.)

mathchakra

2022-01-23 15:43

Exercise 3.4.8

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