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Exercise 1.3.5

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The rows of the 3 by 3 matrix in Problem 4 must also be dependent: $r_{2} = \frac{1}{2} (r_{1} + r_{3})$ . The column and row combinations that produce $0$ are the same: this is unusual. Two solutions to $y_{1} r_{1} + y_{2} r_{2} + y_{3} r_{3} = 0$ are $(Y_{1}, Y_{2}, Y_{3}) = (1, - 2, 1)$ and $(2, - 4, 2)$

mathchakra

2021-11-20 14:14

Exercise 1.3.5

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