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Exercise 2.1.7

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Column $3 =$ Column 1 makes the matrix singular. For $b = (2, 3, 5)$ the solutions are $(x, y, z) = (1, 1, 0)$ or $(0, 1, 1)$ and you can add any multiple of $(- 1, 0, 1) . b = (4, 6, c)$ needs $c = 10$ for solvability (then $b$ lies in the plane of the columns and the three equations add to $0 = 0$ ).

mathchakra

2021-11-20 14:38

Exercise 2.1.7

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