Solutions

1) a) det A = 3

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3)

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4) u $ v = 12 - 12 = 0 so u z v.

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5) 3 + 4b = 0 t b = -3/4

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12)

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c)

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d)

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e)

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f) 3(5 + t) + 4(-2 + 2t) + 5(7 - t) - 60 = 0

so t = 3 which makes the point of intersection (8, 4 , 4)

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back to Linear Algebra MathRoom Index

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