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Unformatted text preview: Solutions to Homework #3, MAT1341A Question # 1: (3+3 = 6 points) (a) Show that 1 , 1 7 , 1 1 1 is (i) linearly independent and (ii) spans R 3 . Is it a basis for R 3 ? Solution: To show that it is linearly independent, we need to solve a 1 + b 1 7 + c 1 1 1 = and show that the only solution is (0 , , 0). To show that it spans R 3 , we need to show that for any ( x, y, z ) R 3 , the equation a 1 + b 1 7 + c 1 1 1 = x y z has at least one solution. Hence, we can combine these two efforts, by solving the second equation, and then specializing to the case ( x, y, z ) = (0 , , 0). (In terms of our discussion about solving general systems of linear equations, we are trying to show that this system always has a unique solution. Then in particular for (0 , , 0) it has only the unique solution, and for general ( x, y, z ) it has at least the unique solution, and all is proven.) Rewrite the equation as: a + b + c = x 7 b + c = y c = z Now remember: we are holding ( x, y, z ) fixed but arbitrary, and need to decide if we can find a, b, c . So we need to solve for a, b, c in terms of x, y, z . Which is easy, by backsubstitution: c = z, b = 1 7 ( y z ) , a = x 1 7 ( y z ) z = x 1 7 y 6 7 z We check that this is indeed a solution. Since there is a solution, we have shown (ii), that the set spans R 3 ; since the solution is unique (no choices for a, b, c besides the one given by the formula), we deduce (i). Since the set is LI and spans R 3 we see it is a basis....
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 Winter '10
 DAIGLE

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