08bAss1sol

# 08bAss1sol - AS1sol/MATH1111/YKL/08-09 THE UNIVERSITY OF...

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AS1sol/MATH1111/YKL/08-09 THE UNIVERSITY OF HONG KONG DEPARTMENT OF MATHEMATICS MATH1111: Linear Algebra Assignment 1 Suggested Solution Remark : You may find that the solutions (especially the proofs) are rather lengthy. It is because we hope to explain every step detailedly. After reading each answer, you should dig out the central ideas and work out the answer again by your words. 1. Substituting x = 3, y = - 1, z = 2 in turn into the equations, we obtain a system of linear equations in unknowns a, b, c : 3 - a + 2 c = 0 3 b - c - 6 = 1 3 a - 2 + 2 b = 5 Its augmented matrix is - 1 0 2 - 3 0 3 - 1 7 3 2 0 7 3 R 1 + R 3 -----------→ - 1 0 2 - 3 0 3 - 1 7 0 2 6 - 2 - 2 3 R 2 + R 3 -----------→ - 1 0 2 - 3 0 3 - 1 7 0 0 20 3 - 20 3 - R 1 , 1 3 R 2 , 3 20 R 3 -----------→ 1 0 - 2 3 0 1 - 1 3 7 3 0 0 1 - 1 . Hence the solution is a = 1 , b = 2 , c = - 1. 2. (a) Let x = ( x y ) T R 2 . To show that x is a linear combination of x 1 and x 2 , we need to prove that x = c 1 x 1 + c 2 x 2 is solvable for some scalars c 1 , c 2 , i.e. x = c 1 c + c 2 a y = c 1 d + c 2 b has a solution (for c 1 , c 2 ). The augmented matrix is c a x d b y . As x 2 6 = 0 , at least one of c or d is nonzero. Case 1. c 6 = 0. Then the row operations 1 c R 1 and - dR 1 + R 2 gives 1 a c x c 0 b - da c y - dx c . 1

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Note that b - da c 6 = 0, for otherwise, x 1 = a c x 2 (contradicting to the condition x 1 6 = λx 2 for any λ R ). Thus the operation ( b - da c ) - 1 R 2 further transforms the augmented matrix into 1 a c x c 0 1 ( b - da c ) - 1 ( y - dx c ) . From it we solve for c 1 and c 2 . Case 2. d 6 = 0. The proof is similar and the details are left to you.
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