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hwLectures 6 and 7 - MAT 1341A These are the solutions to...

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MAT 1341A: These are the solutions to the practice assignment. Don’t read these until you have attempted the questions on your own! Question # 1: (a) Determine whether 4 8 0 1 span 2 1 - 3 5 , 0 2 - 1 3 R 4 . Solution: We need to try to solve: 4 8 0 1 = a 2 1 - 3 5 + b 0 2 - 1 3 for some a, b R . This gives the following 4 linear equations: 2 a = 4 , a + 2 b = 8 , - 3 a - b = 0 , 5 a + 3 b = 1 The first gives a = 2, so the second implies b = 3. But plugging this into the third equation gives a contradiction. So we conclude this system cannot be solved. Hence, our answer is NO. (b) Determine whether 10 6 - 8 22 span 1 2 0 3 , 1 0 - 2 1 M 2 , 2 . Solution: We need to try to solve: 10 6 - 8 22 = a 1 2 0 3 + b 1 0 - 2 1 for a, b R . Equating terms in each component gives the four equations: a + b = 10 , 2 a = 6 , - 2 b = - 8 , 3 a + b = 22 The second gives a = 3 and the third gives b = 4, which fails the first equation, so there is no solution, and again our answer is NO. Question # 2: Let P 1 be the vector space of all polynomials with degree 1. Show that any element of P 1 is in the span of the polynomials x - 2, 2 x + 1, and 2. ( Hint: take an arbitrary element of P 1 , which looks like ax + b for a, b R , and see if you can write it as a linear combination of these vectors. ) Solution: We need to decide if, for any ax + b in P 1
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