solq5 - Solutions to quiz # 5 (October 7) 1. Writing X = 13...

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Unformatted text preview: Solutions to quiz # 5 (October 7) 1. Writing X = 13 31 a + 3c b + 3d ⇐⇒ 3a + c 3b + d a c a c = = = = b , we obtain d b ab 13 a + 3c = ⇐⇒ d cd 31 3a + c a + 3b 3a + b b=c ⇐⇒ c + 3d a = d. 3c + d b + 3d a + 3b 3a + b = 3b + d c + 3d 3c + d Answer: X= ab , ba where a and b can be any numbers. 2. To find a basis of the kernel, we solve the system of linear equations: x1 x2 x3 x4 12 0 3 |0 120 3 |0 2 0 3 | 0 1 −→ 0 0 1 −1 | 0 −→ 0 0 1 −1 | 0 , 1 2 1 2|0 0 0 −1 1 | 0 000 0 |0 1 2 −1 4 | 0 from which the kernel consists of the vectors −2 −3 x1 = −2t2 − 3t1 t2 1 0 x2 = , + t2 = t1 0 1 x3 = t1 0 1 x4 = t1 where t1 and t2 can be any numbers. Hence vectors −2 −3 1 0 and 0 1 0 1 constitute a basis of the kernel. To find a basis of the image, we pick the columns 1 0 1 and 1 1 −1 of the matrix corresponding to the leading variables x1 and x3 . −2 −3 01 4 Answer: The kernel is a plane in R with the basis and the image is , 0 1 0 1 0 1 a plane in R3 with the basis 1 , 1 . 1 −1 1 ...
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This note was uploaded on 12/26/2011 for the course MATH 214 taught by Professor Conger during the Fall '08 term at University of Michigan.

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