Assignment Solutions_a6s

# 0 0 1 2 0 0 1 2 r2 2r3 0000 000 0 03 2 9 1 1 1 0 1

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Unformatted text preview: 4r3 r1 + r2 0 1 2 4 1 0 0 1 0 . 0 0 1 2 −→ 0 0 1 −→ 2 r2 − 2r3 0000 000 0 03 2 9 1 1 −1 0 1 1 1 −1 r1 ↔ r2 0 3 r3 − 2r1 0 0 2 9 2. a) −→ −→ 25 0 −7 25 0 −7 0 1 1 −1 0 r3 − r2 03 2 9 , so the rank of A is R(A) = 3. −→ 00 0 −16 −1 1 b) A = 1 −1 −8 8 −1 1 −→ 00 r3 + 3r2 00 0 4 2 0 1 −1 0 3 2 9 3 2 −7 21 3 r2 + r1 −1 1 2 1 3 1 3 −4 −→ 0 0 3 4 −1 7 −4 36 r3 − 8r1 0 0 −9 −12 12 21 3 3 4 −1 , 00 9 so the rank of 1 −1 c) A = 2 −2 1 0 r3 ↔ r4 −→ 0 0 A is R(A) = 3. (The pivot columns are columns 1, 3 and 5.) 3 −4 r2 + r1 13 −4 1 3 −4 r + 3r2 −1 1 −→ 0 2 −3 3 −→ 0 2 −3 0 0 0 1 r3 − 2r1 0 −6 9 0...
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## This note was uploaded on 02/13/2014 for the course MATH 2050 taught by Professor Dryuan during the Fall '12 term at Memorial University.

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