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Notes1 - Elementary row operations ri(r,s Interchange rows...

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Unformatted text preview: Elementary row operations ri(r,s) Interchange rows r and s rm(m,r) Multiply row r by the scalar m ra(m,r,s) Multiply row r by m and add it to row a. 1 1 #8 example 1 1 6f zz= 1 2 —2 1 zz= 2 —1 1 3 zz= 3 2 1 0 2 5 1 9 1 2 —1 1 -2 1 -1 2 1 3 4 9 —1 2 3 7 2 —1 2 -1 ra(—2,1,2) ra(—1,1,3) ri(1,2) ra(1,2,1) 1 2 -2 1 1 2 —1 1 1 3 0 2 O 1 5 7 2 —1 1 3 —2 1 -1 - 2 0 1 6 B -1 2 3 7 2 —1 2 -1 ra(—1,2,3) ra(—2,1,2),ra(1,1,3) ra(2,1,2),ra(—2,1,3) 1 2 —2 1 1 2 —1 1 1 3 O 2 0 1 5 7 0 —5 3 1 0 7 -1 6 0 0 1 1 0 4 2 B 0 —7 2 —5 ra(-5,3,2),ra(2,3,1) rm(—1,2),ra(—1,3,2) ra(1,2,3) 1 2 O 3 1 2 —1 1 1 3 0 2 0 1 0 2 0 *5 3 1 0 7 —r\ 6 O 0 1 1 0 4 2 8 0 O 1 1 ra(-2,2,1) rm(—1,2),ra(-1,3,2) ra(1,3,2) 1 0 0 —1 1 2 —1 1 1 3 0 2 0 1 0 2 0 1 -5 —9 0 7 0 7 O 0 1 1 O 4 2 8 O 0 1 1 ra(-4,2,3) rm(1/7,2) rref of 22 1 2 -1 1 1 3 0 2 x1 — —1 O 1 -5 —9 O 1 0 1 x2 = 2 0 0 22 44 0 O 1 1 x3 — 1 rm(1/22,3) ra(—3,2,1) 0 1 -5 —9 0 1 0 1 0 0 1 2 0 O 1 1 ra(5,3,2),ra(1,3,1) rref of A 1 2 0 3 0 1 0 1 x1 = —1 0 0 1 2 x2 = 1 ra(—2,2,1) x3 = 1 1 O 0 1 0 1 O 1 0 0 1 2 rref = ech(zz) Example Underdetermined systems zz= 1 2 3 4 5 6 7 8 2 3 4 5 ra(—5,1,2),ra(—2,1,3) 1 2 3 4 0 -4 -8 -12 0 —1 -2 —3 ri(2,3),rm(—1,2) 1 2 3 4 0 1 2 3 0 -4 -8 —12 ra(4,2,3) 1 2 3 4 0 1 2 3 0 0 0 0 ra(-2,2,1) 1 0 —1 —2 0 1 2 3 O 0 0 0 The row of zeros precludes the existence of the third 1 2 o 3 o o 1 2 1 2 -2 —1 —2 «4 1 —4 pv (1 , 1) 1 2 o 3 o o 1 2 o o —2 —4 o o 1 2 pvl (2 , 3) 1 2 0 3 o o 1 2 o o o o o o o o rank(zz)=2 underdetermined system x1 = l — 3x4 -2x2 x2 = r x3 — 3 - 2x4 x4 = 5 x1 = 1 - 33 - 2r x2 = 0 + 1r x3 = 3 - 25 x4 = O + s OOU’H pivot 1 _ Thus there are ==========~============ only two leading variables: x1 = -2 + 1x3 x2 = 3 — 2x3 x3 = anything (= c) thus x1 = —2 + 1c x2 = 3 — 2c x3 = 0 + c we say that the rank of zz is 2 (the numbr of pivot 13. 2 3 —l 1 7 3 4 7 pV(l,1) 1 —l 2 4 O 5 -5 —7 O 10 —10 —21 ra(-2,2,3) 1 —1 2 4 O 5 —5 —7 O 0 O —7 rm(1/5,2) 1 —1 2 4 O 1 —1 -7/5 0 0 0 —7 pV(3,4) 1 —1 2 O O l —1 O 0 0 O 1 The third line corresponds to the equation: 0x1 + 0x2 + 0x3 1 or 0 = 1. Thus there is no solution. ...
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