Preparation problems for the discussion sections on September 30th andOctober 2nd1. Find an explicit description of Nul A, whereA =1 3 5 00 1 4 −2.Solution: We first bring the augmented matrix of the equation Ax1x2x3x4= 0 intoreduced echelon form:(0.1)1 3 5 0 00 1 4 −2 0R1→R1−3R2−−−−−−−→1 0 −7 6 00 1 4 −2 0.Since the first two columns of the augmented matrix are pivot columns, the variablesx1, x2are basic and the variables x3, x4are free. From (0.1), we getx1= 7x3− 6x4, and x2= −4x3+ 2x4.Hence any solution of Ax = 0 is of the formx1x2x3x4=7x3− 6x4−4x3+ 2x4x3x4= x37−410+ x4−6201.HenceNul(A) = Span({7−410,−6201}).2. Let A =1 3 1 22 6 4 80 0 2 4and b =131.Find the set of all solutions to the equation Ax = b, and express it as the sum of aparticular solution and solutions in the null space of A.Solution: Step 1: We start by bringing the augmented matrix of Ax = b into echelonform.1 3 1 2 12 6 4 8 30 0 2 4 1R2→R2−2R1−−−−−−−→1 3 1 2 10 0 2 4 10 0 2 4 1R3→R3−R2−−−−−−−→1 3 1 2 10 0 2 4 10 0 0 0 0.Since columns 1, 3 are the pivot columns, x2, x4are the free variables.12Step 2: Find a particular solution to Ax = b.In order to find a particular solution to Ax = b, set x2= x4= 0. Then by the second rowof the echelon form, 2x3= 1 and hence x3= 0.5. Hence the first row gives1x1+ 0.5 = 1.Hence0.500.50is a particular solution of Ax = b.Step 3: Find an explicit description of Nul(A).The echelon form of the augmented matrix of Ax = 0 is1 3 1 2 00 0 2 4 00 0 0 0 0.We will now bring it to reduced echelon form.1 3 1 2 00 0 2 4 00 0 0 0 0R2→R2/2−−−−−→1 3 1 2 00 0 1 2 00 0 0 0 0R1→R1−R2−−−−−−−→1 3 0 0 00 0 1 2 00 0 0 0 0.we getx1= −3x2, and x3= −2x4.Hence any solution of Ax = 0 is of the formx1x2x3x4=−3x2x2−2x4x4= x2−3100+ x400−21.HenceNul(A) = Span({−3100,00−21}).Step 4: Adding the null space to the particular solution.Every solution of Ax = b is of the form0.500.50+ c1−3100+ c200−21,where c1, c2are real numbers.33. Consider the subspaceW := {abcd: a = 2b + c, 2a = c − 3d}.Find a matrix A and a matrix B such that W = Col (A) and W = Nul(B).Solution: Let B be−1 2 1 0−2 0 1 −3.It is easy to see that Babcd=00iff a = 2b + c and 2a = c − 3d. Hence W = Nul(B).Note that a = 2b + c and 2a = c − 3d hold iff a = 2b + c and d = −43b −13c. Hence forabcdin W , we haveabcd=2b + cbc−43b −13c= b210−43+ c101−13.HenceW = Col(2 11 00 1−43−13).[Note that your answer might look different. In that case, as an exercise, check that thespaces are indeed the same. By the way, if you like recipes, you can always proceed as inthe very first problem to express any null space as a column space.]4. a) For which values of h is v3in the span of v1and v2? b) For which values of h is{v1, v2, v3} linearly dependent?(i) v1=1−5−3, v2=−2106, v3=2−9h,(ii) v1=−73−6, v2=2−14, v3=12h,(iii) v1=6−43, v2=6−122, v3=9h3.Solution: We bring the matrix [v1v2v3] to echelon form.4For (i),1 −2 2−5 10 −9−3 6 hR2→R2+5R1,R3→R3+3R1−−−−−−−−−−−−−−−→1 −2 20 0 10 0 h + 6R3→R3−(h+6)R2−−−−−−−−−−→1 −2 20 0 10 0 0.This can never have three pivots. Hence for every h, {v1, v2, v3} is linearly dependent.Because of R2, the equation x1v1+ x2v2= v3is inconsistent, independent of what h is.In other words, v3is never in the span of v1and v2.For (ii),−7 2 13 −1 2−6 4 hR2→R2+3/7R1,R3→R3−6/7R1−−−−−−−−−−−−−−−−−−→−7 2 10 −1/7 17/70 16/7 h − 6/7R3→R3+16R2−−−−−−−−→−7 2 10 −1/7 17/70 0 h + 38For (a), this system is inconsistent iff h 6= −38. Hence v3is in the span of v1and v2iffh = −38.For (b), the matrix has three pivots iff h 6= −38. Hence {v1, v2, v3} is linearly indepen-dent iff h 6= −38.For (iii),6 6 9−4 −12 h3 2 3R2→R2+2/3R1,R3→R3−1/2R1−−−−−−−−−−−−−−−−−−→6 6 90 −8 h + 60 −1 −3/2R3→R3−1/8R2−−−−−−−−−→6 6 90 −8 h + 60 0−h−188For (a), this system is inconsistent iff h 6= −18. Hence v3in the span of v1and v2iffh = −18.For (b), the matrix has three pivots iff h 6= −18. Hence {v1, v2, v3} is linearly indepen-dent iff h 6= −18.5. Check whether the following sets of vectors are linearly independent. Justify youranswer!a) {4−26,6−39} b) {4−2,60,13,−21}c) {4−26,629,000} d) {40−12,0−5510,31−1−11}5Solution: (a) Linearly dependent, since324−26=6−39.(b) Linearly dependent, because there are four vectors, but only two entries in each vector.(c) Linearly dependent, because it contains the zero vector.(d) Linearly independent. Indeed:4 0 30 −5 1−1 5 −12 10 −11R3→R3+1/4R1,R4→R4−1/2R1−−−−−−−−−−−−−−−−−−→4 0 30 −5 10 5 −1/40 10 −25/2R3→R3+R2,R4→R4+2R2−−−−−−−−−−−−−−−→4 0 30 −5 10 0 3/40 0 −21/2R4→R4+42/3R3−−−−−−−−−→4 0 30 −5 10 0 3/40 0 0There is a pivot position in every column. Hence the vectors are linearly independent.6. True or false? Justify your answers!(a) If three vectors in R3span a plane, then they are
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