Preparation problems for the discussion sections on September 9th and 11th1. Determine if the vector−511−7is a linear combination of1−22,055,208.Solution: We check whether there are x1, x2, x3in R such thatx11−22+ x2055+ x3208=−511−7.For this, it is enough to check whether the system of linear equations with thefollowing augmented matrix is consistent:1 0 2−5−2 5 0 112 5 8 −7.We bring the augmented matrix in echelon form:1 0 2 −5−2 5 0 112 5 8 −7R2→R2+2R1,R3→R3−2R1−−−−−−−−−−−−−−−→1 0 2 −50 5 4 10 5 4 3R3→R3−R2−−−−−−−→1 0 2 −50 5 4 10 0 0 2The system is inconsistent, because in echelon form there is a row of the form0 0 0 x ,where x is non-zero. Hence the vector−511−7is not a linear combination of1−22,055,208.2. Give a geometric description of Span{302,−203}.1Solution: Two nonzero vectors v1and v2span a plane iff (this is short for“if and only if”) there is no real number c such that cv1= v2. Suppose thereis c such thatc302=−203.By the first entry of the two vectors, we have 3c = −2. So c = −23. But by thethird entry, we get 2c = 3. So c =32. This is impossible since −236=32. HenceSpan{302,−203} is a plane.3. True or false? Justify your answers!(a) Let A be an m × n-matrix and B be an m × l-matrix. Then the productAB is defined.(b) The weights c1, ..., cpin a linear combination c1v1+ ... + cpvpcannot allbe zero.(c) Span{u, v} contains the line through u and the origin.(d) Asking whether the linear system corresponding toa1a2a3b isconsistent, is the same as asking whether b is a linear combination ofa1, a2, a3.Solution: (a) This is false. Let A be a m1× m2matrix and B be a n1× n2matrix. Then AB is defined if and only if m2= n1.(b) This is false. The weights can be zero. Check the definition in the lecturenotes!(c) This is correct. The Span{u, v} contains all vectors of the formcu + 0v,where c is in R. These vector form a line through u and the origin.(d) This is correct. Check the definition of being a linear combination in thelecture notes.4. Determine whether2−16is a linear combination of the columns of1 0 5−2 1 −60 2 8.2Solution: We check whether there are x1, x2, x3in R such thatx11−20+ x2012+ x35−68=2−16.For this, it is enough to check whether the system of linear equations with thefollowing augmented matrix is consistent:1 0 5 2−2 1 −6 −10 2 8 6.We bring the augmented matrix in echelon form:1 0 5 2−2 1 −6 −10 2 8 6R2→R2+2R1,−−−−−−−−→1 0 5 20 1 4 30 2 8 6R3→R3−R2−−−−−−−→1 0 5 20 1 4 30 0 0 0The system is consistent, because in echelon form there is a row of the form0 0 0 x ,where x is non-zero. Hence the vector2−16is a linear combination of thecolumns of1 0 5−2 1 −60 2 8.5. Compute AB in two ways: (a) by the definition, where Ab1and Ab2arecalculated separately, and (b) by the row-column rule for computing B.(i) A =4 −2−3 03 5, B =1 32 −1(ii) A=5 1 06 0 1, B =6 10 −10 1Solution: For (i), by the row-column rule we getAB =0 14−3 −913 4.3If we calculated Ab1and Ab2separately, we have4 −2−3 03 512=0−313and4 −2−3 03 531=14−94.For (ii), by the row-column rule we getAB =30 436 7.If we calculated Ab1and Ab2separately, we have5 1 06 0 1600=3036and5 1 06 0 11−11=47.6. Let A =2 14 2.(1) If x =00, what is Ax?(2) If x =1−2, what is Ax?(3) Is Ax = b uniquely solvable: is there for a given b always exactly one x?Solution: For (1):2 14 200=00.For (2):2 14 21−2=00.4For (3): As we have shown in (1) and (2), if b =00, there are two vectors x,namely00and1−2, such that Ax = b (this means that there are actuallyinfinitely many solutions x; find all of them!).57. (Some interesting matrices) Find a matrix A (what size!) such that(i) Axy=xy(ii) Axy=xy + 3x(iii) Axy=yxSolution: For (i):A =1 00 1For (ii):A =1 03 1For (iii):A =0 11
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