Introduction to Linear algebra-Strang-Solutions-Manual_ver13

# Introduction to Linear algebra-Strang-Solutions-Manual_ver13

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Unformatted text preview: es that the ﬁrst n odd numbers add to n2 . " #" # y1 D B1 y1 D B1 1 00 B1 D B2 1 10 B2 3 y1 C y2 gives y2 D B1 CB2 D y1 C y2 C y3 D B3 y3 D B2 CB3 0 11 B3 " # " # 100 100 The inverse of S D 1 1 0 is A D 1 1 0 : independent columns in A and S ! 111 011 4 The combination 0w1 C 0w2 C 0w3 always gives the zero vector, but this problem looks for other zero combinations (then the vectors are dependent, they lie in a plane): w2 D .w1 C w3 /=2 so one combination that gives zero is 1 w1 w2 C 1 w3 : 2 2 5 The rows of the 3 by 3 matrix in Problem 4 must also be dependent: r 2 D 1 .r 1 2 The column and row combinations that produce 0 are the same: this is unusual. " # 135 1 2 4 has column 3 D 2 .column 1/ C column 2 6 cD3 113 " # 101 1 1 0 has column 3 D cD 1 column 1 C column 2 011 " # 000 2 1 5 has column 3 D 3 .column 1/ column 2 cD0 336 C r 3 /. 7 All three rows are perpendicular to the solution x (the three equations r 1  x D 0 and r 2  x D 0 and r 3  x D 0 tell us this). Then the whole plane of the rows is perpendicular to x (the plane is also perpendicular to all multiples c x ). 2 32 3 x1 0 D b1 x1 D b1 1000 b1 x2 x1 D b2 x2 D b1 C b2 6 1 1 0 0 7 6 b2 7 8 D4 D A 1b x3 x2 D b3 x3 D b1 C b2 C b3 1 1 1 0 5 4 b3 5 x4 x3 D b4 x4 D b1 C b2 C b3 C b4 1111 b4 9 The cyclic difference matrix C has a line of solutions (in 4 dimensions) to C x D 0: 2 1 61 40 0 0 1 1 0 0 0 1 1 32 3 2 3 23 1 x1 0 c 0 7 6 x2 7 6 0 7 6c 7 D when x D 4 5 D any constant vector. 0 5 4 x3 5 4 0 5 c 1 x4 0 c Solutions to Exercises z2 z1 D b1 10 z3 z2 D b2 0 z3 D b3 7 z1 D z2 D z3 D b1 b2 b2 b3 b3 b3 D " 1 0 0 1 1 0 1 1 1 #" b1 b2 b3 # D 1 b 11 The forward differences of the squares are .t C 1/2 t 2 D t 2 C 2t C 1 t 2 D 2t C 1. Differences of the nth power are .t C 1/n t n D t n t n C nt n 1 C    . The leading term is the derivative nt n 1 . The binomial theorem gives all the terms of .t C 1/n . 12 Centered difference matrices of even size seem to be invertible. Look at eqns. 1 and 4: 2 32 3 2 3 2 32 3 0 1 00 x1 b1 First x1 b2 b4 0 1 0 7 6 x2 7 6 b2 7 solve 61 6 x2 7 6 b1 7 D D 40 5 1 0 1 5 4 x3 5 4 b3 5 x2 D b1 4 x3 5 4 b4 0 0 10 x4 b4 x3 D b4 x4 b1 C b3 13 Odd size: The ﬁve centered difference equations lead to b1 C b3 C b5 D 0. x2 x3 x4 x5 x1 x2 x3 x4 D b1 D b2 D b3 D b4 D b5 Add equations 1; 3; 5 The left side of the sum is zero The right side is b1 C b3 C b5 There cannot be a solution unless b1 C b3 C b5 D 0. 14 An example is .a; b/ D .3; 6/ and .c; d / D .1; 2/. The ratios a=c and b=d are equal. Then ad D bc . Then (when you divide by bd ) the ratios a=b and c=d are equal! Problem Set 2.1, page 40 1 The columns are i D .1; 0; 0/ and j D .0; 1; 0/ and k D .0; 0; 1/ and b D .2; 3; 4/ D 2i C 3j C 4k. 2 The planes are the same: 2x D 4 is x D 2, 3y D 9 is y D 3, and 4z D 16 is z D 4. The 3 4 5 6 7 8 9 solution is the same point X D x . The columns are changed; but same combination. The solution is not changed! The second plane and row 2 of the matrix and all columns of the matrix (vectors in the colu...
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## This note was uploaded on 09/25/2012 for the course PHY 103 taught by Professor Minki during the Spring '12 term at Korea Advanced Institute of Science and Technology.

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