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9. 18.03 Linear Algebra Exercises Solutions
9A. Matrix Multiplication, Rank, Echelon Form
9A-1.(i) No. The pivots have to occur in descending rows.
(ii) Yes. There's only one pivotal column, and it's as required. (iii) Yes. There's only one pivotal column, and it's as required. (iv) No. The pivots have to occur in consecutive rows. (v) Yes.9A-2(i)41.
(ii) This is reduced echelon already. (iii)1 1 1 1 1 1 0 0 (iv) 242 1 0
12 1 0 123 5 2 41120 12 1 0 123 5 2 4112
0 032
1 0 123 5 2 4112
0 0 123 0 123 5 2
41 013
0 123 0 083 3 5 241 0 0
0 1 00 0 13
5 . (We'll soon see that this is not a surprise.)9A-3There are many answers. For example,1or1work ne. Or [cossin];
etc.9A-4Here's several answers:2
6 6411 0 03 7 75,2
6 641
1 1 33
7 75,2
6 641
2 2 13 7 75
The interesting thing to say is that the answer is any vector in the nullspace of
1 1 1 1. The simplest solution is2
6 6400 0 03 7 75.
9A-5(a) The obvious answer to this question isv=2
6 6400 1 03 7
75; for any matrixAwith
four columns,A2 6 6400 1 03 7
75is the third column ofA.
But there are other answers: Remember, the general solution is any particular solution plus the general solution to the homogeneous problem. The reduced echelon form of 1 Amay be obtained by subracting the last row from the rst row:R=241 0 0 3
0 1 0 2
0 0 1 13
5Rv=0has solutions which are multiples of2
6 6432 1 13 7
75. So for anyt,A2
6 6413t12t 1t 1 +t3 7 75=2
41
0 13 5 (b) For any matrix with three rows, the product [0 0 1]Ais its third row. In this case, the rows are linearly independent|the only row vectorusuch thatuA=0is u=0|so there are no other solutions.
9A-6(a)2
411 13 5
1 01=2
41 011 01 1 013 5 . Rank 1, as always for rowcolumn with both nonzero. (b) 1 212 41
1 13 5 = [2] Rank 1 just because it's nonzero! (c) 1 2 0 0 1 1 2 41 2
0 1 2 33 5 =1 4 2 4 Rank 2: the columns are linearly independent (as are the rows).
9B. Column Space, Null Space, Independence, Basis, Dimension
9B-1First of all, any four vectors inR3are linearly dependent. The question here is
how. The rst thing to do is to make the 34 matrix with these vectors as columns. Now linear relations among the columns are given by the null space of this matrix, so we want to nd a nonzero element in this null space. To nd one, row reduce:241 1 1 2
0 1 1 3
0 0 1 43
5 241 0 01
0 1 1 3
0 0 1 43
5 241 0 01
0 1 01
0 0 1 43
5 The null space does't change under these row operations, and the null space is the space of linear relations among the columns. In the reduced echelon form it's clear that the rst three columns are linearly independent. (This is clear from the original matrix, too, because the rst three columns form an upper triangular matrix with nonzero entries down the diagonal.) The rst three variables are pivotal, and the last is free. Set the last one equal to 1 and solve for the rst two to get a basis vector for the null space:241 1 1 2
0 1 1 3
0 0 1 43
526 641
1 4 13 7 75=2
6 640
0 0 03 7 75
2 Check: the sum of the rst, second, and fourth columns is 4 times the third.