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Unformatted text preview: 18.06 Professor Strang Quiz 1 February 28, 2005
Grading 1 Your PRINTED name is: SOLUTIONS 2 3 4 1 (26 pts.) Suppose A is reduced by the usual row 1 4 R= 0 0 0 0 operations to 0 2 1 2 . 0 0 Find the complete solution (if a solution exists) to this system involving the original A: Ax = sum of the columns of A. Solution The complete solution x = xparticular + xnullspace has 1 1 xparticular = 1 1 4 2 1 0 xnullspace = x2 + x4 . 0 2 0 1 The free variables x2 and x4 can take any values. The two special solutions came from the nullspace of R = nullspace of A. The particular solution of 1's gives Ax = sum of the columns of A. Note: This also gives Rx = sum of columns of R. 1 2 (18 pts.) Suppose the 4 by 4 matrices A and B have the same column space. They may not have the same columns ! (a) Are they sure to have the same number of pivots ? YES NO WHY? (b) Are they sure to have the same nullspace ? YES NO WHY? (c) If A is invertible, are you sure that B is invertible ? YES NO WHY? Solution (a) YES. Number of pivots = rank = dimension of the column space. This is the same for A and B. (b) NO. The nullspace is not determined by the column space (unless we know that the matrix is symmetric.) Example 1 1 1 1 A= 1 1 1 1 with same column spaces 1 1 1 0 1 0 1 1 and B = 1 0 1 1 1 1 1 0 but different nullspaces: 0 0 0 0 . 0 0 0 0 (c) YES. If A is invertible, its column space is the whole space R4 . Since B has the same column space, B is also invertible. 2 3 (40 pts.) (a) Reduce A to an upper triangular matrix U and carry out the same elimination steps on the right side 3 3 A b = 3 5 3 3 b: 1 b1 1 b2  U c 2 b3 Factor the 3 by 3 matrix A into LU = (lower triangular)(upper triangular). (b) If you change the last entry in A from 2 to (what number gives Anew ?) then Anew becomes singular. Describe its column space exactly. (c) In that singular case from part (b), what condition(s) on b1 , b2 , b3 allow the system Anew x = b to be solved ? 3 (d) Write down the complete solution to Anew x = 3 (the first column). 3 Solution = 3 3 1 b1  U c = 0 2 0 b2  b1 3 5 1 b2 3 3 2 b3 0 0 3 b3  3b2 + 4b1 3 3 1 3 3 1 b1 (a) A b Here A = 1 0 0 3 3 1 3 5 1 = LU = 1 1 0 0 2 0 3 3 2 1 3 1 0 0 3 (b) If you change A33 from 2 to 1, the third pivot is reduced by 3 and Anew becomes singular. Its column space is the plane in R3 containing all combinations of the first columns (3, 3, 3) and (3, 5, 3). (c) We need b3  3b2 + 4b1 = 0 on the right side (since the left side is now a row of zeros). 3 (d) Anew gives 3 3 1 3  0 2 0 0 . 3 5 1 3 3 3 1 3 0 0 0 0 3 3 1 3 1 3 Certainly xparticular = 0 . Also xnullspace = x3 0 . 0 1 The complete solution is xparticular + any vector in the nullspace. 4 (16 pts.) Suppose the columns of a 7 by 4 matrix A are linearly independent. (a) After row operations reduce A to U or R, how many rows will be all zero (or is it impossible to tell) ? (b) What is the row space of A ? Explain why this equation will surely be solvable: 1 0 AT y = . 0 0 Solution (a) The rank is 4, so there will be 7  4 = 3 rows of zeros in U and R. (b) The row space of A will be all of R4 (since the rank is 4). Then every vector c in R4 is a combination of the rows of A, which means that AT y = c is solvable for every right side c. 4 ...
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This note was uploaded on 01/14/2011 for the course EECS 18.06 taught by Professor Strang during the Spring '05 term at University of Michigan.
 Spring '05
 Strang

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