quiz1_1806_sol_s05 - 18.06 Professor Strang Quiz 1 February...

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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.

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