Preparation problems for the discussion sections on September 16th and18th1. (1) Find a matrix E such that:ER1R2R3=R1R2− 2R1R3Which matrix E−1undoes the row operation implemented by E? What is E−1E?(2) Find a matrix F such that:FR1R2R3=R2R1R3Which matrix F−1undoes the row operation implemented by F ? What is F−1F ?(3) Find a matrix G such that:GR1R2R3=R13R2R3Which matrix G−1undoes the row operation implemented by G? What is G−1G?Solution:(1) E =1 0 0−2 1 00 0 1. To undo this row operation we have to replace R2→ R2+ 2R1.So E−1=1 0 02 1 00 0 1. It is easy to check that E−1E = I.(2) F =0 1 01 0 00 0 1. To undo this row operation we have to replace R2↔ R1. SoF−1= F =0 1 01 0 00 0 1. It is easy to check that F−1F = I.(3) G =1 0 00 3 00 0 1. To undo this row operation we have to replace R2→13R2. SoG−1=1 0 001300 0 1. It is easy to check that G−1G = I.Note: each row operation on A is equivalent to multiplying A from the left by an invertiblematrix.2. Consider the matrix:2 3 30 5 76 9 81Decompose the matrix A into LU, where L is a lower triangular matrix and U is an uppertriangular matrix. Then use this factorization to solve:2 3 30 5 76 9 8x1x2x3=225That means, find a vector c in R3such that:Lc =225and then find a vector x in R3such that:Ux = cSolution: We start by bringing A to echelon form by multiplying A by elementarymatrices. Let E be the matrix corresponding to subtracting row 1 three times from row3, that is1 0 00 1 0−3 0 1.ThenEA =1 0 00 1 0−3 0 12 3 30 5 76 9 8=2 3 30 5 70 0 −1.Since EA is already upper triangular, we set U := EA. Then E−1U = A. Hence L isE−1. To compute this explicitly, note that the inverse operation to subtracting row 1from row 3 three times is adding row 1 three times to row 3. HenceE−1=1 0 00 1 03 0 1.So the LU-decomposition of A is2 3 30 5 76 9 8=1 0 00 1 03 0 12 3 30 5 70 0 −1.[Note: the above steps include the maximum amount of details (not necessary for theexam). Can you see how to get L and U directly from the one row operation that isperformed here?]We now solve2 3 30 5 76 9 8x1x2x3=225.We first solve1 0 00 1 03 0 1c1c2c3=2252by forward substitution to getc1c2c3=22−1.We then solve2 3 30 5 70 0 −1x1x2x3=22−1by backward substitution to findx1x2x3=1−11.[Important note for the exam: time permitting, take a few moments to check that thesevalues indeed solve the original problem.]3. Let A =2 −1 0 0−1 2 −1 00 −1 2 −10 0 −1 2, L =1 0 0 0−121 0 00 −231 00 0 −341, and U =2 −1 0 0032−1 00 043−10 0 054.(1) Show that A = LU.(2) Let Aibe the i × i matrix introduced by the first i rows and the first i columns ofA, for i = 1, 2, 3. What is an LU decomposition of Ai, for i = 1, 2, 3?Solution:(1) Easy to check.(2) An LU decomposition for Aiis LiUiwhere Li(respectively, Ui) is the matrix in-troduced by the first i rows and the first i columns of L (respectively, U).4. (more challenging) Let A and B be n × n matrices such that AB = I.(1) What is the reduced echelon form of A?(2) Show that BA = I.3Solution:(1) If the reduced echelon form of A has less than n pivot positions, then we get arow of zeros in the echelon form. So, there is a matrix F (namely, the product ofall the elementary matrices used to reduce A) such that F A has a row of zeros.Therefore, F AB = F I = F also has a row of zeros; but this is impossible sincesuch a matrix F cannot possibly correspond to a reversible sequence of row oper-ations.So, the echelon form of A has exactly n pivot positions and we have exactly onepivot position in each row and in each column. These pivots are exactly on themain diagonal. Therefore, in the reduced echelon form every element on the maindiagonal is 1 and every other element is equal to 0. Thus, the reduced echelonform of A is I.(2) From the first part, there is a matrix F such that F A = I. We have:F = F I = F AB = IB = BSo:BA = F A = I5. Answer the following true-false questions. Explain why.(1) If A is invertible then Ax = 0 has exactly one solution, x = 0.(2) If A is invertible then AB is also invertible.(3) If A and B are invertible then A + B is also invertible.(4) If A is invertible then the reduced echelon form of A is equal to I.Solution:(1) True, in this case Ax = b has exactly one solution, x = A−1b.(2) False, let A = I and B = 0 (the zero matrix).(3) False, let A = I and B = −I.(4) True, see the first part of problem (4).6. If G =0 11 2, find G−1. Check that G−1G = I.Solution: Use the following formula:A−1=1ad − bcd −b−c aSo, A−1= −2 −1−1 0=−2 11 0.7. Let A =2 1 24 2 12 1 1. Use the Gauss-Jordan method to either find the inverse of A orto show that A is not invertible.4Solution:2 1 2 1 0 04 2 1 0 1 02 1 1 0 0 1R2→R2−2R1−−−−−−−→2 1 2 1 0 00 0 −3 −2 1 02 1 1 0 0 1R3→R3−R1−−−−−−−→2 1 2 1 0 00 0 −3 −2 1 00 0 −1 −1 0 1R3→R3−R2/3−−−−−−−−→2 1 2 1 0 00 0 −3 −2 1 00 0 0 −1/3 −1/3 1We get a row of zeros in the left hand side, so it is not possible to transform the left handside to the identity matrix. Thus, A is not invertible.8. Calculate the inverse of the matrix2 1 0 −11 −1 1 00 0 0 11 1 1 1.Solution:2 1 0 −1 1 0 0 01 −1 1 0 0 1 0 00 0 0 1 0 0 1 01 1 1 1 0 0 0 1R3↔R4−−−−→2 1 0 −1 1 0 0 01 −1 1 0 0 1 0 01 1 1 1 0 0 0 10 0 0 1 0 0 1 0R3→R3−R4,R1→R1+R4−−−−−−−−−−−−−−→2 1 0 0 1 0 1 01 −1 1 0 0 1 0 01 1 1 0 0 0 −1 10 0 0 1 0 0 1 0R2↔R1−−−−→1 −1 1 0 0 1 0 02 1 0 0 1 0 1 01 1 1 0 0 0 −1 10 0 0 1 0 0 1 0R2→R2−2R1−−−−−−−→1 −1 1 0 0 1 0 00 3 −2 0 1 −2 1 01 1 1 0 0 0 −1 10 0 0 1 0 0 1 05R3→R3−R1−−−−−−−→1 −1
View Full Document
Unlocking...