Name Class TimeMath 2270 Maple Project 2: Linear AlgebraFebruary 2007Due date: See the internet due dates. Maple lab 2 has problems L2.1, L2.2, L2.3.References: Code in maple app e ars in 2270mapleL2-S2007.txt at URL http://www.math.utah.edu/~gustafso/.This document: 2270mapleL2-S2007.pdf.Problem L2.1. (Matrix Algebra)Define A =1 2 34 5 67 8 9, B =2 1 01 2 10 1 2, v =123and w =−141. Create a worksheet in maple whichstates this problem in text, then defines the four objects. The worksheet should contain text, maple code and displays.Continue with this worksheet to answer (1)–(7) below. Submit problem L2.1 as a worksheet printed on 8.5 by 11 inchpaper. See Example 1 for maple commands.(1) Compute AB and BA. Are they the same?(2) Compute A + B and B + A. Are they the same?(3) Let C = A + B. Compare C2to A2+ 2AB + B2. Explain why they are different.(4) Compute transposes C1= (AB)T, C2= ATand C3= BT. Find an equation for C1in terms of C2and C3. Verifythe equation.(5) Solve for X in BX = v by three different methods.(6) Solve AY = v for Y. Do an answer check.(7) Solve AZ = w. Explain your answer using rref theory.Problem L2.2. (Row space)Let A =1 1 1 2 62 3 −2 1 −30 1 −4 −3 −151 2 −3 −1 −9. Find two different bases fo r the row spa c e of A, using the following three methods.1. The method of Example 2, below.2. The maple commandrowspace(A) .3. The rref-method: select rows from rref(A).Two of the methods produce exactly the same basis. Verify that the two bases B1= {v1, v2} and B2= {w1, w2} areequivalent. This means that each vector in B1is a linear combination of the vectors in B2, and conve rsely, that eachvector in B2is a linear combination of the vectors in B1.Problem L2.3. (Matrix Equations)Let A =10 13 5−5 −8 −5−3 −3 2, T =2 0 00 −3 00 0 5. Let P denote a 3 × 3 matrix. Assume the following result:Lemma 1. The equality AP = P T holds if and only if the columns v1, v2, v3of P satisfy Av1= 2 v1,Av2= −3v2, Av3= 5v3. [proved after Example 4](a) Determine three specific columns for P such that det(P ) 6= 0 and AP = P T . Infinitely many answers a re possible.See Example 4 for the ma ple method that determines a column of P .(b) After repo rting the three columns, check the answer by computing AP − P T (it should be zero) and det(P ) (itshould be nonzero).Staple this page on top of the maple work sheets. E xamples and theory on the next page . . .Example 1. Let A =1 2 32 −1 13 0 −1and b =983. Create a maple work sheet. Define and display matrix Aand vector b. Then compute(1) The inverse of A.(2) The augmented matrix C = aug(A, b).(3) The reduced row echelon form R = rref (C).(4) The column X of R which solves AX = b.(5) The matrix A3.(6) The transpose of A.(7) The matrix A − 3A2.(8) The solution X of AX = b by two methods different than (4).Solution: A lab instructor can help you to create a blank work sheet in maple, enter code and print the work sheet.The code to be entered appears below. To get help, enter?linalg into a worksheet, then select commands that matchones below.with(linalg):A:=matrix([[1,2,3],[2,-1,1],[3,0,-1]]);b:=vector([9,8,3]);print("(1)"); inverse(A);print("(2)"); C:=augment(A,b);print("(3)"); R:=rref(C);print("(4)"); X:=col(R,4);print("(5)"); evalm(A^3);print("(6)"); transpose(A);print("(7)"); evalm(A-3*(A^2));print("(8)"); X:=linsolve(A,b); X:=evalm(inverse(A) &* b);Example 2. Let A =1 1 1 2 62 3 −2 1 −33 5 −5 1 −84 3 8 2 3.(1) Find a basis for the column space of A.(2) Find a basis for the row space of A.(3) Find a basis for the nullspace of A.(4) Find rank(A) and nullity(A).(5) Find the dimensions of the nullspace, row space and column space of A.Solution: The theory applied: The columns of B corresponding to the leading ones in rref(B) are independent andform a basis for the column space of B. These columns are called the pivot columns of B. Results for the row spacecan be obtained by applying the above theory to the transpose of the matrix.The maple c ode which a pplies iswith(linalg):A:=matrix([[ 1, 1, 1, 2, 6],[ 2, 3,-2, 1,-3],[ 3, 5,-5, 1,-8],[ 4, 3, 8, 2, 3]]);print("(1)"); C:=rref(A); # leading ones in columns 1,2,4BASIScolumnspace=col(A,1),col(A,2),col(A,4);2print("(2)"); F:=rref(transpose(A)); # leading ones in columns 1,2,3BASISrowspace=row(A,1),row(A,2),row(A,3);print("(3)"); nullspace(A); linsolve(A,vector([0,0,0,0]));print("(4)"); RANK=rank(A); NULLITY=coldim(A)-rank(A);print("(5)"); DIMnullspace=coldim(A)-rank(A); DIMrowspace=rank(A);DIMcolumnspace=rank(A);Example 3. Let A =1 1 1 2 62 3 −2 1 −33 5 −5 1 −84 3 8 2 3. Verify that the following column space bases of A are equivalent.v1=1234, v2=1353, v3=2112,w1=100−3, w2=01017, w3=001−9.Solution: We will use this result:Lemma 2. Bases {v1, v2, v3} and {w1, w2, w3} are equivalent bases if and only if the augmented matrices F =aug(v1, v2, v3), G = aug(w1, w2, w3) and H = aug(F, G) satisfy the rank condition rank(F ) = rank(G) = rank(H) =3.The proof appears below.The maple c ode which a pplies iswith(linalg):A:=matrix([[ 1, 1, 1, 2, 6],[ 2, 3,-2, 1,-3],[ 3, 5,-5, 1,-8],[ 4, 3, 8, 2, 3]]);v1:=vector([1,2,3,4]); v2:=vector([1,3,5,3]); v3:=vector([2,1,1,2]);w1:=vector([1, 0, 0, -3]); w2:=vector([0, 1, 0, 17]); w3:=vector([0, 0, 1, -9]);F:=augment(v1,v2,v3);G:=augment(w1,w2,w3);H:=augment(F,G);rank(F); rank(G); rank(H);We remark that the two bas e s in the example were discovered from the maple coderref(A); # pivot cols 1,2,4v1:=col(A,1); v2:=col(A,2); v3:=col(A,4);B:=rref(transpose(A)); # pivot cols 1,2,3w1:=row(B,1); w2:=row(B,2); w3:=row(B,3);Proof of Lemma 2.Proof: The test appears in the online pdf documents at the course web site. Let’s justify the test here, independently,showing only half the proof: rank(F ) = rank(G) = rank(H) = n implies the bases are equivalent.The equation rref(F ) = EF holds for E a product o f elementary ma trices. Then EH has to have n lead variables,because of F in the first n columns, and the r emaining rows are zero, because rank(H) = n. Therefore, the first ncolumns of H are the pivot columns of H. The non-pivots of H are just the co lumns of G, and by the pivot theory, theyare linear combinations of the