1A. MillerM340 Exam 1Feb 23, 1990Put your answers on this exam. Put your work on separatesheets of paper to be handed in with the exam. There are a totalof 50 points.Name ScoreA ="1 1 + i 0−1 2i 3#B =01 + i1C =i1iCompute the following or say that it is undefined.1. (1 point(s)) AB=2. (1 point(s)) B + C=3. (1 point(s)) BA=4. (1 point(s)) iA=5. (1 point(s)) A0=6. (1 point(s)) hB, Ci=7. (1 point(s)) |hB, Ci|=8. (1 point(s)) ||C||=29. (1 point(s)) If the norm of a vector Q is 6 what is the norm of iQ?10. (2 point(s)) If hU, V i = 2 + 3i what is hV, 2Ui?11. (2 point(s)) If B is the inverse of the matrix A, then what is the inverseof the matrix 6A?12. (2 point(s)) Give the de finition of the inverse of an n × n matrix A.(Your answer should consist of a grammatically correct sentence.)13. (2 point(s)) Give an example other than the identity of an orthogonalsymmetric 2 × 2 matrix.14. (1 point(s)) Give an example of a trivial 3 × 1 vector.15. (1 point(s)) Give an example of a nontrivial 2 × 1 vector.16. (2 point(s)) For what values of a, b, c does the equation1 1 0 1 00 0 1 0 10 0 0 0 0X =abchave(a) no solutions?(b) a unique solution?(c) infinitely many solutions?17. (2 point(s)) What are the slack columns of the matrix A?A =0 1 0 1 00 0 0 0 10 0 0 0 0318. (4 point(s)) Find the inverse of A or show that the inverse does not existby exhibiting a nontrivial vector X such that AX = [0]. (Rememb er: Putyour work on a separate sheet of paper.)(a)A =0 −1 0 21 0 0 00 0 1 00 1 0 −1(b)A =1 2 10 1 11 3 219. (2 point(s)) Find an elementary matrix E such that EA is in row echelonform or say that no such E exists.A =0 1 0 2 01 0 0 0 20 0 1 0 −1420.(1 point)(Circle one: True False.) If U and V are n × n uppertriangularmatrices then so is UV .21.(1 point)(Circle one: True False.) If D is a diagonal n × n matrix, thenso is cDn+1where c is a scalar.22.(1 point)(Circle one: True False.) If A, B, C and D are n × n matrices,then (AB)(CD) = A(BC)D.23.(1 point)(Circle one: True False.) If A and B are n × n matrices and c isa scalar, then A(cB) = B(cA).24.(1 point)(Circle one: True False.) If A, B, and C are n × n matrices andA + C = C + B, then A = B.25.(1 point)(Circle one: True False.) If A is an n × n invertible matrix, thenthe matrix equation AX = B has a unique solution.26.(1 point)(Circle one: True False.) If A and B are symmetric n×n matrices,then AB is a symmetric matrix.27.(1 point)(Circle one: True False.) If A and B are symmetric n×n matrices,then A + B is a symmetric matrix.28.(1 point)(Circle one: True False.) If A, B, and C are n × n matrices andAC = BC, then either A = B or C is the zero matrix.29.(1 point)(Circle one: True False.) It is possible to find 5 × 1 vectors Uand V such that ||U||=2, ||V || = 3 and ||U + V || = 4.30.(1 point)(Circle one: True False.) It is possible to find 17 × 1 vectors Uand V such that ||U||=4, ||V || = 2 and ||U + V || = 7.31.(1 point)(Circle one: True False.) It is possible to find 3 × 1 vectors Uand V such that hU, V i = 5 and ||U|| = ||V || = 2.32.(1 point)(Circle one: True False.) For any n there exists an n × n elemen-tary matrix E such that EA = ATfor any n × n matrix A.33.(1 point)(Circle one: True False.) For any m × n matrix A it is possibleto find an invertible m × m matrix M such that M A is in row echelon form.34.(1 point)(Circle one: True False.) The sum of two n×n invertible matricesis invertible.35. (6 point(s)) Supp ose A is an n ×n matrix such that kAXk = kXk for alln ×1 vectors X. On the back of this page prove that hAX, AY i = hX, Y ifor all n × 1 vectors X and Y .5exam 1 solutions by minimat** Diary file: C:\CLASS\340JAN90\EX1opened at Date/Time: 90-02-23 10:38:24**(Type DIARY OFF to close the diary.)#> i=sqrt(-1)i =( 0.000 + 1.000*i)#> a=[1 1+i 0; -1 2*i 3]a =( 1.000 + 0.000*i) ( 1.000 + 1.000*i) ( 0.000 + 0.000*i)( -1.000 + 0.000*i) ( 0.000 + 2.000*i) ( 3.000 + 0.000*i)#> b=[0;1+i;1]b =( 0.000 + 0.000*i)( 1.000 + 1.000*i)( 1.000 + 0.000*i)#> c=[i;1;i]c =( 0.000 + 1.000*i)( 1.000 + 0.000*i)( 0.000 + 1.000*i)#> a*bans =6( 0.000 + 2.000*i)( 1.000 + 2.000*i)#> b+cans =( 0.000 + 1.000*i)( 2.000 + 1.000*i)( 1.000 + 1.000*i)#> b*a[71] : Incompatible sizes for multiplication#> i*aans =( 0.000 + 1.000*i) ( -1.000 + 1.000*i) ( 0.000 + 0.000*i)( 0.000 - 1.000*i) ( -2.000 + 0.000*i) ( 0.000 + 3.000*i)#> a’ans =( 1.000 + 0.000*i) ( -1.000 + 0.000*i)( 1.000 - 1.000*i) ( 0.000 - 2.000*i)( 0.000 + 0.000*i) ( 3.000 + 0.000*i)#> b*c[71] : Incompatible sizes for multiplication#> inner(b,c)ans =1.0007#> norm(c)ans =1.732#> sqrt(3 )ans =1.732#> q=6q =6.000#> norm(i*q)ans =6.000#> u=2+3*i, v=1u =( 2.000 + 3.000*i)v =1.000#> inner(u,v)ans =( 2.000 + 3.000*i)#> inner(v,2*u)8ans =( 4.000 - 6.000*i)#> a=rand(3,3)a =0.598 0.882 0.2950.189 0.486 0.1610.476 0.125 0.469#> b=inv(a)b =3.967 -7.197 -0.027-0.225 2.667 -0.774-3.966 6.595 2.366#> inv(6*a)ans =0.661 -1.200 -0.004-0.037 0.444 -0.129-0.661 1.099 0.394#> a=[0 -1 0 2; 1 0 0 0 ; 0 0 1 0 ; 0 1 0 -1]a =0.000 -1.000 0.000 2.0001.000 0.000 0.000 0.0000.000 0.000 1.000 0.0000.000 1.000 0.000 -1.000#> inv(a)ans =0.000 1.000 0.000 0.0001.000 0.000 0.000 2.00090.000 0.000 1.000 0.0001.000 0.000 0.000 1.000#> randsoln(a)ans =0.0000.0000.0000.000#> a=[ 1 2 1; 0 1 1; 1 3 2 ]a =1.000 2.000 1.0000.000 1.000 1.0001.000 3.000 2.000#> inv(a)[82] : Vanishing determinant#> randsoln(a)ans =0.483-0.4830.483#> a=[0 1 0 2 0 ; 1 0 0 0 2; 0 0 1 0 -1]a =0.000 1.000 0.000 2.000 0.0001.000 0.000 0.000 0.000 2.0000.000 0.000 1.000 0.000 -1.00010#> [b m]=refm(a)b =1.000 0.000 0.000 0.000 2.0000.000 1.000 0.000 2.000 0.0000.000 0.000 1.000 0.000 -1.000m =0.000 1.000 0.0001.000 0.000 0.0000.000 0.000 1.000#> round(rand(15,1))ans =0.0000.0000.0000.0001.0000.0000.0000.0001.0000.0000.0001.0001.0000.0001.000#> quit11** Diary file: C:\CLASS\340JAN90\EX1 closed atDate/Time: 90-02-23