# Homework 1 Solutions # 1. Find all solutions to the given linear systems. # a) x + 3y = 5 # 3x -4y = 2 > read("/afs/eos.ncsu.edu/users/k/kaltofen/www/courses/LinAlgebra/Maple/ > RefPkg/InitPkg.mpl"): > with(LinearAlgebra): > with(RefPkg): > A := Matrix(2,3,[[1,3,5],[3,-4,2]]); [1 3 5] A := [ ] [3 -4 2] > Ref(A); [1 3 5] [ ] [0 -13 -13] # Backsolving we can see that -13y=-13, so y = 1 and x = 5-3y = 5-3(1) = # 2. So the only solution is x=2 and y=1. # b) x + 4y -z = 2 # -x -4y +2z = 0 # x + 4y = 4 > B := <<1,-1,1>|<4,-4,4>|<-1,2,0>|<2,0,4>>; [ 1 4 -1 2] [ ] B := [-1 -4 2 0] [ ] [ 1 4 0 4] > Ref(B); [1 4 -1 2] [ ] [0 0 1 2] [ ] [0 0 0 0] # Back solving we see that z=2, y=y is a free variable and x = 2 +z - # 4y = 2+2-4y =4 - 4y. So the solution set S = {(4-4y, y, 2) | where y # is any real number}. # 2. Find all the values of a for which the linear system has a) no # solutions, b) a unique solution and c) infinitely many solutions. # x + y + 4z = 1 # x + 3y + 4z = 0 # x + (a^2)y + 4z = a # First I will put the augmented matrix into row echelon form as above. > C := <<1,1,1>|<1,3,a^2>|<4,4,4>|<1,0,a>>; [1 1 4 1] [ ] C := [1 3 4 0] [ ] [ 2 ] [1 a 4 a] > Ref(C); [1 1 4 1 ] [ ] [0 2 0 -1 ] [ ] [ 2] [0 0 0 a - 3/2 + 1/2 a ] # a) For the system to have no solutions, mean that the system is # inconsistent. A system is inconsistent if there is a pivot element in # the last column. So if the system above will have # any kind of solutions ( infinite or unique) then the 3,4 entry, must # be equal to zero. This gives us a quadratic equation in the variable a # and so we solve for a. > solve((1/2)*a^2+a -(3/2)=0,a); 1, -3 # Thus for there to be a solution a=1 or -3. Therefore the system has # no solutions for any real number a, except a=1 and a=-3. # b) We have ruled out all possibilities for a except 1 and -3. If a=1 # or a=-3 then there is a solution, but also we see that in either case # the variable z is a free variable. # So if there is a solution then there must be infinitely many # solutions. Therefore the system never has a unique solution, that is # ther is no value of a which gives the # system a unique solution. # c) We have already said above that if a=1 or a=-3 then there are # infinitely many solutions. # 3. Write the augmented matrix for the following system and write it in # row echelon form. Is the system consistent? # x + 4y - z = 2 # 3x + 10y - 2z = 14 # 2x + 6y - z = 16 > Augmented := Matrix(3,4,[[1,4,-1,2],[3,10,-2,14],[2,6,-1,16]]); [1 4 -1 2] [ ] Augmented := [3 10 -2 14] [ ] [2 6 -1 16] > refmat := Ref(Augmented); [1 4 -1 2] [ ] refmat := [0 -2 1 8] [ ] [0 0 0 4] # The system is not consistent because there is a pivot element in the # last column. # 4. Find all conditions of r,s, and t such that the matrix > <|||<0,t,s,r>|<0,0,t,s>>; [r s t 0 0] [ ] [0 r s t 0] [ ] [0 0 r s t] [ ] [0 0 0 r s] # a) is in reduced row echelon form. # To be in reduced row echelon form then the pivot points must be equal # to 1 and everything above and below a pivot point is zero. # Thus the answer is (r=1 and s=0 and t=0) or (r=0 and s=1 and t=0) or # (r=0 and s=0 and t=1) or (r=0 and s=0 and t=0). # b) the matrix is in reduced row echelon form and the matrix is # consistent. # For the matrix to be consistent then there can be no pivot element in # the last column. So we start with the possibilities from # above and see that s and t cannot be equal to 1 since then there would # be a pivot element in the last column. # So the only possible answers are (r=1 and s=0 and t=0) or (r=0 and s=0 # and t=0).