2250 Midterm 2 sample exam answer checks Fall 2004

 1(a)
 A:=matrix([[1,2,1,b],[3,1,2,2*b],[4,3,3,1+b]]);
  A1:=addrow(A,1,3,-1);
  A2:=addrow(A1,2,3,-1);
  1-2b == 0 gives infinitely many solutions

 1(b)
 with(linalg):
 A:=matrix([[1,2,1,a],[5,1,2,3*a],[6,3,b,1+a]]);
 A1:=addrow(A,1,3,-1);
 A2:=addrow(A1,2,3,-1);
 A3:=addrow(A2,1,2,-5);
 A4:=mulrow(A3,2,-1/9);
 A5:=addrow(A4,2,1,-2);
 -3+b == 0 and 3a-1 == 0 gives one free variable and infinitely many
 solutions

 2(a) V={ c1 + c2 t } and W = { c1 + c2 t + c3 t^2 } are vector spaces
 of polynomials with dim(V)=2 and dim(W)=3.

 2(b) All functions y in S look like y=c1 exp(-2t). Adding two such
      functions gives a function in S and multiplying such a function
      by a scalar gives a function in S. So S is closed under addition
      and scalar multiplication. Therefore, S is a subspace of V.
 2(c)
 with(linalg):
 A:=matrix([[1,2,-1,0],[1,1,-2,0],[0,1,1,0]]);
 rref(A);
 x = 3t, y = -t, z = t. basis = vector([3,-1,1])


 3(a) A:=matrix([[1,2,1],[-1,1,2],[1,0,-1]]);rref(A);
  The rref has a row of zeros, so the vectors are dependent. 
 3(b)
  A:=matrix([[1,2,3,0,1],[-1,-2,-1,2,1],[0,0,0,0,0],
            [-1,-2,1,4,3]]);rref(A); 
  position of leading ones identifies a,c as indep.


 4(a) det(B)=det(E1)det(E2)det(A) by the product rule for
      determinants. Each swap rule has determinant -1. So
      det(B)=(-1)(-1)det(A)=det(A).
 4(b) det(AB)=0 because the determinant has two duplicate rows. Then
      det(A)det(B)=0 by the product theorem for determinants. Hence
      either det(A)=0 or det(B)=0.


 5(a) A:=matrix([[1,2,0],[2,0,-3],[0,x,1]]);det(A); x \ne 4/3
 5(b) A:=matrix([[1,2,0],[3,0,2],[2,-2,1]]);
      b:=vector([1,0,-1]);linsolve(A,b); then x=0, y=1/2, z=0.