There were two versions of Test 1: in one version (Version A) the
first problem was a system with variables a, b, c, d and in the other
version (Version W) the variables in the first problem were
w, x, y, and z.
Answers to Version A:
Problem 1: infinitely many solutions, one solution for each value of d:
general solution: a = 1 - d, b = -6 + 3d, c = 3 - d, d arbitrary
two solutions: a = 1, b = -6, c = 3, d = 0
a = 0, b = -3, c = 2, d = 1
Problem 2: det(A) = 9, det(A') = 9, det(A^{-1}) = 1/9,
and because det(A) is not 0, A is invertible, so
X = 0 is the only solution of AX = 0, that is "one solution"
Problem 3: A^{-1} = [ 1 -3 -2; -3 9 5; 1 -2 -1]
Problem 4: (a) Problem asks for LINEARLY INDEPENDENT solutions of (H):
(0, 2, -2, 2, 2, 0) and (4, 1, -2, 1, 0, 0) work
also
(0, -3, 3, -3, -3, 0) and (-8, -2, 4, -2, 0, 0) work,
and many other solutions are possible.
(b) Possible solutions of (N) are
(1, 3, 0, 1, 1, 1)
(1, 1, 2, -1, -1, 1)
(-3, 2, 2, 0, 1, 1)
(-3, 0, 4, -2, -1, 1) and many other solutions are possible.
Problem 5: (a) w = 2 v1 + 3 v2
(b) u is NOT a linear combination of v1 and v2
(c) the dimension of span{ v1, v2, w, u } is 3:
v1, v2, and u are linearly independent and span the
subspace, so are a basis for it
Problem 6: (a) the given system is the system that needs to be solved
to find the nullspace of B, so v = (2, -3, 1) is
a linearly independent vector in the nullspace of B.
Since nullspace(B) is 1 dimensional, v is a basis.
(b) Let C1, C2, and C3 be the columns of B. The system
of equations given shows that
2 C1 - 3 C2 + C3 = 0
so C1, C2, and C3 are linearly dependent: they are not
a basis. However, C1 and C2 are linearly independent
and span the column space of B, so they are a basis
for the range of B.
(c) Rank(B') = rank(B). By part (b), rank(B) = 2, so
we know rank(B') = 2, that is, the range of B' is
2 dimensional. (1, 1, 1) and (1, 2, 4) are linearly
independent and therefore form a basis for range(B').
Answers to Version W:
Problem 1: infinitely many solutions, one solution for each value of z:
general solution: w = 2 + z, x = -1 - 3z, y = -2 - z, z arbitrary
two solutions: w = 2, x = -1, y = -2, z = 0
w = 3, x = -4, y = -3, z = 1
Problem 2: det(A) = -14, det(A') = -14, det(A^{-1}) = -1/14,
and because det(A) is not 0, A is invertible, so
X = 0 is the only solution of AX = 0, that is "one solution"
Problem 3: A^{-1} = [ 9 5 -3; -3 -2 1; -2 -1 1]
Problem 4: (a) Problem asks for LINEARLY INDEPENDENT solutions of (H):
(0, 2, -2, 2, 2, 0) and (4, 1, -2, 1, 0, 0) work
also
(0, -3, 3, -3, -3, 0) and (-8, -2, 4, -2, 0, 0) work,
and many other solutions are possible.
(b) Possible solutions of (N) are
(1, 2, -1, 1, 2, 1)
(1, 0, 1, -1, 0, 1)
(5, 1, -1, 0, 0, 1) and many other solutions are possible.
Problem 5: (a) w = 3 v1 - 2 v2
(b) u is NOT a linear combination of v1 and v2
(c) the dimension of span{ v1, v2, w, u } is 3:
v1, v2, and u are linearly independent and span the
subspace, so are a basis for it
Problem 6: (a) the given system is the system that needs to be solved
to find the nullspace of B, so v = (-3, 2, 1) is
a linearly independent vector in the nullspace of B.
Since nullspace(B) is 1 dimensional, v is a basis.
(b) Let C1, C2, and C3 be the columns of B. The system
of equations given shows that
-3 C1 + 2 C2 + C3 = 0
so C1, C2, and C3 are linearly dependent: they are not
a basis. However, C1 and C2 are linearly independent
and span the column space of B, so they are a basis
for the range of B.
(c) Rank(B') = rank(B). By part (b), rank(B) = 2, so
we know rank(B') = 2, that is, the range of B' is
2 dimensional. (1, 1, 1) and (1, 2, -1) are linearly
independent and therefore form a basis for range(B').
Back to Math 351 Page