midsem_sol

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DEPARTMENT OF MATHEMATICS, IIT GUWAHATI
MA101: Mathematics I Mid Semester Exam (Maximum Marks: 30)
Date: September 20, 2011 Time: 2 pm - 4 pm
Model Solutions
1. (a) Prove or disprove: If A and B are two matrices of the same size such that the linear system of equations
Ax = a and Bx = b have the same set of solutions then the matrices [A | a] and [B | b] must be
row-equivalent. 2
Solution: Consider the matrices A =
[
1 0
0 0
]
and B =
[
0 1
0 0
]
and vectors a = b = [0, 1]t.
Clearly, the systems Ax = a and Bx = b are inconsistent, and hence they have the same solution
set ∅. However, the matrices [A | a] and [B | b], being in different reduced row echelon form, are not
row-equivalent. Thus, the given statement is disproved by this counterexample.
(b) Find all real values of k for which the following system of equations has (i) no solution, (ii) unique
solution, and (iii) infinitely many solutions: 3
kx+ y + z = 1, x+ ky + z = 1, x+ y + kz = 1.
Solution: The augmented matrix of the given system is
[A | b] =


k 1 1 1
1 k 1 1
1 1 k 1

 R1 ↔ R3−−−−−−→


1 1 k 1
1 k 1 1
k 1 1 1


R2 → R2 −R1−−−−−−−−−−→


1 1 k 1
0 k − 1 1− k 0
k 1 1 1


R3 → R3 − kR1−−−−−−−−−−−→


1 1 k 1
0 k − 1 1− k 0
0 1− k 1− k2 1− k


R3 → R3 +R2−−−−−−−−−−→


1 1 k 1
0 k − 1 1− k 0
0 0 (1− k)(2 + k) 1− k

 .
If k = −2 then rank([A | b]) = 3 6= 2 = rank(A). Hence the system has no solution if k = −2.
If k = 1 then rank([A | b]) = rank(A) < 3. Hence the system has infinitely many solutions if k = 1.
If k 6= 1,−2 then rank([A | b]) = rank(A) = 3. Hence the system has a unique solution if k 6= 1,−2.
2. (a) Let u,v,w be three linearly independent vectors in Rn, where n ≥ 3. For what real values of k, are
the vectors v − u, kw − v and u−w linearly independent? 2
Solution: Let a(v − u) + b(kw − v) + c(u−w) = 0 for some a, b, c ∈ R. Then we have
a(v − u) + b(kw − v) + c(u−w) = 0
⇒ (c− a)u+ (a− b)v + (kb− c)w = 0
⇒ c− a = 0, a− b = 0, kb− c = 0; since u,v and w are linearly independent
⇒ a = b = c = 0 if k 6= 1.
Also if k = 1, then 1.(v−u) + 1.(w− v) + 1.(u−w) = 0. Thus the vectors v−u, kw− v and u−w
are linearly independent if and only if k 6= 1.
(b) Find a basis for the subspace V , where V = {[x1, x2, . . . , x6]
t ∈ R6 : xi = 0 if i is even}. 3
Solution: We have V = {[x1, 0, x3, 0, x5, 0]
t : x1, x3, x5 ∈ R}. Let e1 = [1, 0, 0, 0, 0, 0]
t, e3 =
[0, 0, 1, 0, 0, 0]t and e5 = [0, 0, 0, 0, 1, 0]
t. Then we have
[x1, 0, x3, 0, x5, 0]
t = x1e1 + x3e3 + x5e5.
Thus the set {e1, e3, e5} spans V . Also
ae1 + be3 + ce5 = 0⇒ [a, 0, b, 0, c, 0]
t = 0⇒ a = b = c = 0.
Thus the set {e1, e3, e5} is also linearly independent. Hence {e1, e3, e5} is a basis for V .
3. (a) Prove or disprove: There exist 2× 2 matrices A and B such that AB −BA = I2. 2
Solution: Let A =
[
a b
c d
]
and B =
[
x y
z w
]
. Then we have
AB −BA = I2 ⇒
[
ax+ bz ay + bw
cx+ dz cy + dw
]
−
[
ax+ cy bx+ dy
az + cw bz + dw
]
=
[
1 0
0 1
]
⇒
[
bz − cy (a− d)y + b(w − x)
c(x− w) + (d− a)z cy − bz
]
=
[
1 0
0 1
]
.
Comparing the diagonal entries on both the sides, we have bz − cy = 1 and cy − bz = 1. Adding
corresponding sides of these two equations, we get 0 = 2, which is absurd. Hence there cannot exist
2× 2 matrices A and B such that AB −BA = I2.
(b) Let A be an invertible matrix with integer entries. Show that A−1 has all entries integer if and only
if det(A) = ±1. 3
Solution: Since A is an invertible matrix with integer entries, we have that det(A) is an integer.
First assume that A−1 has all entries integer. Then det(A−1) is also an integer. Now AA−1 = I ⇒
det(A)det(A−1) = 1. Hence, each of det(A) and det(A−1) being integers, we find that det(A) = ±1.
Conversely, assume that det(A) = ±1. Since entries of A are integers, we find that entries of
adj(A) are also integers. Now A−1 = 1det(A)adj(A) = (±1)adj(A), which gives that A
−1 has all entries
integer.
4. Let A be an n × n real matrix and let {u1,u2, . . . ,un} be a basis for R
n. Show that rank(A) = n if and
only if {Au1, Au2, . . . , Aun} is a basis for R
n. 5
Solution: Given that A is an n × n real matrix and S = {u1,u2, . . . ,un} is a basis for R
n. Let T =
{Au1, Au2, . . . , Aun}.
First assume that rank(A) = n so that A is an invertible matrix. Let a1(Au1)+a2(Au2)+. . .+an(Aun) = 0
for some a1, a2, . . . , an ∈ R. Then we have
a1(Au1) + a2(Au2) + . . .+ an(Aun) = 0
⇒ A(a1u1 + a2u2 + . . .+ anun) = 0
⇒ a1u1 + a2u2 + . . .+ anun = 0, since A is invertible
⇒ a1 = a2 = . . . = an = 0, since S is linearly independent
⇒ the set T is linearly independent
⇒ T , being a linearly independent set of n vectors in Rn, is a basis for Rn.
Conversely, assume that T is a basis for Rn. Then for any b ∈ Rn, we have b = a1(Au1) + a2(Au2) +
. . .+ an(Aun) for some a1, a2, . . . , an ∈ R. This gives that b = A(a1u1 + a2u2 + . . .+ anun) and hence the
system Ax = b is consistent for every b ∈ Rn. Thus Rn = {b : Ax = b is consistent} = col(A), and hence
n = dim(col(A)) = rank(A).
Aliter (proving the contrapositive statements)
First assume that T is not a basis for Rn. Hence T is linearly dependent, and so there exists real
numbers a1, a2, . . . , an, not all zero, such that a1(Au1) + a2(Au2) + . . . + an(Aun) = 0. This gives that
A(a1u1 + a2u2 + . . .+ anun) = 0. Then u = a1u1 + a2u2 + . . .+ anun 6= 0, since this a non-trivial linear
combination of vectors from a basis for Rn. Thus u is a non-zero solution of the system Ax = 0, and hence
rank(A) < n.
Conversely, assume that rank(A) < n. Then we can find a non-zero vector u ∈ Rn such that Au = 0. Since
S is a basis for Rn there exist real numbers a1, a2, . . . , an, not all zero, such that u = a1u1+a2u2+. . .+anun.
Then we have Au = 0 ⇒ a1(Au1) + a2(Au2) + . . . + an(Aun) = 0, which gives that the set T is linearly
dependent and hence T is not a basis for Rn.
5. (a) Let A be a diagonalizable matrix such that every eigenvalue of A is either 0 or 1. Show that A2 = A. 2
Solution: Since A is diagonalizable there exists an invertible matrix P such that P−1AP = D, where
D is a diagonal matrix whose diagonal entries are the eigenvalues of A. Now since every eigenvalue of A
is either 0 or 1, we find that D2 = D. Therefore (P−1AP )(P−1AP ) = D2 = D ⇒ P−1A(PP−1)AP =
D ⇒ P−1A2P = D ⇒ A2 = PDP−1 = A.
(b) Let λ1 and λ2 be two distinct eigenvalues of a matrix A and let u1 and u2 be eigenvectors of A
corresponding to λ1 and λ2, respectively. Show that u1 + u2 is not an eigenvector of A. 3
Solution: The vectors u1 and u2, being eigenvectors corresponding to distinct eigenvalues of A, are
linearly independent. Suppose, if possible, u1 +u2 is an eigenvector corresponding to an eigenvalue µ
of A. Then we have
A(u1 + u2) = µ(u1 + u2)⇒ Au1 +Au2 = µ(u1 + u2)
⇒ λ1u1 + λ2u2 = µ(u1 + u2)
⇒ (λ1 − µ)u1 + (λ2 − µ)u2 = 0
⇒ λ1 − µ = 0 = λ2 − µ, since u1 and u2 are linearly independent
⇒ λ1 = µ = λ2,which contradicts the fact that λ1 6= λ2.
Hence we conclude that u1 + u2 is not an eigenvector of A.
6. (a) Let W be a subspace of R5 and v ∈ R5. Suppose that w and w′ are orthogonal vectors with w ∈ W
and that v = w + w′. Is it necessarily true that w′ ∈ W⊥? Either prove that it is true or find a
counterexample. 2
Solution: Consider the subspace W = {[x, y, 0, 0, 0]t ∈ R5 : x, y ∈ R} of R5. Consider v =
[1, 1, 0, 0, 0]t,w = [1, 0, 0, 0, 0]t and w′ = [0, 1, 0, 0, 0]t. Then w ∈ W, v = w + w′ and, w and w′
are orthogonal to each other. However, w′ /∈W⊥ since w′ ∈W, w′ 6= 0 and W ∩W⊥ = {0}.
(b) Find a basis for M⊥, where M = {[x, y, z]t : x = s, y = −s, z = 3s, s ∈ R}. 3
Solution: Since [s,−s, 3s]t = s[1,−1,3]t, we find thatM is spanned by the vector [1,−1, 3]t. Therefore
[x, y, z]t ∈ M⊥ if and only if [x, y, z]t.[1,−1, 3]t = 0 ⇐⇒ x − y + 3z = 0 ⇐⇒ x = y − 3z. Thus
M⊥ = {[x, y, z]t ∈ R3 : x = y − 3z}. We have [y − 3z, y, z]t = y[1, 1, 0]t + z[−3, 0, 1]t. Thus the set
{[1, 1, 0]t, [−3, 0, 1]t} spans M⊥. Also
a[1, 1, 0]t + b[−3, 0, 1]t = 0⇒ [a− 3b, a, b]t = 0⇒ a = b = 0.
Thus the set {[1, 1, 0]t, [−3, 0, 1]t} is linearly independent. Hence the set {[1, 1, 0]t, [−3, 0, 1]t} is a basis
for M⊥.