Download Sample examinations Linear Algebra (201-NYC-05) Autumn 2010 1. Given

Sample examinations
Linear Algebra (201-NYC-05)
Autumn 2010
12. Let
1. Given

1
A = −4
−5

8
−26
−30
−1
1
1
1
−3
−3

−4
b =  0 .
2
and
a. Solve the equation Ax = b. Express your answer in parametric vector form.
b. Find a particular solution of Ax = b whose first entry is 6.
c. Give a basis of Nul A.
2. Given the matrix


1 0 2 1
1 1 1 2

A=
1 2 0 b  .
0 3 a b
Under what conditions on a and b is rank A equal to a. 4?
3. Let

3
A = 1
0
−2
−1
4

−4
−3
21
b. 3?
c. 2?
 
3
b = 1 .
8
and
0
M = A
0
B
C
0
a.
b.
c.
d.
Does H contain the zero vector of R4 ?
Is H closed under addition? Justify your answer.
Is H closed under scalar multiplication? Justify your answer.
Is H a subspace of R4 ? Justify your answer.
13. Find a specific example of each of the following, if possible. If there is no such
example, explain why not.
a. Find A−1 .
b. Solve Ax = b using your answer to Part a.
4. Assume that the partitioned matrix


 
a





 
b

H =   : ad = bc ⊂ R4 .
c






d

a.
b.
c.
d.
e.
A non-zero 2 × 2 matrix A such that Col A = Row A.
A 2 × 2 matrix A such that Nul A = Row A.
A lower triangular 3 × 3 matrix A such that A and A + I are both singular.
A square matrix A such that x
Ax is surjective but not injective.
A unit vector that is orthogonal to both
 
 
1
2
−1
 1 .
and
−4
−2
14. You are given points A(2, 7, −1), B(3, 3, −1), C(3, 7, −4) and D(5, 5, 5).
−−→
−−→
a. Find the cosine of the angle between AB and AC .
−−→
−−→
b. Find proj−−→ AB and perp−−→ AB .

0
0
D
AC
is invertible, and that A, B, C and D are square.
a. Express M −1 as a partitioned matrix.
b. Which of the matrices A, B, C, D must be invertible for M −1 to exist?
5. Set up an augmented matrix for balancing the chemical equation
Ca(OH)2 + H3 PO4 → Ca3 (PO4 )2 + H2 O
6. Let A be a 4 × 4 symmetric matrix with det A = −5. For each part, either
provide an answer or write “not enough information”.
a. What is the value of det(−4A−1 )?
b. What is the value of det(2AT − A)?
c. What is the value of det(A − I)?
AC
c. Find the distance from the point B to the line AC.
d. Find a normal equation of the plane containing the points A, B and C.
−−→ −−→
−−→
e. Find the volume of the parallelepiped with edges AB , AC and AD .
15. Let $1 and $2 be, respectively the planes with standard equations
4x − 2y + 5z = 3
− 2x + y + kz = 0.
and
a. For which values of k, if any, are $1 and $2 parallel?
b. For which values of k, if any, are $1 and $2 orthogonal?
c. For which values of k, if any, does the point (−1, −1, 1) lie on $1 and $2 ?
16. Let A and B be matrices of the same size. Suppose that x is in both Nul A
and Nul B. Show that x must be in Nul(A + B).
17. Find a basis of the subspace
7. Let A, B and C be n × n matrices and suppose that AB T C −1 = In .
H = { X ∈ M2×2 : AX = XA }
a. Use determinants to explain why A and B must be invertible.
b. Does A commute with B T C −1 ? Explain your answer.
c. Find B −1 .
of M2×2 , where
8. Express
18. Let U = Span{p, q} and V = Span{u, v}, where
 
 
 
0
1
1
1
0
1





and
p =  ,
q =  ,
u= 
1
1
0
1
2
1


2
5

A = −2 −8
8
2
as a product LU , where L is lower triangular and U is upper triangular.
9. Find elementary matrices E1 and E2 such that
−5 6
E2 E1
= I2 .
0
1
10. Let P1 be the parallelogram with vertices (0, 0), (1, −2), (4, −1) and (3, 1),
let P2 be the parallelogram with vertices (0, 0), (−1, −1), (1, −4) and (2, −3),
and let A be a 2 × 2 matrix, such that the linear transformation x
Ax maps P1
onto P2 .
a. What are the possible values of det A?
b. Give a specific matrix B such that the linear transformation x
Bx maps P2
onto P1 .
11. Simplify the expression
u · (v − u) × (w − u)
as much as possible, where u, v and w are vectors in R3 .
A=
1
.
−1
1
0


2
−1

v=
 3 .
1
a. Find a basis of U + V .
b. Find a basis of U ∩ V .
c. What is the distance between the line p + tq and V ?
d. Find the point(s) on U which minimize the distance to the line u + tv, and find
the distance between U and this line.
19. a. Give the (precise) definition of a basis of a vector space.
b. Suppose that V and W are vector spaces, that B = {b1 , . . . , bn } is a basis of
V , and that T : V → W is a linear transformation such that
T (bi ) = 0W
for i = 1, . . . , n.
Prove that T is the zero transformation.
c. Suppose that V is a vector space, and that
A = {a1 , . . . am }
are bases of V . Prove that m = n.
and
B = {b1 , . . . bn }
Sample examinations (solutions)
Linear Algebra (201-NYC-05)
1. a. Reducing the augmented matrix of the given equation yields

 
1
1
−1
8
−4
1 1 −1
8
−4 −3
1
−26
0  ∼ 0 1 −3
6
−5 −3
1
−30
2
0 2 −4 10

1 1 −1
8
∼ 0 1 −3
6
0 0
2
−2

1 1 0
7
∼ 0 1 0
3
0 0 1 −1

1 0 0
4
∼ 0 1 0
3
0 0 1 −1

−4
−16
−18

−4
−16
14

3. a. Reducing

3 −2

1 −1

0
4
Autumn 2010
A I3 gives
−4
1
0
−3
0
1
21
0
0
3
5
7

−2
5 .
7
Therefore, the general solution of Ax = b consists of all vectors of the form
p + tu, where t is a real number,
 
 
−4
−2
−3
 5 



.
and
u= 
p= 
1 
7
1
0
b. p − 2u, i.e.,


   
−2
−4
6
 5 
   
  − 2 −3 =  11 
 7 
 1   5 
0
1
−2
is a solution of Ax = b whose first entry is 6.
c. The null space of A, i.e., the solution set of Ax = 0, consists of all vectors of
the form tu, where t is a real number. So {u} is a basis of Nul A
2. Reducing the matrix A gives

 
1 0 2 1
1
 1 1 1 2  0

 
1 2 0 b  ∼ 0
0 3 a b
0

1
0

∼
0
0

1
0

∼
0
0
0
1
2
3
2
−1
−2
a
0
1
0
0
2
−1
0
a+3
0
1
0
0
2
−1
a+3
0

1
1 

b − 1
b

1
1 

b − 3
b−3

1
1 
.
b − 3
b−3
a. The rank of A is equal to 4 if a 6= −3 and b 6= 3.
b. The rank of A is equal to 3 if a = −3 and b 6= 3, or a 6= −3 and b = 3.
c. The rank of A is equal to 2 if a = −3 and b = 3.
and therefore
 
0
3
 
0
∼
0
 
0
1

3

∼
0
0

3

∼
0
0

3


∼ 0
0

1

∼
0
0
−2

0

− 53 − 31 1 0

21
0
0 1

−4
1
0
0

− 53 − 31
1
0

1
−4 12 1

0 −15 48 4

0
−7
21 53 

1
−4
12 1

27 −78 −6

21 −63 −5

−4
12
1

9
−26 −2

21 −63 −5
,
−4
12
1
−4
− 13
4
−2
− 31
0
−2
− 13
0
0
0
1
0
0
1
0
0
1
0
0
1
1
0


9
−26 −2

A
= 21 −63 −5 .
−4
12
1
b. Since A is invertible, the unique solution to Ax = b is

  

9
−26 −2
3
−15
−1
x = A b =  21 −63 −5 1 = −40 .
−4
12
1
8
8
−1
4. a. Since the inverse of a nonsingular block
the inverse of M , if it exists, has the form

P Q
N = R S
0
0
diagonal matrix is block diagonal,

0
0 ,
T
with square blocks. Comparing the blocks of N M ,


 
P Q 0
0 B 0
QA
 R S 0  A C
0  =  SA
0
0 T
0
0 D
0
P B + QC
RB + SC
0

0
0 ,
TD
to those of the block diagonal identity matrix, gives
QA = I,
P B + QC = 0,
SA = 0,
RB + SC = I
and T D = I. The equations QA = I and T D = I, since the factors on the
left sides are square, give Q = A−1 and T = D−1 . Then SA = 0 implies that
S = 0. Since S = 0, the equation RB + SC = I becomes RB = I and, again
since the factors on the left side are square, it follows that R = B −1 . Finally, the
equation P B + QC = 0 becomes P B + A−1 C = 0, or P B = −A−1 C, i.e.,
P = −A−1 CB −1 . Therefore, the inverse of M , if it exists, is


−A−1 CB −1 A−1
0
−1

B
0
0 .
0
0
D−1
b. If M is invertible then A, B and D are invertible.
5. If (taking Calcium, Oxygen, Hydrogen and Phosphorus, in order)
 
 
 
 
1
0
3
0
2
4
8
1







s1 =   ,
s2 =   ,
s3 =  
and
s4 =  
2
3
0
2
0
1
2
0
then positive integer solutions of the vector equation x1 s1 + x2 s2 = x3 s3 + x4 s4
balance the chemical equation in
question. The corresponding augmented matrix
is s1 s2 −s3 −s4 0 .
Sample examinations (solutions)
Linear Algebra (201-NYC-05)
c. If x belongs to H and α is a scalar, then x1 x4 = x2 x3 , and therefore
(αx1 )(αx4 ) = α2 x1 x4 = α2 x2 x3 = (αx2 )(αx3 ), so αx belongs to H .
This shows that H is closed under scalar multiplication.
d. Since H is not closed under addition, it is not a subspace of R4 .
6. Using that det A = −5 and AT = A, one finds that
a. det(−4A−1 ) = (−4)4 (det A)−1 = − 256
and
5
b. det(2AT − A) = det A = −5, but that
c. det(A − I) is not determined, for example if A = e1 e2 e3 −5e4 then
5
.
det(A − I) = 0, and if A = 2e1 2e2 2e3 − 8 e4 then det(A − I) = − 13
8
13. a. If A is any symmetric
matrix then Col A
such
= Row A. Simple examples
matrices are I2 , e1 0 , 0 e2 , e2 e1 , e1 + e2 e1 + e2 , & c.
AB T C −1
7. a. If In =
then, since the determinant preserves products and is
invariant under transposition, (det A)(det B)(det C)−1 = 1, which implies that
det A 6= 0 and det B 6= 0, so A and B are nonsingular.
b. Since A and B T C −1 are n × n matrices and AB T C −1 = In , it follows that
A−1 = B T C −1 , which implies that A commutes with B T C −1 .
c. By Part b, B T C −1 A = In , which implies (since B T and C −1 A are n × n
matrices) that (B T )−1 = C −1 A, and therefore B −1 = (C −1 A)T .
8. Here is an LU factorization of A

 
2
5
1
−2 −8 = −1
8
2
4

0
2
0 0
1
0
0
1
6
b. If y ∈ Nul A and y ∈ Row A = Col AT then y = AT x for some x and so
yT y = yT AT x = (Ay)T x = 0T x = 0, which implies that y = 0. Hence,
if Nul A ⊂ Row A then Nul A = {0}, which implies that A is invertible and
therefore Row A = R2 . This shows that there is no 2 × 2 matrix A such that
Nul A = Row A.
c. If A = 0 0 −e3 then det A = 0, and A + I3 = e1 e2 0 , so
det(A + I3 ) = 0, as required.
d. If x
Ax is surjective then A is invertible (by the Invertible Matrix Theorem),
which implies that x
Ax is injective. So there is no matrix as requested.
e. The cross product
     
1
2
6
−1 ×  1  = −6 ,
−4
−2
3

5
−3 ,
0
via the rough work
−3
−18
0
is orthogonal to each of the given vectors, so its normalization,
 
2
1 
−2 ,
3
1
9. By inspection, one has
− 15
0
!
0
1
1
0
!
−6
1
!
6
−5
0
.
is a vector as required
1
10. The parallelogram P1 is formed by u1 and u2 , and P2 is the parallelogram
formed by the vectors v1 and v2 , where
3
−1
2
1
,
u2 =
,
v1 =
and
v2 =
.
u1 =
1
−1
−3
−2
14. In this problem one has
 
 
 
1
1
3
−−→
−−→
−−→
u = AB = −4 , v = AC =  0  and w = AD = −2 .
0
−3
6
−−→
−−→
a. The cosine of the angle between u = AB and v = AC is equal to
a. The area of the parallelograms P1 and P2 are, respectively,
−1
1
2 3
and
−1 −3 = 5,
−2 1 = 7
1
uT v
1 √
= 170
= √ √
170.
10 17
kukkvk
−−→
−−→
b. The orthogonal projection of u = AB onto v = AC is equal to
 
1
vT u
1 
projv u = uv = T v = 10
0 ,
v v
−3
so the determinant of a linear transformation which maps P1 onto P2 is equal to
± 75 (depending on whether or not the transformation preserves or reverses orientation).
b. One such matrix B will satisfy B v1 v2 = u1 u2 , i.e.,
−1
B = u1 u2 v1 v2
−1
1
3
−1
2
=
−2 1
−1 −3
1
3
−3 −2
= 51
−2 1
1
−1
0 −5
= 15
.
7
3
and the orthogonal component of the projection is
 
 
1
1
1 


perpv u = u − uv = −4 − 10
0 =
0
−3
v2 v1
−1
= − 51
8
5

1
10

9
−40 .
3
c. The distance from the point B to the line AC is equal to the length of the or−−→
−−→
thogonal component of the projection of u = AB onto v = AC , i.e.,
√
√
13
1
1690 = 10
10.
kperpv uk = 10
The other such matrix is
u1 u2
Autumn 2010
Alternatively, the distance between the point B and the line AC is equal to the area
−−→
−−→
of the parallelogram formed by u = AB and v = AC , divided by the length of
−−→
(the base) v = AC . The area of the parallelogram formed by u and v is the is the
length of the cross product
     
1
1
12
n = u × v = −4 ×  0  =  3  ,
0
−3
4
√
√
i.e., 13, and the length of v is 10 which, as expected, gives a distance of 13
10.
10
d. The vector n (from Part c) is a normal to the plane containing the points A, B
and C, so a normal equation of this plane is nT =nT p, i.e.,
7
.
0
11. The bilinearity of the dot and cross product, the algebraic characterization of
the cross product and the alternating property of the determinant, gives
u · (v − u) × (w − u)
=u· v×w−v×u−u×w+u×u
= det u v w − det u v u − det u u w
= det u v w .
12x1 + 3x2 + 4x3 = 41
−−→ −−→ −−→
where for example p is OA , OB or OC .
−−→
−−→
−−→
e. The volume of the parallelepiped with edges u = AB , v = AC , w = AD
T
is equal to, 54, i.e., (the absolute value of) det u v w = (u × v) w.
12. a. Since 0 · 0 = 0 · 0, the zero vector of R4 belongs to H .
b. e2 and e3 belong to H , since each side of the equation defining H is zero, but
e2 + e3 does not belong to H , because the left side of the equation defining H
is zero and the right side is 1. Therefore, H is not closed under addition.
2
Sample examinations (solutions)
Linear Algebra (201-NYC-05)
d. Let A = p q ; then, since v is parallel to U , the points on U which minimize
the distance to the line u + tv are of the form x + tv, where
15. Let n1 be the given normal to the plane $1 , and let n2 be the given normal to
the plane $2 .
a. The planes $1 and $2 are parallel if their normal vectors are so; i.e., if
n2 = − 12 n1 , or k = − 52 .
b. The planes $1 and $2 are orthogonal if their normal vectors are so; i.e., if
0 = nT
1 n2 = −10 + 5k, or k = 2.
c. The point P (−1, −1, 1) lies on the planes $1 and $2 if its coordinates
satisfy the equations of both planes. The point P is on the plane $1 since
4(−1) − 2(−1) + 5(1) = 3, and it is also on $2 if −2(−1) + (−1) + k(1) = 0,
i.e., k = −1.
x = A(AT A)−1 AT u


1 1 0 1 3 3 −1 1 0 1

=
1 0 3 6
1 1 0
1 2


1 1 
0 1
−1
2
 2
= 31
= 31 
1 0 −1
1
3
1 2
 
2
 
1 1
= 3 
1
3
16. If x is in both Nul A and Nul B, then (A + B)x = Ax + Bx = 0 + 0 = 0,
which shows that x belongs to Nul(A + B).
17. If
X=
x11
x21
x12
x22
,
then
AX =
1
0
1
x11
−1
x21
x12
x22
=
x11 + x21
−x21
x12 + x22
−x22
x11
x21
x12
x22
1
0
1
x11
=
−1
x21
x11 − x12
x21 − x22
,
and so by comparing entries one finds that AX = XA if, and only if, x21 = 0
and x11 = 2x12 + x22 . Therefore, X ∈ H if, and only if,
x22 + 2x12 x12
X=
= x22 M1 + x12 M2 ,
0
x22
where
 
0

1 
1
2 1
1


1 1 0 1 1


1 0 1
1 2
is the orthogonal projection of u onto U . The distance between U and the line
u + tv is equal to the length of
 
−2
 
1  2 
u − x = 3  ,
2
0
√
which is 32 3.
and
XA =
Autumn 2010
19. a. An indexed set B of vectors in a linear space V is a basis of V if B is
linearly independent and Span B = V .
b. If x ∈ V then there are scalars β1 , . . . , βn such that x = β1 b1 + · · · + βn bn ,
and so
1 0
2 1
.
and
M2 =
0 1
0 0
Since {M1 , M2 } spans H , and is clearly linearly independent, it is a basis of H .
T (x) = T (β1 b1 + · · · + βn bn )
M1 =
= β1 T (b1 ) + · · · + βn T (bn )
= 0W ,
18. a. Since U + V = Span{p, q, u, v}, and since


1 1 0
2
0 1 1 −1

p q u v =
1 0 1
3 
1 2 1
1


1
1
0
2
0
1
1 −1

∼
0 −1 1
1 
0
1
1 −1


1 1 0
2
0 1 1 −1

∼
0 0 2
0 
0 0 0
0


1 1 0
2
0 1 0 −1

∼
0 0 1
0 
0 0 0
0


1 0 0
3
0 1 0 −1
,
∼
0 0 1
0 
0 0 0
0
since T (bj ) = 0W for j = 1, . . . , n., and so T is the zero transformation.
c. Since Span
A = V (because A is a basis of V ), there is an n × m matrix
A = αji such that
bj = αj1 a1 + · · · + αjm am =
m
X
αji ai
i=1
for j = 1, . . . , n. Likewise, there is an m × n matrix B = βij such that
ai = βi1 b1 + · · · + βin bn =
n
X
βij bj
j=1
for i = 1, . . . , n (since B is a basis of V and hence Span B = V ). So for
k = 1, . . . , n,
X
m
m
n
X
X
bk =
αki ai =
αki
βij bj
i=1
=
i=1
n X
m
X
j=1
j=1
αki βij bj ,
i=1
which implies that
m
X
i=1
(
αki βij =
1 if k = j, and
0 if k 6= j,
i.e., AB = In , since B is linearly independent. By a similar computation, the fact
that A is linearly independent implies that BA = Im . It follows that m = n
and that A and B are inverses (e.g., because AB = In implies that B has linearly
independent columns, so n 6 m, and likewise BA = Im implies that A has
linearly independent columns, so m 6 n).
from which it follows that p, q, u are linearly independent and v = 3p − q, the
list {p, q, u} is a basis of U + V .
b. From the solution to Part a it follows that every vector in U ∩ V is a multiple of
v = 3p − q. Therefore, {v} is a basis of U ∩ V .
c. By the solution to Part a, 31 v = p − 31 q belongs to V and to the line p + tq,
from which it follows that the distance between V and this line is zero.
Note: A more compact solution is available if you know about coordinate vectors
and the list-vector product (the product of a list of abstract vectors and a column
vector): Since Span A = V there is a m × n matrix P such that A P = B. If
P x = 0 then Bx = A P x = 0, and so, since B is linearly independent, x = 0.
Hence, P has linearly independent columns, which implies that n 6 m. Likewise,
since Span B = V and A is linearly independent, it follows that m 6 n.
3