Download Sample examinations Linear Algebra (201-NYC-05) Winter 2012

Sample examinations
Linear Algebra (201-NYC-05)
1. Given the homogeneous system with matrix equation Ax = 0:
 

 
 x1
−1 0
2
−1
0
0
x 
1
−5
5
1   2
 1
0
x3  =   .
 2

2 −10
10
3
0
x 
4
2
1
−7
6
−1
0
x5
13. Let
2. Find a quadratic polynomial p such that p(2) = 0, p(−2) = 32 and p0 (1) = −7.
 
 


0
1
k





0 .
3. Let S = {v1 , v2 , v3 }, where v1 = 1 , v2 = 0 and v3 =
2
k
2k + 3
3
x1
2
∈ R : x1 = 0 or x2 = 0 .
x2
Is 0 ∈ W ? Justify your answer.
Is W closed under scalar multiplication? Justify your answer.
Is W closed under addition? Justify your answer.
Is W a subspace of R2 ? Explain.
W =
a. Write the solution of the system in parametric vector form.
b. Write the zero vector as a non-trivial linear combination of the columns of A.
3
Winter 2012
3
For which value(s) of k is Span S: a. R ? b. a plane in R ? c. a line in R ?
x
−x + 2y
4. Let T1 : R2 → R2 be a linear transformation given by T1
=
.
y
2x − 3y
a. Find the standard matrix of T1 .
b. Find all values of k for which T1 (`) a horizontal line, where ` has parametric representation
2
1
+t
.
0
k
1 −2 −3
c. If T1 ◦ T2 is a linear transformation with standard matrix
, then
−3 5
7
i. identify the domain and codomain of T2 , and ii. find the standard matrix of T2 .
5. Suppose that {u, v, w} is linearly independent, and that x = 2u+3w and y = v+2w.
Prove that {u, x, y} is linearly independent.


1 6
2 7
6. Find an LU factorization of 
.
3 8
4 9




a d g
a + 2b + 4c d + 2e + 4f g + 2h + 4k
7. Let A =  b e h and B = 3a + 4b + 7c 3d + 4e + 7f 3g + 4h + 7k.
c f k
5a + 7b + 8c 5d + 7e + 8f 5g + 7h + 8k
a.
b.
c.
d.
14. Let
1 1
.
1 1
Find a 2 × 2 matrix B such that AB = 0 but BA 6= 0.
A=
15. Suppose that A and B are n × n matrices such that AB = 0 and BA 6= 0.
a. Show that each column of B is in the null space of A.
b. What that even though BA 6= 0, it must be true that (BA)2 = 0.
c. Show that B is not invertible.
16. Let V be the set of all 2 × 2 matrices X such that AX = 0, where
1 1
A=
.
1 1
Explain why V is a subspace of M2×2 , find a basis of V .
17. Let V = { p ∈ P2 : p(2) = 0 }. Explain why V is a subspace of P2 , and find a basis
of V and the dimension of V .
18. Let T : Rn → Rm be a linear transformation.
a. What is the dimension of the range of T if T is injective? Explain.
b. What is the dimension of the kernel of T if T is surjective? Explain.
c. Prove that if the rank of T is m then there is a linear transformation S : Rm → Rn such
that T ◦ S = idRm .
19. Given this picture
u
v
0
a. Find a matrix C such that B = CA.
b. Find a scalar λ such that det B = λ det A for all possible choices of A.
of two vectors in Rn , draw a picture of each of the following.
a. { (1 − t)u + tv : 0 6 t 6 1 }
b. { su + tv : 0 6 s 6 1, 0 6 t 6
8. Let A be a 3 × 3 matrix withdet A = −2.
a. Find det AT A2 (−2A)−1 .
b. Find det adj(2A) .
c. projv u + proju v
d. projv u − perpv u
20. Let
9. a. Given matrices A, B and C with A and B invertible, find matrices W , X, Y and Z
such that
0
A
W
X
I
0
=
.
B C
Y
Z
0 I
b. Use the above result to find D −1 , where

0 0
0 0

D = 2 1
1 0
0 0
0
0
0
3
1
10. Use Cramer’s Rule to solve the linear system:
3
1
2
1
0

4
2

0 .
3
1
11. Simplify the matrix expression B(B + I)
−B
−1
For which value of n is Row A a subspace of Rn ?
Without calculation, give a basis of Row A.
For which value of m is Col A a subspace of Rm ?
Without calculation, give a basis of Col A.
What is the rank of A?
What is the dimension of Nul AT ?
−1
2
+t
:t∈R .
3
1
a. Plot ` (i.e., draw a rough picture of `).
b. Find the distance between ` and the origin.
c. For which values of a and b, if any, is the line with parametric representation
1
1
+t
a
b
21. Let V = Span{u, v}, where
 
1
u = −1
1
.
12. You are given a the matrix A and its reduced echelon form R:



1 2 0
3 6 2 1 5 2 2
1 2 1 0 2 0 −3
0 0 1



A = 1 2 0 1 1 1 4  ; R = 0 0 0
0 0 0
1 2 1 0 2 1 1 
1 2 0 0 3 0 −1
0 0 0
a.
b.
c.
d.
e.
f.
`=
equal to `?
d. Where does ` intersect the x1 -axis?
e. What is the cosine of the angle between ` and the x1 -axis.
7x − 9y = 11
4x + 5y = −2.
−1 −1
1
2
0 3
0 −1
1 −2
0 0
0 0

0 −1
0 −2

0 1 .
1 4 
0 0
and
 
2
v = 1 .
0
a. Find an implicit equation defining V .
b. Find a parametric representation of the line orthogonal to V which contains the origin.
c. For which values of α, if any, does i + 2j + αk belong to V ?
22. Give two different unit vectors which are orthogonal to both
 
 
2
3
3
1 .
and
4
1
23. What are the possible values of the angle between vectors u and v in R3 if
1
(u × v)
u·v
is a unit vector?
}
Sample examinations (solutions)
Linear Algebra (201-NYC-05)
1. Below is the coefficient matrix A of the system, and its reduced echelon form.
 


−1 0
2
−1
0
1 0 −2 1 0
 1


1
−5
5
1  0 1 −3 4 0


∼
 2
2 −10 10
3  0 0
0
0 0
2
1
−7
6
−1
0 0
0
0 1
a. The first, second and fifth columns of A are pivot columns, and the non-pivot
columns of A are given by
a3 = −2a1 − 3a2
where s and t are real numbers.
b. Either of the relations displayed in Part a (or any combination thereof) expresses the zero vector as a non-trivial linear combination of the columns of A;
e.g., 2a1 + 3a2 + a3 = 0.
2. The polynomial p(x) = a0 + a1 x + a2 x2 satisfies p(2) = 0, p(−2) = 32
and p0 (1) = −7 if, and only if,
a0 + 2a1 + 4a2 = 0,
a0 − 2a1 + 4a2 = 32 and
a1 + 2a2 = −7,
where the last equation uses p0 (x) = a1 + 2a2 x. Reducing the coefficient matrix
of this linear system gives

 

1
2
4
0
1 0 0 14

 

1 −2 4 32  ∼ 0 1 0 −8 ,

 

1
0
1
2 −7
0 0 1
2
and therefore, p(x) = 14 − 8x + 12 x2 .
3. Row reduction gives

A = v2 v1 v3
1
∼ 0
0
0
1
0

k
,
0
−k2 + 2k + 3
and therefore: a. Span S = R3 if, and only if, A has three pivot columns, i.e.,
0 6= −k2 + 2k + 3 = (3 − k)(1 + k), or k 6= −1, 3; b. Span S is a plane in R3
if, and only if, A has two pivot columns, i.e., 0 = (3 − k)(1 + k), or k = −1, 3;
c. there are no values of k such that Span S is a line in R3 since in any case A has
at least two pivot columns.
−1
2
4. a. The standard matrix of T1 is A = T1 (e1 ) T1 (e2 ) =
.
2
−3
b. T1 (`) is a horizontal line if, and only if, there is a non-zero scalar α such that
1
−1 + 2k
α
T1
=
=
;
k
2 − 3k
0
i.e., if, and only if, k =
6. One has
 
1
6

7
 = 2
8 3
4
9

1
2

3
4
via the rough work
1
2
3
4
a4 = a1 + 4a2 ,
and
so the general solution of Ax = 0 is the set of all vectors of the form
 
 
2
−1
3
−4
 
 

 
s
1 + t  0  ,
0
 1 
0
0
Winter 2012
2
.
3
R3 .
c. i. The domain of T2 is the domain of T1 ◦ T2 , which is
The codomain of
T2 is the domain of T1 , which is R2 . ii. Since T1 : R2 → R2 is invertible (for
example, because det A 6= 0) and T1 ◦ T2 : R3 → R2 is linear, it follows that
T2 = T1−1 ◦ (T1 ◦ T2 ) : R3 → R2 is linear, and its standard matrix is
3 2
1
−2 −3
−3 4 5
A−1 T1 ◦ T2 =
=
.
2 1
−3
5
7
−1 1 1
5. If α, β and γ are scalars and αu + βx + γy = 0, then the expressions defining
x and y in terms of u, v and w give αu + β(2u + 3w) + γ(v + 2w) = 0, or
(α + 2β)u + γv + (3β + 2γ)w = 0, which implies that α + 2β = 0, γ = 0
and 3β + 2γ = 0 because {u, v, w} is linearly independent. Since γ = 0 and
3β + 2γ = 0 it follows that β = 0, and therefore, since α + 2β = 0, that α = 0.
This shows that αu + βx + γy = 0 implies that α = 0, β = 0 and γ = 0; i.e.,
that {u, x, y} is linearly independent, as required.
0
1
2
3

1
0

0
 0
0 0
0
1
0
0
1
0
6
7
8
9

6
−5

0 
0
−5
−10
−15.
7. Since B = CA, where

1
C = 3
5

4
7
8
2
4
7
it follows that det B = (det C)(det A), so that
1 2 4 1
0
0 2
λ = det C = 3 4 7 = 3 −2
−5 = 3 1
5 7 8 5 −3 −12
5
=9
4
(using column reduction, multilinearity and then cofactor expansion).
8. Multilinearity, product preservation and invariance under transposition of the
determinant gives
a. det AT A2 (−2A)−1 = (det A)(det A)2 (−2)−3 (det A)−1 = − 21 , and
b. det adj(2A) = det det(2A)(2A)−1 = (23 )3 (det A)3 2−3 (det A)−1
= 256.
9. a. Expanding the product on the left side of the given equation yields
AY
AZ
I 0
=
.
BW + CY BX + CZ
0 I
Since A is invertible, comparing the upper blocks gives Y = A−1 and Z = 0.
Next, comparing lower left blocks gives BW + CY = 0, i.e., BW = −CY ,
or W = −B −1 CY since B is invertible. Finally, comparing lower right blocks
gives I = BX + CZ = BX (since Z = 0), and so X = B −1 . Therefore,
−1 0 A
−B −1 CA−1 B −1
.
=
−1
B C
A
0
b. Partition the given matrix D as in Part a so that


2 1 0
3 4

A=
,
B = 1 0 3
and
1 2
0 0 1

2
C = 1
0

0
3 .
1
The formula for the inverse of a 2 × 2 matrix, and row reducing B I3 , gives


0
1
−3
2
−4
−1
−1
1
A
= 2
and
B
= 1 −2
6 ;
−1
3
0
0
1
and so

−B
−1
−1
CA
by a direct calculation. Therefore,

−1

 0

1
D−1 = 
 2

 1
− 12
=
− 21

−4
0 
3
2
 0
−1
2
0
1
0
1
−2
− 32
0
0
−2
0
0
3
2
0
0

−3

6 

1 


0 
0
by the result from Part a.
10. Cramer’s Rule gives
11 −9
−2
5 =
x= 7 −9
4
5 37
71
and
7
4
y= 7
4
11 −2
= − 58
.
71
−9
5 Sample examinations (solutions)
Linear Algebra (201-NYC-05)
18. a. If T is injective then the dimension of the kernel of T is zero. Since
dim ker T + rank T = n by the rank formula, it follows that the dimension of the
range of T (i.e., the rank of T ) is n.
b. If T is surjective then its rank is m, so (as in Part a) the rank formula implies
that the dimension of the kernel of T is n − m.
c. If the rank of T is m then T is surjective, and so there are xi ∈ Rn such that
T (xi ) = ei for i = 1, . . . , m. Let S : Rm → Rn be the unique linear transformation such that S(ei ) = xi for i = 1, . . . , m (i.e., column i of the standard
matrix of S is xi ). If y = y1 e1 + · · · + ym em is any vector in Rm , then
11. Since the inverse of a product of two matrices is the product of their inverses,
taken in the opposite order, and the inverse of the inverse of a matrix is the original
matrix, one has
−1
−1 −1
B(B + I)−1
= (B + I)−1
B
= (B + I)B −1 .
Therefore, the expression in question is equal to
(B + I)B −1 − B −1 = (B + I − I)B −1 = BB −1 = I,
by distributivity and the definition of the inverse of a matrix.
12. a. Row A is a subspace of R7 .
b. A basis of Row A is the list of transposes of the non-zero rows of the reduced
echelon form (or any echelon form) of A:
       
1
0
0
0 


 2   0   0  0




       



 0   1   0  0


       
 0  ,  0  ,  1  , 0 .










       


 3  −1 −2 0




 0   0   0  1






−1
−2
1
4
T ◦ S(y) = y1 T ◦ S(e1 ) + · · · + ym T ◦ S(em )
= y1 T (x1 ) + · · · + ym T (xm )
= y1 e1 + · · · + ym em
= y,
by the linearity of T ◦ S (since S and T are linear, so is T ◦ S) and the definition
of S. This shows that T ◦ S = idRm , as required.
19. a. { (1−t)u+tv : 0 6 t 6 1 } is the segment joining u and v (parametrized
in the direction going from u to v).
u
c. Col A is a subspace of R5 .
d. A basis of Col A is the list of pivot columns of A:
       
3
2
1
2 





      


1 1 0 0
1 , 0 , 1 , 1 .
       


1 1 0 1






1
0
0
0
0
b. { su + tv : 0 6 s 6 1, 0 6 t 6 12 } is the (interior and boundary of the)
parallelogram formed by the segments joining the origin to u and to 21 v.
u + 12 v
13. a. The zero vector belongs to W .
b. If x ∈ W and α is any scalar, then whichever entry of x is zero remains so
when multiplied by α, so αx ∈ W . Thus, W is closed under scalar multiplication.
c. Since e1 ∈ W and e2 ∈ W but e1 + e2 ∈
/ W , W is not closed under addition.
d. By Part c, W is not closed under addition, so W is not a subspace of R2 .
14. If B = e1 − e2 0 , then
1 1
1
0
0 0
AB =
=
1 1
−1 0
0 0
BA =
1
−1
1
0
1
0
1
1
=
1
−1
{ (1 − t)u + tv : 0 6 t 6 1 }
v
e. The rank of A is four.
f. dim Nul AT = 5 − rank A = 5 − 4 = 1.
but
Winter 2012
1
0
6=
−1
0
u
v
1
v
2
0
c. The projections and their sum are displayed below (together with u and v).
projv u + proju v
proju v
0
,
0
u
as required.
15. a. If bj denotes column j of B then column j of AB = 0 is Abj = 0, which
implies that bj ∈ Nul A.
b. One has (with all of the details)
(BA)2 = (BA)(BA) = (BA)B A = B(AB) A = (B0)A = 0A = 0,
v
projv u
0
as required.
c. If B is invertible then 0 6= BA = BABB −1 = B0B −1 = 0, which is
impossible.
d. The projections (the orthogonal component translated by projv u) and their
difference are displayed below (along with u and v).
u
16. V is a subspace of M2×2 because V is the kernel of the linear transformation
T : M2×2 → M2×2 defined by T (X) = AX. Next, if AX = 0 then each
column of X belongs to Nul A = Span{e1 − e2 }; therefore V is the span of
1
0
0
1
B=
,
.
−1 0
0 −1
perp v
u
v
0
Since B is clearly linearly independent, it forms a basis of V .
17. V is a subspace of P2 because it is the kernel of the linear transformation
T : P2 → R defined by T (p) = p(2). Next, since { p ∈ P2 : p(0) = 0 } =
{ αt + βt2 : α, β ∈ R } has basis {t, t2 }, and since replacing t by t − 2 defines
an isomorphism P2 → P2 , it follows that {t − 2, (t − 2)2 } is a basis of V , and
the dimension of V is two.
pro j v
u
projv u − perpv u
2
Sample examinations (solutions)
Linear Algebra (201-NYC-05)
20. a. Below is a rough sketch of ` = { p + tv : t ∈ R }, where
−1
2
p=
and
v=
;
3
1
Winter 2012
must be consistent, which gives a = 4.
d. ` intersects the x1 -axis at the point (−7, 0).
e. The cosine of the angle between ` and the x1 -axis is
vT e1
=
kvkke1 k
the x1 intercept (not displayed) is at −7.
x2
2√
5.
5
21. a. A normal vector to V is given by
     
2
1
1
n = v × u = 1 × −1 = −2 ,
0
1
−3
and an implicit equation defining V is nT x = 0, or x1 − 2x2 − 3x3 = 0.
b. The line which is orthogonal to V and contains the origin is the set of all vectors
of the form tn, where t is a real number and n is the normal vector found in Part a.
c. w = i + 2j + αk belongs to V if, and only if, nT w = 0, or α = −1.
7
2
−1
3
22. The unit vectors ±ˆ
u are orthogonal to the given vectors, where
     
 
2
3
−1
−1
1
1 √ 






ˆ=
u = 3 × 1 = 10 , and so u
u = 30 6 10  .
kuk
4
1
−7
−7
x1
b. The distance from p to the origin is equal to the area of parallelogram formed
by p and v, divided by the length of v, which is given by
|det p v |
√
= 75 5.
kvk
(Note: The distance is also given by the length of perpv p.)
c. For the given line to be equal to `, its direction vector must be parallel to v,
which gives b = 12 , and the equation
−1
2
1
+t
=
3
1
a
23. First of all, if
1
(u × v)
u·v
is a unit vector then neither u nor v is 0 (otherwise, u · v would be zero and so w
would be undefined). Next, observe that if kwk = 1 then u · v = ±ku × vk, i.e.,
kukkvk cos ϑ = ±kukkvk sin ϑ, where ϑ is the angle between u and v. Since u
and v are non-zero vectors, it follows that kukkvk 6= 0, and so cos ϑ = ± sin ϑ.
Therefore, ϑ = 41 π or ϑ = 34 π (because 0 6 ϑ 6 π).
w=
3