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Ad. Math. Mtds. for Economists Homework 1
October 10, 2019
Question 1
Find the general solution of the systems whose augmented matrix is given by
Answer
R 2 − (3)R 1 =
R 3 − (2)R 1 =
Swap R 2 and R 3 =
; (−1)R 2 =
R 1 + (4)R 2 =
R 3 =
R 2 − (3)R 3 =
R 1 − (13)R 3 =
Therefore x 1 = 3; x 2 = -1 and x 3 = 1
Question 2
Is b in the span
a 1 , a 2 , a 3
a 1 =
(^) , a 2 =
(^) , a 3 =
(^) , b =
Ad. Math. Mtds. for Economists Homework 1
October 10, 2019
Answer
For b to be in the Span{a 1 , a 2 , a 3 }, there should be c 1 , c 2 , c 3 ∈ R that gives c 1 a 1 +c 2 a 2 +c 3 a 3 = b, making the solution consistent.
swap R 1 and R 2 =
3
R 1 =
R 3 + (2)R 1 =
R 3 − R 2 =
R 3 =
R 2 − R 3 =
R 1 + R 3 =
R 1 − (2)R 2 =
Since there is a c 1 , c 2 , c 3 that makes the system of solution consistent, it means b is in the Span{a 1 , a 2 , a 3 }
Question 3
Do the columns of A span R^3.
A =
Answer
R 2 + R 1 =
R 3 − R 1 =
Ad. Math. Mtds. for Economists Homework 1
October 10, 2019
R 1 + (5)R 2 =
R 3 − (6)R 2 =
R 3 =
R 2 − (2)R 3 =
R 1 + (5)R 3 =
The set are linearly independent. Because non of the vectors in the set can be defined as a linear combination of the others b (^) [ 2 10 1 5
]
R 1 =
[
]
R 2 − R 1 =
[
]
The set are linearly dependent. Because one of the vectors in the set can be defined as a linear combination of the others.
Question 5
For A =
[
]
, define T : R^2 → R^2 by T (x) = A(x).
Find the image under T of U =
[
]
Answer
T (x) = A(x)
T (U ) = A(U ) =
[
] [
]
[
]
[
]
Therefore the image under T of U is
[
]
Ad. Math. Mtds. for Economists Homework 1
October 10, 2019
Question 6
Let A =
a. Find all x ∈ R^2 that are mapped into the zero vector by the transformation x → Ax.
b. Is b =
(^) in the range of the linear transformation x → Ax.?
Answer
a (^)
R 2 + (2)R 1 =
R 3 − R 1 =
R 3 + (2)R 2 =
R 2 =
R 1 + (2)R 2 =
R 1 − (
)R 3 =
R 2 + (^1
)R 3 =
x 1 = 53 x 4 , x 2 = − 32 x 4 , x 3 = -3x 4 , x 4 is free variable
x = x 4
5 −^32 3 − 3 1
b (^)
R 2 + (2)R 1 =
Ad. Math. Mtds. for Economists Homework 1
October 10, 2019
Answer
det
(^) = (1)det
[
]
-(3)det
[
]
+(2)det
[
]
= 1[−1 + 9] − 3[−3 + 6] + 2[−9 + 2]
Since the determinate of the matrix is not equal to zero, it means the matrix is invertible.
Question 9
Find the inverse of the matrix A=
[
]
, if it exists.
Answer
det(A) = 8 − 9 = − 1
Since the det(A) 6 = 0, the inverse of A exist.
[ 2 9 1 0 1 4 0 1
]
; swap R 1 and R 2 =
[
]
R 2 − (2)R 1 =
[
]
R 1 + (4)R 2 =
[
]
A−^1 =
[
]
Question 10
Let A=
[
]
. Write 3A. Is det(3A) = 3detA?
Ad. Math. Mtds. for Economists Homework 1
October 10, 2019
Answer
3 A = 3
[
]
[
]
[
]
det(3A) =det
[
]
3 det(A) =det
[
]
= 3[8 − 9] = 3(−1) = − 3
Therefore det(3A) 6 = 3det(A)
Question 11
Let A=
[
a b c d
]
and k as a scalar. Find a formula that relates det(kA) to k and det(A)
Answer
kA = k
[
a b c d
]
[
k(a) k(b) k(c) k(d)
]
[
ka kb kc kd
]
det(kA) =det
[
ka kb kc kd
]
= k^2 ad − k^2 bc = k^2 (ad − bc)
Question 12
Consider
A =
Put A in row echelon form and calculate the determinant of A.
Answer
)R 1 =
R^2 −^ (3)R^1 =
Ad. Math. Mtds. for Economists Homework 1
October 10, 2019
det
det
x =
y =
z =
Question 14
Let A=
a. Is u=
(^) in Nul A?
b.Is v=
(^) in Nul A?
Answer
A vector x is in Nul A if Ax = 0
a.
Since Au = 0, u is in Nul A.
b.
Since Av 6 = 0, v is not in Nul A.
Ad. Math. Mtds. for Economists Homework 1
October 10, 2019
Question 15
let A=
. Find the vectors that span the null space of the matrix
A.
Answer
swap R 1 and R 3 =
R 2 + (−1)R 1 =
(3)R 2 =
(2)R 3 =
R 3 + (−1)R 2 =
R 1 =
R 2 =
R 3 =
R 2 + (−1)R 3 =
R 1 + (−2)R 3 =
R 1 + (3)R 2 =
x 1 = 2x 3 − 3 x 4 + 24x 6 x 2 = 2x 3 − 2 x 4 + 7x 6 x 3 and x 4 are free variables x 5 = − 4 x 6 x 6 is free variables 2 2 1 0 0 0
Ad. Math. Mtds. for Economists Homework 1
October 10, 2019
b
c
[
]
[
]
Answer
A subset U of a vector space V is called a basis if i. U is linearly independent, and ii. U is a spanning set. a. (^)
R 3 + R 1 =
R 1 − R 2 =
R 3 − R 2 =
R 1 + R 3 =
R 2 − R 3 =
It fellows that the solution is x 1 = x 2 = x 3 = 0. Hence U is linearly independent. Therefore the set is in basis of R^3. b. (^)
R 2 − (4)R 1 =
R 3 − (7)R 1 =
R 2 =
R 1 − (2)R 2 =
R 3 + (6)R 2 =
It fellows that the general solution is x 1 = x 3 , x 2 = − 2 x 3. Where x 3 is a free variable. Hence U is linearly dependent. Therefore the set is not in basis of R^3. c. (^) [ 2 0 0 2 2 0
]
R 2 =
[
]
Ad. Math. Mtds. for Economists Homework 1
October 10, 2019
R 1 =
[
]
R 2 − R 1 =
[
]
It fellows that the general solution is x 1 = x 2 = 0. Hence U is linearly independent. Therefore the set is in basis of R^2.
Question 19
We know that A =
be row equivalent to^ B^ =
Find the bases for ColA and NulA.
Answer
Basis for ColA:
, and
Basis for NulA:
Question 20
Find a basis and state the dimension of the following sets:
a.
r − 2 s + t 5 r − 3 s − 2 t r + 3t
where, r,s,t∈ R
b.
2 s + t 3 s − 2 t s + t
where r,s,t∈ R
Ad. Math. Mtds. for Economists Homework 1
October 10, 2019
R 2 + R 3 =
R^4 −^ (2)R^3 =
The dimension is 3.
Question 22
The matrix A, is row equivalent to B. Without calculation, list rank of A and dim NulA. Then find bases for ColA, RowA, and NulA.
A =
,^ B^ =
Answer
Basis for ColA:
, and
Basis for RowA:
, and
Basis for NulA:
RankA = 3 and dim NulA = 2
Question 23
Is λ = 4 and eigenvalue of A =
Ad. Math. Mtds. for Economists Homework 1
October 10, 2019
Answer
If λ = 4 is an eigenvalue of A, then det(A − 4 I) = 0
det(A − 4 I) = det
Therefore λ = 4 is an eigenvalue of A
Question 24
Find the eigenvalues and the basis for the eigenspace of the eigenvalues you have find.
A =
Answer
A − λI =
λ 0 0 0 λ 0 0 0 λ
1 − λ − 3 3 3 − 5 − λ 3 6 − 6 4 − λ
Set det(A − λI) = 0
(1 − λ) det
[
− 5 − λ 3 − 6 4 − λ
]
[
6 4 − λ
]
[
3 − 5 − λ 6 − 6
]
(1 − λ)((− 5 − λ)(4 − λ) - (3)(-6)) + 3(3(4 − λ) - (3)(6)) + 3((3)(-6) - (− 5 − λ)(6)) = 0 16 + 12λ - λ^3 = 0
λ = 4, λ = -2 and λ = - Therefore the eigenvalues are 4 and -
For λ = 4
A − 4 I =
R 1 =
R 2 − (3)R 1 =
Ad. Math. Mtds. for Economists Homework 1
October 10, 2019
Answer
First
det(A − λI) =
[
7 − λ 8 − 4 − 5 − λ
]
= (7 − λ)(− 5 − λ) + 32
= λ^2 - 2λ -35+32 = λ^2 - 2λ - 3 = (λ − 3)(λ + 1) = λ = 3 , -
D =
[
]
For λ = 3
(A − (3)I) =
[
]
R 1 =
[
]
R 2 + (4)R 1 =
[
]
x 1 = − 2 x 2 ; x 2 is a free variables
Therefore the eigenvector for λ = 3 is
[
]
For λ = -
(A − (−1)I) =
[
]
swapR 1 andR 2 =
[
]
R 1 =
[
]
R 2 − (8)R 1 =
[
]
x 1 = −x 2 ; x 2 is a free variables
Therefore the eigenvector for λ = -1 is
[
]
Ad. Math. Mtds. for Economists Homework 1
October 10, 2019
P =
[
]
[
]
; swap R 1 and R 2 =
[
]
R 2 + (2)R 1 =
[
]
R 1 − R 2 =
[
]
P −^1 =
[
]
A = P DP −^1 =
[
] [
] [
]
A^100 = P D^100 P −^1 =
[
] [
] [
]
Question 26
Which of the sets are orthogonal? a.
√^1
b.
Answer
a.
[ 1 0 − 1
]
×
√^1
[
]
×
