1. A set πΆ is defined to be set difference between two set π΄ and π΅: πΆ = π΄ β π΅. And set π΄ and π΅ are defined as like below. π΄: π ππ‘ ππ πππ ππππ ππ’πππππ π΅: π ππ‘ ππ πππ πππ‘πππππ ππ’πππππ ππ is an element of the set πΆ. Check the statements below by each and verify whether they are TRUE or FALSE, and shortly explain why. a. For any element ππ β πΆ, it holds ππ 2 > 0 b. There is an element ππ β πΆ, satisfying this condition; ππ ππ π ππππππ‘πππ πππππππ c. π β πΆ d. One of the two solutions of equation (π₯ β 2) (π₯ + 7 9 ) = 0 is an element of the set πΆ e. πΆ ππ ππ πππππππ‘π π ππ‘ f. π΅βπΆ = β g. For any two different elements ππ and ππ in the set πΆ, their product is also an element of πΆ: ππππ β πΆ (ππ β ππ) 2
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Problem set 1
Due date: February 13th Tuesday 14:00 (before beginning of the class)
You can write your answers by hand or in a word (or Latex) file. You can submit your
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for calculation.
1. A set πΆ is defined to be set difference between two set π΄ and π΅: πΆ = π΄ − π΅. And set π΄ and
π΅ are defined as like below.
π΄: π ππ‘ ππ πππ ππππ ππ’πππππ π΅: π ππ‘ ππ πππ πππ‘πππππ ππ’πππππ
ππ is an element of the set πΆ. Check the statements below by each and verify whether they are
TRUE or FALSE, and shortly explain why.
a. For any element ππ ∈ πΆ, it holds ππ 2 > 0
b. There is an element ππ ∈ πΆ, satisfying this condition; ππ ππ π ππππππ‘πππ πππππππ
c. π ∈ πΆ
d. One of the two solutions of equation (π₯ − 2) (π₯ + 7
9 ) = 0 is an element of the set πΆ
e. πΆ ππ ππ πππππππ‘π π ππ‘
f. π΅βπΆ = ∅
g. For any two different elements ππ and ππ in the set πΆ, their product is also an element of πΆ:
ππππ ∈ πΆ (ππ ≠ ππ)
2
2. Textbook exercise 2.3.4. from (a) to (g)
3. Textbook exercise 2.4.6.
4. Check the statements below by each and verify whether they are TRUE or FALSE, and
shortly explain why.
a. A relation π₯2 + π¦2 = 9 (π₯| − 3 ≤ π₯ ≤ 3) is a function.
b. There is a function π¦ = 3π₯ + 1 and domain of π₯ is all integers. Then, image of π¦ includes
some irrational numbers.
c. π₯3 ÷ π₯7 = π₯10
d. π₯√2(π₯2√2 + π₯√3) = π₯4+√6
e. π₯π × π₯π × π₯π = π₯πππ
f. ππ₯ × ππ₯ × ππ₯ = (πππ)π₯
5. Find equilibrium market price and quantity the market given below
ππ = ππ ππ = 30 − 2π ππ = −6 + 4π
6. Textbook exercise 3.2.2.(b)
7. Textbook exercise 3.4.3.
8. Find solution of system of equations. If solution is not unique, then shortly explain why.
a. π₯ − 4π¦ = −3 2π₯ + 3π¦ = 7 b. 3π₯ + 2π¦ = 5 4π₯ + 8
3 π¦ = 6
3
9. Answer below matrix operations. If it cannot be defined, shortly explain why.
a. [ 1 2 3 4
] + [ −1 −2 5 0
] b. 3 [ 0 −3 3 0
] − 2 [ −1 −2 1 7
]
c. [ 1 2 3 4
] [ −2 3 0 5 −8 1
] d. [ 5 0 1 2 1 −3
] [ 3 0 2
−3 2 8 ]
e. [ −2 3 0 5 −8 1
] [ 1 2 3 4
]
10. Textbook exercise 4.2.6.
11. Check linear dependence of three vectors.
π’ = (−2, 3), π£ = (5, 1), π€ = (1, −2)
12. For two vectors π’ = (3,2) and π£ = (−1, 7), find scalars π, π, π, and π satisfying below
equations. If you cannot find those scalars, then shortly explain why.
a. ππ’ + ππ£ = (1,0) b. ππ’ + ππ£ = (0,1)
13. Prove (π΄′)−1 = (π΄−1)′. You can prove this statement by utilizing matrix multiplication,
and properties of transpose and inverse. You do not have to introduce dimension of matrix π΄.
14. Find determinant of below matrices
a. [7] b. [ 1 2 3 4
]
c. [ 1 0 2
−1 5 5 −3 0 7
] d. [ 1 2 3 4 5 6 0 0 0
] e. [ 1 2 7
−6 5 0 −6 2 2
]
4
15. Find rank of below matrices
a. [ 6 8 3
4 1] b. [
1 2 5 1 3 9 2 0 4
] c. [ 1 0 0 0 1 0 0 0 1
]
16. Check the statements below by each and verify whether they are TRUE or FALSE, and
shortly explain why.
a. If π΄ is a square matrix, then both of π΄π΄′ and π΄′π΄ are well defined.
b. If π΄ is a (2 × 2) nonsingular matrix, then its rank is equal to 2.
c. If determinant of (3 × 3) matrix π΄ is zero, then its rank cannot be 3.
d. For any (3 × 3) symmetric matrix π΄, π΄2 is also a symmetric matrix.
e. If π΄π΅ is well defined, then π΄′π΅′ is also well defined.
f. For a (2 × 2) matrix π΄, if π΄2 = πΌ then π΄ = πΌ
17. Textbook exercise 5.4.2. only (a) and (b)
18. Textbook exercise 5.4.5. only (a) and (b)
19. Find π₯2 ∗ by applying Cramer’s rule. If you cannot find a unique value of π₯2
∗, then shortly
explain why.
a.
3π₯1 + 2π₯2 + 4π₯3 = 2
2π₯1 − 4π₯2 + π₯3 = 3
2π₯1 − 1π₯2+ = −4
b.
π₯1 + 2π₯2 + 3π₯3 = 1
5π₯1 − 1π₯2 + π₯3 = −3
2π₯1 − 1π₯2 + 3π₯3 = 2
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