Matrices #2 - Questions and Answers

Question 1 Let u = [ 3 ] and v = [ 8 ] [1 ] [ 6 ] Compute the distance from y to the line through u and the origin. Answer Formula for orthogonal projection: ŷ = (y * u) / (u * u) * u Formula for distance: ||y - ŷ|| y * u = (3, 1) * (8, 6) = 24 + 6 = 30 u * u = (8, 6) * (8, 6) = 64 + 36 = 100 ŷ = (30/100) * [8] = [24/10] [6] [18/10] y - ŷ = [3] - [24/10] = [3/5] [1] [18/10] [-4/5] || y - ŷ|| = √ ((3/5)² + (-4/5)²) = √(9/25 + 16/25) = √25/25 = √1 = 1 Question 2 Show that if A ⊆ C and B ⊆ D, then A×B ⊆ C×D Answer Ok you know that A × B just means A × B = {(a, b) | a ∈ A and b ∈ B}. And we can assume that A ⊆ C and B ⊆ D. Now we want to show A×B ⊆ C×D. For the proof I would begin by saying "Suppose A ⊆ C and B ⊆ D. Let x ∈ A×B where x = (a, b) where a ∈ A and b ∈ B. We want to show x ∈ C×D. From here you should be able to complete the proof just from the assumption." Question 3 Determine whether each statement is true or false. If a statement is true, give a reason or cite an appropriate statement from the text. If a statement is false, provide an example that shows the statement is not true in all cases or cite an appropriate statement from the text. (a) Addition of matrices is not commutative. (b) The transpose of the sum of matrices is equal to the sum of the transposes of the matrices. Answer a) False. The addition of Matrices is Commutative. For example: Consider two matrices of order 2x2 A = [[1,3],[2,4]], B = [[0,1],[-1,4]] Matrices A and B are commutative under addition, if A+B = B+A. [[1,3],[2,4]]+[[0,1],[-1,4]] = [[0,1],[-1,4]]+[[1,3],[2,4]] = [[1,4],[1,8]] = [[1,4],[1,8]] Hence Commutative. b) True. Yes, The transpose of the sum of matrices is equal to the sum of the transposes of the matrices. Consider two matrices of order 2 x 2 A = [[1,3], [2,4]], B = [[0, 1], [-1, 4]] A + B = [[1,4], [1, 8]] (A + B)ᵀ = [[1, 1], [4, 8]] (1) Aᵀ = [[1, 2], [3,4]] Bᵀ = [[0, -1], [1, 4]] Aᵀ + Bᵀ = [[1, 2], [3, 4]] + [[0, -1], [1, 4]] Aᵀ + Bᵀ = [[1, 1], [4, 8]] (2) From (1) and (2), (A+B)ᵀ = Aᵀ + Bᵀ Hence proved Question 4 Quesuuns Exercise 0.27 Migo (exponential Propaommy Distribution) Consider the following exponential probability density function. for z ≥ 0 f(x) = 1/4* e^ (-x/4) a. Which of the following is the formula for P(Z <2)> 1P(z <= z₀) = e^(-kappa₀/2) 2 P(x <= x₀) = 1 e^(-x₀) 3 P(x <= x₀) = 1-e^(-n₀/2) 9. 10. Select your answer b. Find P(x < 1) (to 4 decimals). c. Find P(2≤ z≤ 4) (to 4 decimals). d. Find P(z < 6) (to 4 decimals). Answer Let's solve the given problem step by step. a. To find the formula for P (z < 2), we need to integrate the given probability density function from 0 to 2. P(z < 2) = ∫₀² f(x) dx = ∫₀² (1/4)e⁻ˣ/⁴ dx To solve this integral, we can use u-substitution, where u = -x/4 and du = -dx/4 P(z < 2) = ∫₀² (1/4)e⁻ˣ/⁴ dx = -4∫ eᵘ du Now, we can solve the integral using the fundamental theorem of calculus: P(z < 2) = -4 [e⁻ˣ/⁴]²₀ = -4 [e⁻²/⁴ - e⁰] = -4[e⁻¹/² - 1] = 4(1-e⁻¹/²) ≈ 1.5364 Therefore, the correct formula for P (Z < 2) is not among the options given. b. To find P (x < 1), we need to integrate the probability density function from 0 to 1: P(x < 1) = ∫₀¹ f(x)dx = ∫₀¹ (1/4)e⁻ˣ/⁴ dx Again, we can use u-substitution with u = -x/4 and du = -dx/4 P(x < 1) = ∫₀¹ (1/4) e⁻ˣ/⁴ dx = -4∫ eᵘ du = -4 [e⁻ˣ/⁴]¹₀ = 4(1 - e⁻¹/⁴) ≈ 0.2846 Therefore, the probability that x is less than 1 is approximately 0.2846. c. To find P(2≤ z≤ 4), we need to integrate the probability density function from 2 to 4: P(2 ≤ z ≤ 4) = ∫₂⁴ f(x) dx = ∫₂⁴ (1/4)e⁻ˣ/⁴ dx Using u-substitution with u = -x/4 and du = -dx/4 P(2 ≤ z ≤ 4) = ∫₂⁴ (1/4)e⁻ˣ/⁴ dx = -4∫eᵘ du = -4 [e⁻ˣ/⁴]₄² Solving the integral: P(2< z < 4) = -4(e⁻¹ − e⁻¹/²) = 4(e⁻¹/² − e⁻¹) ≈ 0.0903 Therefore, the probability that z is between 2 and 4 is approximately 0.0903. d. To find P (z < 6), we need to substitute a = x/4 into the probability density function: f(x) = (1/4)e⁻ˣ/⁴ Then, we can integrate from 0 to 24: P(a < 6) = ∫₀²⁴ f(a) da = ∫₀²⁴ (1/16)e⁻ᵃ/¹⁶ da Using u-substitution with u = -a/16 and du = -(1/16) da P(a < 6) = ∫₀²⁴ (1/16)e⁻ᵃ/¹⁶ da = -16∫ eᵘ du = -16[e⁻ᵃ/¹⁶]²⁴₀ Solving the integral: P(a < 6) = -16[e⁻¹·⁵ − 1] ≈ 3.7622 Therefore, the probability that a is less than 6 (or x is less than 24) is approximately 0.3762. Question 5 Construct a 4 x 3 matrix with rank 1 Answer 1 [ a λa λa] [ λa a λa] [ λa λa a] [λa λa λa] the rest of the matrix just has to be a scalar multiple of the first row. Answer 2 A rank 1 matrix has a one-dimensional column space. Every column is a multiple of some fixed vector. To construct a 4 x 3 matrix, choose any nonzero vector in R⁴, and use it for one column. Choose any multiples of the vector for the other two columns.