Matrices #6 - Questions and Answers

Question 1 Determine if b is a linear combination of the vectors formed from the columns of the matrix A. [ 1 -4 2 ] [ 3 ] A = [ 0 3 5], b = [-7] [-2 8 -4] [-3] Answer Step 1 We must ascertain if [3] b= [-7] [-3] is a linear combination of the vectors obtained from the A matrix's columns. [ 1 -4 2 | 3 ] [A | b] = [0 3 5 | -7 ] [-2 8 -4 | -3] [ 1 -4 2 | 3 ] ~ [0 3 5 |-7] R₃ + R₁→ R₃ [-2 8 -4 | -3] [ 1 -4 2 | 3] ~ [ 0 3 5 |-7] R₃ + 2R₁ → R₃ [ 0 0 0 | 3 ] The third formula is 0 = 3, therefore, we draw the conclusion that the system cannot be solved. The vectors created from the columns of matrix A are therefore not linearly combined to form b. Result: The vectors created from the columns of the matrix A are not linearly combined to yield the vector b. Question 2 Explain why each of the following algebraic rules will not work in general when the real numbers a and b are replaced by n x n matrices A and B. a) (a + b)² = a² + 2ab + b² b)(a+b)(a - b) = a²-b² Answer 1 (a) (A + B)² = (A + B)(A + B) (A + B)² = A(A + B) + B(A + B) (A + B)² = AA + AB + BA + BB (A + B)² = A² + AB + BA + B² As nxn matrices are not always commutative, the formula (A + B)² = A² + 2AB + B² will not always work in general for all matrices. (b) (A + B)(A - B) = A(A - B) + B(A - B) (A + B)(A - B) = AA - AB + BA - BB (A + B)(A - B) = A² - AB + BA - B² As nxn matrices are not always commutative, the formula (A+B)(A - B) = A² - B² will not always work in general for all matrices. Answer 2 Step 1: We need to consider the properties and operations of matrices. a) (A + B)² = A² + 2AB + B² This rule does not hold for matrices in general. When expanding (A + B)², we get: (A + B)² = (A+B)(A + B) = A(A + B) + B(A + B) Expanding further, we have: A(A + B) + B(A + B) = A² + AB + BA + B² In general, matrix multiplication is not commutative, so AB and BA are not necessarily equal. Therefore, we cannot simplify the expression to A² + 2AB + B², making the rule invalid for matrices. Step 2: b) (A + B)(A - B) = A² - B² This rule also does not hold for matrices in general. When expanding (A + B)(A - B), we have: (A + B)(A - B) = A(A - B) + B(A - B) Expanding further, we get: A(A - B) + B(A - B) = A² - AB + BA - B² Similarly to the previous case, matrix multiplication is not commutative, so AB and BA are not necessarily equal. Therefore, we cannot simplify the expression to A² - AB + BA - B², making the rule invalid for matrices. In conclusion, these algebraic rules that hold true for real numbers a and b do not hold in general when matrices A and B are used instead. Answer 3 To explain why the given algebraic rules do not work in general when the real numbers 'a' and 'b' are replaced by nxn matrices 'A' and 'B,' we will analyze each rule separately. a) (A + B)² = A² + 2AB + B² Expanding the left-hand side of the equation, we have: (A + B)² = (A + B)(A + B) Using the distributive property, we can expand this expression as follows: (A + B)(A + B) = A(A + B) + B(A + B) Applying the distributive property again, we get: A(A + B) + B(A + B) = A² + AB + BA + B² Notice that in the above expression, we have the terms AB and BA, which represent matrix multiplication between A and B in different orders. In general, matrix multiplication does not commute, meaning that AB and BA can yield different results. Therefore, we cannot simplify A² + AB + BA + B² to A² + 2AB + B², as we would in the case of real numbers. b) (A + B)(A - B) = A² - B² Expanding the left-hand side of the equation, we have: (A + B)(A - B) Using the distributive property, we can expand this expression as follows: (A + B)(A - B) = A(A - B) + B(A - B) Applying the distributive property again, we get: A(A - B) + B(A - B) = A² - AB + BA - B² Similar to the previous case, we have the terms AB and BA, which represent matrix multiplication between A and B in different orders. Since matrix multiplication does not generally commute, we cannot simplify A² - AB + BA - B² to A² - B². In conclusion, the algebraic rules (A + B)² = A² + 2AB + B² and (A + B)(A - B) = A² - B² do not hold in general when real numbers 'a' and 'b' are replaced by nxn matrices 'A' and 'B' due to the non-commutativity of matrix multiplication. Question 3 Find the value(s) of h for which the vectors are linearly dependent. Justify each answer. [1] [-2] [3] v₁= [5] , v₂= [-9] , v₃ = [h] [3] [6] [-9] Answer 1 Step 1 Definition: A set of vectors {v₁, ......, vₙ} is linearly dependent if x₁v₁ + x₂v₂ + ... + xₙvₙ = 0 is NOT the only solution to the equation Of course, x₁ = x₂ = . . = xₙ = 0 will always be one solution to this equation, but it may or may not be the only solution. Step 2 Notice that in this case, equation (1) is equivalent to the equation Ax = 0 Step 3 [ 1 -2 3 | 0] A = [ 5 -9 h | 0] [-3 6 -9 | 0] x = [ x₁ ] [ x₂ ] [ x₃ ] We can find the solution to Ax = 0 by row reduction: [ 1 -2 3 | 0] [ 5 -9 h | 0] R₂ - 5R₁ → R₂ [-3 6 -9 | 0] R₃ + 3R₁ → R₃ [ 1 -2 3 | 0] ~ [ 0 1 h - 15 | 0] [ 0 0 0 | 0] [ 1 0 -27 + 2h | 0 ] R₁ + 2R₂ ~ [ 0 1 h - 15 | 0 ] [ 0 0 0 | 0 ] Step 4 From here we see that for any value of h, the possible solutions of Ax = 0 are [ x₁ ] [(27 - 2h)x₃] [ x₂ ] = [(h - 15)x₃ ] [ x₃ ] [ x₃ ] Since the last coordinate can be anything and is independent of h, we see that regardless of the value of h, this set of vectors is always linearly dependent. Result: The vectors are linearly dependent for all h Answer 2 Step 1: Let's set up the following equation: c₁ [1] + c₂ [-2] + c₃ [3] = [0] [5] [-9] [h] [0] [3] [6] [-9] [0] Expanding this equation, we get the following system of equations: c₁ - 2c₂ + 3c₃ = 0 5c₁ - 9c₂ + hc₃ = 0 -3c₁ + 6c₂ - 9c₃ = 0 We can solve this system of equations to find the values of h. First, let's eliminate c₁ from the second and third equations by subtracting the first equation multiplied by 5 from the second equation and by subtracting the first equation multiplied by 3 from the third equation: -9c₂ + hc₃ -5(-2c₂ + 3c₃) = 0 => -9c₂ + 5hc₃ - 15c₃= 0 => c₁ = (1/5)(9c₂ - (h-5)c₃ ) 6c₂ - 9c₃ + 3(-2c₂ + 3c₃) = 0 => 6c₂ - 9c₃ - 6c₂ + 9c₃ = 0 Step 2: Now, let's express c₂ and c₃ in terms of t as follows: c₂ = (5/9)c₃ c₃ = t Substituting these expressions into the equations, we get: (1/5)(9c₂ - (h-5)c₃) = -2((5/9)t) + 3t 5( (5/9)t ) - 9t + ht = 0 6(5/9)t - 9t = 0 Simplifying, we have: -t + ht = 0 -t - 9t = 0 -t + t = 0 10/9t - 9t = 0 The first equation is satisfied for any value of t. For the second equation, we can find the value of t that makes it true: 10/9t- 9t = 0 0/10t - 81/9t = 0/10 -71t = 0 / 10 t=0 The equation 0 = 0 is true for any value of t. Therefore, there is no value of t that satisfies the second equation. Since the second equation cannot be satisfied, there are no values of h that make the given vectors linearly dependent. Answer 3 Answer: There are no values of h that make the given vectors linearly dependent. Explanation: To find the value(s) of h for which the vectors are linearly dependent, we need to determine when one of the vectors can be written as a linear combination of the other two vectors. [1] [-2] [3] Let's set up the equation [5] = a [-9] + b [h], where a and b are scalars. [3] [6] [-9] Expanding the equation, we get the following system of equations: -2a + 3b = 1 (1) -9a + hb = 5 (2) 6a - 9b = 3 (3) To determine when the vectors are linearly dependent, we need to find values of h that satisfy this system of equations. Let's solve this system of equations. By manipulating (1) and (3), we can eliminate b: -2a + 3b = 1 (1) 6a - 9b = 3 (3) Multiplying (1) by 3 and (3) by 1, we obtain: -6a + 9b = 3 (4) 6a - 9b = 3 (3) Adding (4) and (3), we eliminate b: 0a + 0b = 6 The equation has no variables, and the left side is nonzero. Therefore, there is no solution for h that makes the vectors linearly dependent. Question 4 Show that A and B are not similar matrices [1 1 0] A = [2 2 0], B = [1 0 1] [3 0 3] [2 2 0] [0 1 1] Answer Step 1 If the matrices A and B are not similar, then their eigenvalues are not the same. Step 2 Consider the matrix A and find its eigenvalues as follows. [ 1 1 0 ] A = [ 2 2 0 ] [ 3 0 3 ] |A - λI| = 0 [1 1 0 ] [2 2-λ 0 ] [3 0 3 -λ] = 0 (1 – λ)[(2 – λ)(3 - λ) - 0] – 1 [2(3 – λ) - 0] + 0 = 0 (1 - λ)(6 -5λ + λ²) - 6 + 2λ = 0 6 - 5λ + λ² - 6λ + 5λ² + λ³ - 6 + 2λ = 0 -λ³ + 4λ² - 3λ = 0 λ(λ² + 4λ -3)= 0 λ(λ(λ - 4) = 0 λ = 0, λ = 0, λ = 4 Consider the matrix B and find its eigenvalues as follows. [1 1 0] B = [2 2 0] [0 1 1] |B - λI| = 0 [1 - λ 1 0 ] [2 2 - λ 0 ] = 0 [0 1 1 - λ ] (1 – λ)[(2 – λ)(1 – λ) - 0] – 1[2(1 – λ) - 0] = 0 (1-λ)(2-3λ+λ²)-2 + 2λ= 0 2-3λ+λ² -2λ+3λ² -λ³ -2 + 2λ= 0 -λ³ + 4λ² - 3λ= 0 λ(-λ² + 4λ - 3) = 0 λ(λ - 1)(λ - 3) = 0 λ₁ = 0, λ₂ = 1, λ₃ = 3 Observe that the eigenvalues of the matrices A and B are not the same, and thus these matrices are not similar. Question 5 (square roots of the identity matrix) For how many 2 x 2 matrices A is it true that A² = I ? Now answer the same question for n × n matrices where n > 2 Answer Step 1 To find how many matrices A of order 2 exist such that A² = I Here we have to find A² = A × A = I Let us consider any arbitrary matrix of order 2. [a b ] A = [c d] Then A² is given by A² = [a b ] [a b] = [a²+bc ab+bd] [c d] [c d] [ac+dc bc+d²] Now from A² = I, we get [a²+bc ab+bd] = [1 0] [ca+dc bc+d²] [0 1] (1) a² + bc = 1 (2) d² + bc = 1 (3) b(a + d) = 0 (4) c(a + d) = 0 Step 2 Case 1: If (a + d) ≠ 0 From (3), we get : b = 0 From (4), we get : c = 0 From (1), we get : a² = 1 ⇒ a = ±1 From (2), we get : d² = 1 ⇒ d = ±1 Therefore, we get the matrix of the form [a 0] where a=±1 and d =±1 A = [0 d] Therefore, there are only 4 matrices that exist such that A² = I from case 1. Step 3 Case 2: If (a + d) = 0 subcase a: If a = 0, that implies d = 0 Then from (1) and (2), we get : bc = 1 ⇒ c = b⁻¹ = 1/b in real numbers. Therefore, we get the matrix of the form [0 b] where b ∈ R A = [b⁻¹ 0] Therefore, there are infinitely many matrices that exist if the field is the set of real numbers such that A² = I Step 4 We can also consider further subcases to find the type of involutory matrices, but as the question is asked to find how many, we have already got infinitely many 2 x 2 matrices of such type. Similarly, for any n > 2, we can find infinitely many matrices satisfying this condition on the field of real numbers.