Matrices #1 - Questions and Answers

Question 1 Determine if the system has a nontrivial solution. Try to use as few row operations as possible. 2x₁ - 5x₂ + 8x₃ = 0 -2x₁ - 7x₂ + x₃ = 0 4x₁ + 2x₂ + 7x₃ = 0 Answer Step 1 The given system is 2x₁ - 5x₂ + 8x₃ = 0 -2x₁ - 7x₂ + x₃ = 0 4x₁ + 2x₂ + 7x₃ = 0 The augmented matrix is [2 -5 8 | 0] [-2 -7 1 | 0] [4 2 7 | 0] [-2 -7 1 | 0] R₂ + R₁ → R₂ [0 -12 9 | 0] [4 2 7 | 0] [4 2 7 | 0] [2 -5 8 | 0] [0 -12 9 | 0] R₃ - 2R₁ → R₃ [0 12 -9 | 0] [4 2 7 | 0] [2 -5 8 | 0] [0 -12 9 | 0] R₃ + R₂ → R₃ [0 0 0 | 0] [0 0 0 | 0] [1 -5/2 4 | 0] [0 1 -3/4 | 0] -1/12 R₂ → R₂ [0 1 -3/4 | 0] [0 0 0 | 0] [1 0 17/8 | 0] [-1/4 R₁ - 1/4 R₂→ R₁ [0 1 -3/4 | 0] [0 0 0 | 0] Thus we get x₁ + (17/8)x₃ = 0 → x₁ = -(17/8)x₃ x₂ - (3/4)x₃ = 0 → x₂ = (3/4)x₃ Hence, the given system has a nontrivial solution because variable x₃ is a free variable. Result The system has a nontrivial solution because variable x₃ is a free variable. Question 2 Find a basis for the space of 2 × 2 diagonal matrices. Basis = { [ ] [ ] } Answer 1 Step 1 Let the space of 2 x 2 diagonal matrices be as [ a 0 ] [ 0 b ] we can choose values of a and b randomly Step 2 A basis of matrix [ a 0 ] [ 0 b ] can be taken as a=1, b=0 [ 1 0 ] [ 0 0 ] a = 0, b = 1 [ 0 0 ] [ 0 1 ] Basis = { [ 1 0 ] [ 0 0 ] } { 0 0 ] [ 0 1 ]} Answer 2 We need to identify a collection of linearly independent matrices that cover the space of 2×2 diagonal matrices in order to construct a basis for the space. D = [ d₁ 0] [ 0 d₂ ] where d₁ and d₂ are arbitrary constants. To form a basis, we need to find distinct diagonal matrices that are linearly independent. Here's a possible basis for the space of 2×2 diagonal matrices: B = { [1 0] , [0 0] } {[0 0] , [0 1]} We have two distinct diagonal matrices, and we can see that no linear combination of one matrix can produce the other. Therefore, these matrices are linearly independent. To confirm that these matrices span the space, we can take an arbitrary 2×2 diagonal matrix: D = [d₁ 0] [0 d₂] We can rewrite this matrix as a linear combination of the basis matrices: D = [d₁ 0] = d₁ [1 0] + d₂ [0 0] [0 d₂] [0 0] [0 1] This shows that any 2×2 diagonal matrix can be expressed as a linear combination of the basis matrices. Therefore, the set B is a basis for the space of 2x2 diagonal matrices. Answer 3 To find a basis for the space of 2x2 diagonal matrices, we can start by considering the general form of a diagonal matrix: [ a 0 ] [ 0 b ] where a and b are scalars. We can see that any 2x2 diagonal matrix can be written as a linear combination of the following matrices: [ 1 0 ], [0 1], [0 0] [0 0] , [0 0], [0 1] Let's call these matrices D₁, D₂, and D₃, respectively. We can express any 2x2 diagonal matrix as: aD₁ + bD₂ + cD₃ where a, b, and c are scalars. Therefore, the set {D₁, D₂, D₃} forms a basis for the space of 2x2 diagonal matrices. Alternatively, we can express the basis as: [ 1 0], [0 0], [0 1] [0 0], [1 0], [0 1] Question 3 Write the solution set of the given homogeneous system in parametric vector form. x₁ + 3x₂ - 5x₃ = 0 x₁ + 4x₂ - 8x₃ = 0 -3x₁ - 7x₂ + 9x₃ = 0 Answer The given system is x₁ + 3x₂ - 5x₃ = 0 x₁ + 4x₂ - 8x₃ = 0 -3x₁ - 7x₂ + 9x₃ = 0 The augmented matrix is [ 1 3 -5 | 0] [ 1 4 -8 | 0] [-3 -7 9 | 0] Reduce matrix to row echelon form [ 1 3 -5 | 0] [0 1 -3 | 0] R₂ - R₁→ R₂ [-3 -7 9 | 0] [ 1 3 -5 | 0] [0 1 -3 | 0] R₃ + 3R₁ → R₃ [0 2 -6 | 0] [ 1 3 -5 | 0] [ 0 1 -3 | 0] R₃ - 2R₂ → R₃ [ 0 0 0 | 0] [ 1 0 4 | 0] [ 0 1 -3 | 0] R₁ - 3R₂ → R₁ [ 0 0 0 | 0] Thus we get x₁ + 4x₃ = 0 → x₁ = -4x₃ x₂ - 3x₃ = 0 → x₂ = 3x₃ Hence, the given homogeneous system in parametric vector form is [ x₁ ] = [ -4 ] x₃ [ x₂ ] [ 3 ] [ x₃ ] [ 1 ] Question 4 Find A such that the given set is Col A. { [2s + 3t] [r + s -2t ] [4r+ s ] [3r - s - t] : r, s, t real } Answer An element in this set may be written as [2s + 3t] [0] [2] [3] [r + s -2t ] = r [1] + s[1] + t [-2] [4r + s] [4] [1] [0] [3r - s - t] [3] [-1] [-1] where r, s, and t are any real numbers. So, the set is Col A where A = [0 2 3] [1 1 -2] [4 1 0] [3 -1 -1] Result: A = [0 2 3] [1 1 -2] [4 1 0] [3 -1 -1] Question 5 Find x such that the matrix is equal to its inverse. [ 7 x] A = [ -8 -7] Answer 1 Step 1 Given matrix, [ 7 x] A = [-8 -7] And, A = A⁻¹ Formula for an inverse of a matrix is A⁻¹ = 1/det(A) adj(A) Step 2 Now, det(A) = 7(-7) - x(-8) det(A) = -49 + 8x det(A) = 8x - 49 And, [ -7 -x ] adj(A) = [ 8 7 ] So, A⁻¹ = 1/(8x-49) [ -7 -x ] [ 8 7 ] Step 3 Now, as given A = A⁻¹ [ 7 x] = 1/(8x-49) [ -7 -x ] [-8 -7] [ 8 7 ] On comparing the matrices, 7 = -7 / (8x - 49) and x = -x / (8x - 49) 7(8x - 49) = -7 56x - 343 = -7 56x = 343 - 7 56x = 336 x = 336 / 56 x = 6 Step 4 Therefore, The value of x such that the matrix is equal to its own inverse is x = 6 Answer 2 Explanation: Let's represent the given matrix A as: [ 7 x ] A = [-8 -7] To find the inverse of A, we can use the formula for a 2x2 matrix: A⁻¹ = 1 / det(A) [ d -b] [-c a] where det(A) represents the determinant of A, and a, b, c, and d are the elements of the matrix A. The determinant of A can be calculated as: det(A) = ad - bc In this case, we have: a = 7, b = x, c = -8, d = -7 Substituting the values into the determinant formula, we get: det(A) = (7 * -7) - (x * -8) Simplifying further: det(A) = -49 + 8x For A to be equal to its inverse, we need to set A equal to its inverse and solve for x. This gives us the following equation: [ 7 x ] = 1 / (8x-49) [ -7 -x] [-8 -7] [ 8 7 ] To find x, we can equate the corresponding elements of the matrices: 7 = -7 / (8x - 49) and x = -x / (8x - 49) Solving these equations will give us the value(s) of x that satisfy the condition. Answer 3 To find x such that the matrix A is equal to its inverse, we can set up the equation as follows: A = A⁻¹ Given matrix A: A = [ 7 x] [-8 -7] Let's find the inverse of matrix A: A⁻¹ = 1 / (ad-bc) [ d -b] [-c a] where a, b, c, and d are the elements of matrix A. Plugging in the values: A⁻¹ = 1/((7)(-7)-(x)(-8)) [-7 -x] [8 7] Simplifying the determinant: (7)(-7)-(x)(-8) = 49 + 8x To find x such that A is equal to its inverse, we equate the matrices: [ 7 x ] = 1/(49+8x) [-7 -x] [-8 -7] [8 7] Now, we can equate the corresponding elements of the matrices: 7 = -7 / (49 + 8x) x = -x / (49+8x) -8 = 8 / (49 + 8x) -7 = 7 / (49 + 8x) Solving these equations will give us the value of x.