Matrices #3 - 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] [0 0 0 | 0] [ 2 -5 8 | 0 ] [ 0 -12 9 | 0 ] -1/12R₂ → R₂ [ 0 0 0 | 0 ] [ 1 -5/2 4 | 0 ] [ 0 1 -3/4 | 0] -1/4 R₁ → R₁ [ 0 0 0 | 0 ] [ 1 0 17/8 | 0 ] [ 0 1 -3/4 | 0 ] R₁ - 5/2R₂ → R₁ [ 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 [1 0] + d₂ [0 0] [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 [-3 -7 9 | 0] [ 1 3 -5 | 0] [ 0 1 -3 | 0] R₃ + 3R₁ [ 0 2 -6 | 0] [ 1 3 -5 | 0] [ 0 1 -3 | 0] R₃ - 2R₂ [ 0 0 0 | 0] [ 1 0 4 | 0] [ 0 1 -3 | 0] R₁ - 3R₂ [ 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 ] = [ -3x₂ + 4x₃ ] [ x₂ ] [ x₂ ] = x₂[ -3 ] + x₃ [ 4 ] [ x₃ ] [ 0 ] [ 0 ] [ x₄ ] [ 0 ] [ 1 ] Answer 2 Row operations: Replace R2 with R2 - 2R1. The resulting row-echelon form is [ 1 3 -4 | 0 ] [ 0 0 0 | 0 ] Let's denote the variables as x₁, x₂, x₃, and x₄ for the corresponding columns. Since the second row consists of all zeros, it represents the equation 0x₁ + 0x₂ + 0x₃ + 0x₄ = 0, which simplifies to 0 = 0. This equation is always true, indicating that we have free variables. Therefore, the system of equations has infinitely many solutions, and we can express the solutions in parametric vector form as follows: x₁ = -3x₂ + 4x₄ x₂ = x₂ x₃ = 0 x₄ = x₄ Using the parametric variable t for x₂ and s for x₄, the solutions can be written as: x₁ = -3t + 4s , x₂ = t , x₃ = 0 , x₄ = s where s and t are arbitrary real numbers. Answer 3 To find all the solutions of the equation Ax = 0, where A is row equivalent to the given matrix [1 3 0 -4] [2 6 0 -8] we can perform row operations on A to obtain its row echelon form. The row echelon form of A can then be used to describe the solutions in parametric vector form. Performing row operations on A: [1 3 0 -4] [2 6 0 -8] [0 0 0 0] R₂ - 2R₁ The resulting row echelon form shows that the second row consists of all zeros. This means that the second equation in the system is redundant and can be ignored. We are left with a single equation x₁ + 3x₂ - 4x₄ = 0. To express the solutions in parametric vector form, we introduce parameters for the free variables. In this case, x₂ and x₄ are free variables. We can choose them to be x₂ = t and x₄ = s, where t and s are arbitrary real numbers. Now, we can express the solution vector x in terms of the parameters t and s: x₁ = -3t + 4s x₂ = t x₃ = 0 x₄ = s Therefore, the solutions to the equation Ax = 0 in parametric vector form are: [ x₁ ] = [-3t + 4s] [ x₂ ] [ t ] [ x₃ ] [ 0 ] where t and s are arbitrary real numbers. [ x₄ ] [ s ] Question 4 Determine the null space of each of the following matrices: [1 3 -4] [2 -1 -1] [-1 -3 4] Answer Step 1 Look at the matrices provided: [1 3 -4] [2 -1 -1] [-1 -3 4] To discover the null spaces, first solve the following system of equations: [1 3 -4] [x₁ ] [0] [2 -1 -1] [x₂ ] = [0] [-1 -3 4] [x₃ ] [0] Use the Gauss Elimination method to solve the aforementioned matrix. Find the row echelon form: R₂ = R₂ - 2R₁ [ 1 3 -4 | 0] [ 0 -7 7 | 0] [-1 -3 4 | 0] R₃ = R₃ + R₁ [ 1 3 -4 | 0 ] [ 0 -7 7 | 0 ] [ 0 0 0 | 0 ] Step 2 We get the equations: -7x₂ + 7x₃ = 0 x₂ = x₃ And x₁ + 3x₂ - 4x₃ = 0 x₁ + 3x₂ - 4x₂ = 0 x₁ - x₂ = 0 x₁ = x₂ Therefore, the null space is given by: [ x₁ ] [ 1] [ x₂ ] = x₂ [ 1] [ x₃ ] [ 1] Question 5 Find a basis for the space of 2 × 2 lower triangular matrices Basis = Answer 1 S = set of lower triangular matrices. To find the basis of s Let [ a 0 ] A = [ c b ] = a [1 0] + c [0 0] + b [0 0] [0 0] [1 0] [0 1] ∴ A can be written as the linear combination of vectors [1 0 ] , [0 0] and [0 0] [0 0] , [1 0], [0 1] and B = { [1 0] , [0 0] , [0 0] } {[0 0], [1 0] , [0 1] } is linearly independent set ∴ B is the basis for the space of 2 × 2 lower triangular matrices Finally, the answer is: B = { [1 0], [0 0], [0 0] } {[0 0], [1 0], [0 1] } Answer 2 Step 1: To find a basis for the space of 2x2 lower triangular matrices, we need to determine a set of linearly independent matrices that span this space. A 2×2 lower triangular matrix has the form: [ a 0 ] A = [ c b ] where a, b, and c are elements of a field, such as real numbers or complex numbers. Let's denote the matrix above as matrix A. To find a basis, we need to consider different values of a, b, and c that give distinct matrices. We'll start with a = 1, b = 0, and c = 0: A₁ = [ 1 0 ] [ 0 0 ] Step 2: Next, we'll consider a = 0, b = 1, and c = 0: A₂ = [ 0 0 ] [ 0 1 ] Lastly, we'll consider a = 0, b = 0, and c = 1: A₃ = [0 0] [1 0] These three matrices, A₁, A₂, and A₃, are linearly independent and span the space of 2x2 lower triangular matrices. Therefore, they form a basis for this space. The basis is given by: Basis = {A₁, A₂, A₃} Alternatively, we can express the basis as a set of column vectors: Basis = { [1 0], [0 0], [0 0] } { [0 0] [0 1] [1 0] } Each of these column vectors represents one of the matrices A₁, A₂, and A₃. Answer 3 [a 0] [c b] To form a basis, we need linearly independent vectors that span the space of lower triangular matrices. We can choose the following matrices as a basis: [1 0] , [0 0], [0 0] v₁ = [0 0] v₂ = [0 1] v₃ = [1 0] Therefore, the basis for the space of 2x2 lower triangular matrices is given by: Basis = {v₁, v₂, v₃}.