Matrices #7 - Questions and Answers

Question 1 Find three 2 x 2 idempotent matrices. (Recall that a square matrix A is idempotent when A² = A) Answer Step 1 Given data: The order of the given matrix is: m × n = 2 × 2. The expression for the idempotent matrix is, A² = A The expression of the identity matrix of 2 × 2 is, A = [1 0] [0 1] The expression for the square of the given matrix is, A² = [1 0] * [1 0] = [1 0] [0 1] [0 1] [0 1] Substitute the given matrix in the above expression. A² = A Step 2 Assume the given matrix of order 2 x 2 is, B = [0 0] [0 1] The expression for the square of the given matrix is, B² = [0 0] * [0 0] = [0 0] [0 1] [0 1] [0 1] Substitute the value in the expression. B² = B Step 3 Assume the given matrix of order 2 × 2 is, C = [1 1] [0 0] The expression for the square of the given matrix is, C² = [1 1] * [1 1] = [1 1] [0 0] [0 0] [0 0] Substitute the value in the expression. C² = C and [ 1 1 ] [1 0] C = [ 0 0 ] and [ 0 1] Question 2 Determine the null space of each of the following matrices: [ 1 2 -3 -1 ] [-2 -4 6 3] Answer Step 1 Given: [ 1 2 -3 -1 ] A = [-2 -4 6 3] Step 2 The first step is to find the reduced row echelon form of the matrix, Swap matrix rows: R₁ ↔ R₂ [ -2 -4 6 3 ] A = [ 1 2 -3 -1] Cancel the leading coefficient in row R₂ by performing R₂ ↔ R₂ + (½)R₁ [ -2 -4 6 3 ] A = [ 0 0 0 1/2] Multiply matrix row by constant: R₂ → 2 * R₂ [ -2 -4 6 3 ] A = [ 0 0 0 1 ] Cancel the leading coefficient in row R₁ by performing R₁ ← R₁ - 3R₂ [ -2 -4 6 0 ] A = [ 0 0 0 1] Multiply matrix row by constant: R₁ ← (-½) R₁ A = [ 1 2 -3 0] [ 0 0 0 1] Step 3 [ 3 -2 ] A = [1 0 ] [0 1 ] [0 0 ] Question 3 Compute the LU factorization of each of the following matrices. [ 2 4 ] A= [-2 1 ] Answer Step 1: Given: A matrix A = [2 4] [-2 1] We have to compute the LU factorization of this matrix. Step 2: Formula Used LU Factorization: If we have a matrix A, then an upper triangular matrix U obtained without pivoting under Gaussian elimination method, and there exists a lower triangular matrix L s.t. A = LU. Step 3: Calculation ⇒ [2 4] [-2 1] Use Gaussian Elimination Method, R₂ → R₂ - (-1)R₁ [2 4] [0 5] :. U = [2 4] [0 5] Now, use Gaussian elimination with 1st column, we get [1 0] = [-1 1] [1 0] L= [-1 1] Hence, [2 4] [1 0] [2 4] A = [-2 1] = [-1 1] [0 5] = LU This is the LU factorization of the given matrix. Question 4 Which of the following matrices is an elementary matrix? [0 3] a) [1 0] [2 0] b) [0 1] [1 0] c) [0 1] [2 0] Use code with caution. d) [0 2] Answer Step 1 Since your question has multiple subparts, we have solved the first three subparts for you. Identify the elementary matrices An elementary matrix is a matrix that differs from the identity matrix by one single elementary row operation. a) [0 3] [1 0] Apply minimum elementary row operations on the identity matrix to form this matrix. [1 0] R₁ ↔ R₂ [0 1] [0 1] [1 0] [0 1] R₁ → (1/3)R₁ [0 3] [1 0] [1 0] Since the matrix is formed by applying two elementary row operations, therefore it is not an elementary matrix. Step 2 Similarly check other matrices [2 0] b) [0 1] [1 0] R₁ ↔ R₁ [1 0] [0 1] [0 1] Since [ 1 0] differs the identity matrix by one elementary operation, hence it is an [0 1] elementary matrix. Step 3 Similarly check other matrices [1 0] c) [0 2] [1 0] [0 1] R₂ → 2R₂ [1 0] [0 2] Since [1 0] differs the identity matrix by one elementary operation, hence it is an [0 2] elementary matrix. Question 5 Find a basis for the eigenspace corresponding to each listed eigenvalue of A below. [1 0] A = [-1 2] λ = 2,1 Answer Given matrix A is [1 0] A = [-1 2] For its eigenvalue λ = 2, 1 Next find the eigenvector corresponding to each eigenvalue. For λ = 2 ⇒ Ax = λx ⇒ [1 0] [x₁] = 2 [x₁] [-1 2] [x₂] [x₂] ⇒ x₁ = 2x₁ -x₁ + 2x₂ = 2x₂ ⇒ -x₁ = 0 ⇒ x₁ = 0 ⇒ x₂ = x₂ [0] ⇒ x = [1] For λ = 1 ⇒ Ax = λx ⇒ [1 0] [x₁] = 1 [x₁] [-1 2] [x₂] [x₂] ⇒ x₁ = x₁ -x₁ + 2x₂ = x₂ ⇒ x₁ = x₂ x = [1] [1] is a basis for the eigenspace corresponding to λ = 1