MAT 343 LAB 3 - MAT 343 MATLAB Assignment#3 Exercise 1...

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MAT 343 MATLAB Assignment #3 % Exercise 1 %Declaring all the matrices as per the exercise >> A = floor(10*rand(4,3)) % generating a random 4x3 matrix A = 8 6 9 9 0 9 1 2 1 9 5 9 >> E1 = eye(4); % generating matrix E1 >> E1([1,3],:) = E1([3,1],:) E1 = 0 0 1 0 0 1 0 0 1 0 0 0 0 0 0 1 >> E2 = eye(4); % generating matrix E2 >> E2(2,2) = 5 E2 = 1 0 0 0 0 5 0 0 0 0 1 0 0 0 0 1 >> E3 = eye(4); % generating matrix E3 >> E3(4,1) = 7 E3 = 1 0 0 0 0 1 0 0 0 0 1 0 7 0 0 1
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% Computing the Matrix products % Product of matrix A with E1 >> E1*A ans = 1 2 1 9 0 9 8 6 9 9 5 9 Observations – Clearly by multiplying the matrix A by E1 the rows 3 and 1 are inter changed. This is confirmed by looking at the matrix A before and after the multiplication with matrix E1. % Product of matrix A with E2 E2*A ans = 8 6 9 45 0 45 1 2 1 9 5 9 Observations – Clearly by multiplying the matrix A by E2 the 2 nd row is multiplied by 5. This is confirmed by looking at the second row in the matrix A before and after the multiplication with matrix E1. Earlier the row was [9 0 9] and after the multiplication it is [45 0 45]. % Product of matrix A with E3
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Unformatted text preview: E3*A ans = 8 6 9 9 0 9 1 2 1 65 47 72 Observations – Clearly by multiplying the matrix A by E3 the Fourth row is replaced with the sum oF Fourth row and 7 times the frst row. % Exercise 2 % Part A % Declaring the input matrix >> A = [1, -2, 3; 2, -6, 5; -1, -4, 0] A = 1 -2 3 2 -6 5-1 -4 0 % Defining matrix E1 >> E1 = eye(3); >> E1(3,1) = 1 E1 = 1 0 0 1 1 0 1 % Defining matrix E2 >> E2 = eye(3); >> E1(2,1) = -2 E2 = 1 0 -2 1 0 0 1 % Defining matrix E3 >> E3 = eye(3); >> E1(2,1) = -2 E3 = 1 0 0 1 0 -3 1 % Computing the product E3*E2*E1*A to give matrix U >> U = E3*E2*E1*A ans = 1 -2 3 0 -2 -1 0 0 6 Clearly, the above matrix is an upper triangular matrix % Part B % Computing the product L = E1-1 *E2-1 *E3-1 to give matrix U >> L = inv(E1)*inv(E2)*inv(E3) L = 1 0 2 1 -1 3 1 Clearly, the matrix L is a lower triangular matrix with diagonal entries as 1 % Verifying the product A = LU >> L*U ans = 1 -2 3 2 -6 5-1 -4 The result above is same as the matrix A....
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  • Fall '08
  • ringhofer
  • matlab, Matrices, Characteristic polynomial, Diagonal matrix, Triangular matrix

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