Determine elementary matrices E1, E2, E3 of Type III such that E3E2E1A = U with U an upper triangular matrix. The matrix E1 should turn the element in position (2,1) into a 0. Enter this matrix in MATLAB as E1 using commands similar to the ones in Example 1. The matrix E2 should turn the element in position (3,1) into a zero. Enter this matrix in MATLAB as E2. Note that to zero out the entries in column 1, you need to add or subtract a multiple of row 1. Once you have found the matrices E1 and E2, compute the product E2E1A in MATLAB. Use format rat so that the entries will be given as fractions. Based on the result, determine the matrix E3 that turns the element in position (3,2) into a zero. Enter this matrix as E3 in MATLAB and compute U=E3*E2*E1*A.

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Step-by-step explanation:

MATLAB Code:

close all

clear

clc

% Part (a)

format rat

A = [-3 -2 4; -6 4 12; 6 -8 -11]

% Note: For Elementary Row Operation: Ri <- Ri + b*Rj

% E = eye(size(A)); E(i,j) = b;

% Elementary Row Operation: R2 <- R2 - 2*R1

E1 = eye(size(A)); E1(2,1) = -2

% Elementary Row Operation: R3 <- R3 + 2*R1

E2 = eye(size(A)); E2(3,1) = 2

E2E1A = E2*E1*A

% Elementary Row Operation: R3 <- R3 + (12/8)*R2

E3 = eye(size(A)); E3(3,2) = 12/8

U = E3*E2*E1*A

% Part (b)

L = inv(E1)*inv(E2)*inv(E3)

fprintf('Hence, L is a lower triangular matrix with ones on the diagonal.\n\n')

format short

A_LU = A - L*U;

disp('A - LU ='), disp(A_LU)

fprintf('Hence, A = LU.\n')

Kindly check the image below to see the Output.

the photograph is photographer and- in the process- ourselves.

step-by-step explanation:

the answer is a) y= 2 csc [0.5 (x+pi/2)]-1

step-by-step explanation: