5. Make a control flow that calculates the area of a rectangle based on the lengths and widths from your struct, RECTANGLE, using for-end loop (10 times). In doing so, make a condition that if the calculated area is bigger than 80, the area should be changed into its half size. Save the calculated area as a column vector using a concatenation (either using an empty array to store the temporary results or a cat() command). This should a separate M-file and this M-file should be able to call your previously built area() function. 6. Given a matrix-vector notation for a linear system of equations, 1 2 3 41rX11 30 1 3 5 7X250 1-2 5 6x334 2 3 78J LX4 Write a MATLAB code that does "Forward Elimination" and "Back Substitution." The goal of "Forward Elimination" is to transform the coefficient matrix into an Upper Triangular Matrix 25 5 2551 64 8 1110-4.8-1.56 144 12 1 0.7 "Back Substitution" starts with the last equation because it has only one unknown and it can be solved very easily (e.g., [ 0 0 0.7] [x3] = 0.735, hence x3 is 1.050). After all the results are obtained, compare the results with those obtained by Laplacian Expansion and Cramer's rule 5. Make a control flow that calculates the area of a rectangle based on the lengths and widths from your struct, RECTANGLE, using for-end loop (10 times). In doing so, make a condition that if the calculated area is bigger than 80, the area should be changed into its half size. Save the calculated area as a column vector using a concatenation (either using an empty array to store the temporary results or a cat() command). This should a separate M-file and this M-file should be able to call your previously built area() function. 6. Given a matrix-vector notation for a linear system of equations, 1 2 3 41rX11 30 1 3 5 7X250 1-2 5 6x334 2 3 78J LX4 Write a MATLAB code that does "Forward Elimination" and "Back Substitution." The goal of "Forward Elimination" is to transform the coefficient matrix into an Upper Triangular Matrix 25 5 2551 64 8 1110-4.8-1.56 144 12 1 0.7 "Back Substitution" starts with the last equation because it has only one unknown and it can be solved very easily (e.g., [ 0 0 0.7] [x3] = 0.735, hence x3 is 1.050). After all the results are obtained, compare the results with those obtained by Laplacian Expansion and Cramer's rule