% PS1_1.m is the m-file answer for Problem set #1, question 1.
% Written by T. Kenna 9809041140.

% NOTE: Any information appearing after a "%" sign will be ignored as a 
% matlab command it is a good idea to use this feature to make comments 
% about your program so in a few weeks, months, years??, when you want to 
% try and figure out what you did, your comments will be right there. Also

% blank spaces like the one above will also be ignored and will make your 
% programs easier to follow.

% For more information on any matlab command type help and the command. 
% For other ways to search or get help type help help

% Now for the answer

% The display format I am using makes the answers come out in matlab
% as a readable paragraph.  What I found works best when writing text is to 
% define each line of the paragraph as an individual text string. By using 
% the command disp(string1),disp(string2),etc. after you have defined each
% line, you get a well-formatted paragraph without, headers, funny line-wraps
% or spaces. A clc command before the display clears the screen before 
% displaying the answer and a pause command after the display suspends the 
% program so the answer can be read; the program is resumed by striking any 
% key.
 
clc             		% clears the previous screen
ans1=['Start by loading in the data and solving the equation x=A\b.'];
disp(ans1)  			% displays the answer strings
load A.dat               	% use the load command       
load b.dat               	% use it again
x=A\b                    	% solve for x
ans1=['Above is the solution for x, hit any key to see the Determinant and'];
ans2=['the Rank.'];
disp(ans1),disp(ans2)		% displays and centers the answer string
pause           		% suspends program execution.
   
% Now get the determinant and the rank as follows. Remember to leave off 
% the ";" so the m-file will print the values when it runs. <return>'];

clc
Determinant=det(A)       	% use the det() command          
Rank=rank(A)             	% use the rank() command
ans1=['The rank value of 7 says that our matrix is full; and that the'];
ans2=['equations are solvable uniquely (remember 7 equations and 7'];
ans3=['unknowns. Now we calculate the singular values for A by issuing the'];
ans4=['command [U S V]=svd(A,0); <return>'];
disp(ans1),disp(ans2),disp(ans3),disp(ans4)
pause                 

% Now do a singular value decomposition of matrix A by issuing the command
% [U S V]=svd(A,0); remember to use the semicolon unless you want a face full
% of numbers']    

clc
[U S V]=svd(A,0);        	% the svd command.
ans1=['By defining [U S V] as the output of svd(A,0), we end up with 3 new'];
ans2=['variables (see your notes for the definition of each matrix). What we'];
ans3=['are interested in for this exercise is the diagonal of matrix S which'];
ans4=['are the singular values of A. To obtain this, we use the command'];
ans5=['singular_values_of_A=diag(S). The long variable name is used for'];
ans6=['clairity. <return>'];
disp(ans1),disp(ans2),disp(ans3),disp(ans4),disp(ans5),disp(ans6)
pause   

clc
singular_values_of_A=diag(S)     % takes the diagonal of matrix S.
ans1=['The key to the matrix being well behaved is that the range of'];
ans2=['singular values is relatively small (about 1.5 orders of magnitude).'];
ans3=['Although not required, you can obtain the condition of this matrix by'];
ans4=['typing cond(A) or by taking the ratio of the largest to the smallest'];
ans5=['singular values; what do you get? <return>'];
disp(ans1),disp(ans2),disp(ans3),disp(ans4),disp(ans5)
pause

clc
Condition_of_A=cond(A)
ans1=['mmm 22.1819, that is actually not too bad considering that a truly'];
ans2=['ill-conditioned matrix steps towards positive infinity. <return>'];
disp(ans1),disp(ans2)
pause

clc
ans1=['End of Problem 1'];
disp(ans1)