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The command rref ( is ) returns the reduced row echelon form of a matrix A 1 The command , pivcols ] = rref (

The command rref (is) returns the reduced row echelon form of a matrix A1
The command , pivcols]=rref (A) returns B and pivcols, where matrix B=rref(A) and pivcols indicates pivot
columns of A2
The command rank (A) returns the rank of a matrix A2
You also need two m-fle functions rank_comp (given) and Ls_solut ion (template only). In these functions, you need to know:
if-else statement
If an expression is true, it executes statements 1 block. Otherwise, it executes statements 2 block.
if-elseif statement
If expression 1 is true, it executes statements 1 block. If not, check expression 2. If expression 2 is true, it executes statements 2 block.
If not, execute statements 3 block.
if expression 1
statements 1 block
elseif expression 2
statements 2 block
eise
statements 3 block
xa'
following in the Live Editor.
Display the reduced row echelon form and the pivot columns of the augmented matrix Alb.
Write a report to explain if there is a solution of x=b based on rref [(,b]).
B. Use the Rouch-Capelli Theorem by comparing the rank() and rank({:[ib]).
Call the function rank_comp and determine if x=b is consistent.
Compare the result with Part AL. You should have the same conclusion.
Note: rank_comp gives you a new command to compare if rank(1)=rank([1b]). When you need to use the command, read the
the two input matrices.
C. Open the template Ls_solut ion. Use the Rouch-Capelli Theorem as a guide and code the function using an if-elseif statement ido
not use nested if-else statements). When Ls_solut ion is called it should return the correct system type based on whether x=b
has a solution and how many. The function should output/return/set systern_type to:
inc ir the system is inconsistent and has no solution, or
con_with_one_sol if the system is consistent and has a unique solution, or
con_with_inf_sols if the system is consistent and has infitely many solutions.
Note: You should include three input arguments ) for Ls_solut ion, where n is the number of variables in the system of
equations. Code I have so far:%--- Part A [10 Points]---%
Ab =[A, b]; %(WRITE COMMAND BEFORE SEMI-COLON)
%(COMMENTS FOR B AND pivcols)
[B, pivotcol]= rref(Ab);
disp('Reduced Row Echelon Form (RREF):');
disp(B);
disp('Pivot Columns:');
disp(pivotcol);
%(EXPLAIN WHY SYSTEM IS INCONSISTENT)
%--- Part B [10 Points]---%
rank_A = rank(A);
rank_Ab = rank(Ab);
rank_comp(rank_A, rank_Ab);
if rank_A == rank_Ab
disp('The system Ax = b is consistent.');
if rank_A == n
disp('The system has a unique solution.');
else
disp('The system has infinitely many solutions.');
end
else
disp('The system Ax = b is inconsistent.');
end
%(STATE THEOREM)
%(APPLY/USE THEOREM & COMPARE TO RESULT FROM PART A)
%--- Part C [10 Points]---%
LS_solution(n, A, Ab);
[m, n]= size(E); %(UNCOMMENT LINE AND COMPLETE)
solution_type =NaN;
end I am getting an error that there is no variable n as it is called in the function. How would I implement n? Here is the function I wrote. LS_solution(n, A, Ab)%(PURPOSE OF FUNCTION)
% n = number of variables
% A = coefficient matrix
% Ab = augmented matrix [A b]
inc = "Inconsistent";
con_with_one_sol = "Consistent with One Solution";
con_with_inf_sols = "Consistent with Infinite Solutions";
rank_A = rank(A);
rank_Ab = rank(Ab);
% Add your code below
if rank_A rank_Ab
system_type = inc;
elseif rank_A == rank_Ab
if rank_A == n
system_type = con_with_one_sol;
else
system_type = con_with_inf_sols;
end I need help with the parts under the comments and anything that says NaN or to explain something. In essence, if it has a comment then I need help.
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