H.W. #5 The code below is a working Gaussian elimination function. It returns the solution vector X to a system of linear equations AX=B. You are to create a new function that employs matrix inversion to find the inverse of any size square coefficient matrix A. Use this function to solve the following system:

51 + 22 43 = 10

101 82 + 63 = 51

151 + 52 73 = 33

by finding the inverse of A and then using X=A-1B in the main program. Present both the inverse of A and the solution for X1,X2,X3 as output. Also show that your solution for the inverse of A is correct by showing the result of A -1A.

function X = Gauss_Elim(A,B) % %make sure B is in column form B=B(:); %determine size of coefficient matrix [nr,nc]=size(A); %build Augmented matrix Aug=[A B]; %Perform Gaussian Elimination without any error reducing row-swapping for col = 1:nc-1 %current column value being reduced to zero for row = col+1:nr %current row value being reduced to zero %identify factor to multiply by pivot row factor = -Aug(row,col)/Aug(col,col); %replace whole row with new reduced row Aug(row,:)=Aug(col,:)*factor+Aug(row,:); end end %Reduce Augmented matrix back into A and B A=Aug(:,1:nc); B=Aug(:,nc+1); %Perform back substitution on A and B for i = nr:-1:1 summ=0; for k = (i+1):nr summ=summ+A(i,k)*X(k); end X(i)=(B(i)-summ)/A(i,i); end X=X(:); %convert solution to column form

Also Attach a Pseudocode with this.