***** ********** Back substitution ********** &Defining [%] as a vector. x-zeros (1,n); Solving for the nth equation as it has only one unknown. x(n)=RHS1 (n)/A1 (n,n) Solving for the remaining (n-1) unknowns working backwards from *the (n-1) th equation to the first equation. for 1=(n-1):-1:1 Setting the series sum equal to zero. slunm-0 ; for 1-1+1:n summas unm+Al(i))*(j); end Generating the summation term, at which time the unknowns that have been solved for we used to calculate Xi. (1)=(RHS1 (1) -summ) /A1 (1,1) end Question 4: What are the values taken by X matrix? Answer: ***** ********** Back substitution ********** &Defining [%] as a vector. x-zeros (1,n); Solving for the nth equation as it has only one unknown. x(n)=RHS1 (n)/A1 (n,n) Solving for the remaining (n-1) unknowns working backwards from *the (n-1) th equation to the first equation. for 1=(n-1):-1:1 Setting the series sum equal to zero. slunm-0 ; for 1-1+1:n summas unm+Al(i))*(j); end Generating the summation term, at which time the unknowns that have been solved for we used to calculate Xi. (1)=(RHS1 (1) -summ) /A1 (1,1) end Question 4: What are the values taken by X matrix