Assignment #1, Problem 5
Contents
Solving linear system (i)
Gaussian elimination fails, it does not give matrix U with nonzero diagonal elements. Hence there is either no solution or infinitely many solutions.
Note that the backslash command in machine arithmetic gives a warning "matrix is singular to working precision". But it does not return any solution (despite the fact that there are solutions).
Using symbolic computation in Matlab we obtain a particular solution xpart, and a basis V for the null space (one vector). Hence the solution is given by a line xpart+t*V with arbitrary t.
A = [2 -1 -1; -1 2 -1; -1 -1 2] [L,U,p] = lu(A,'vector'); U % matrix U from Gaussian elimination b = [1;-3;2] % Using Matlab with machine arithmetic x = A\b % Matlab gives NaN (not a number) % Using symbolic computation: xpart = sym(A)\sym(b) % symbolic \ gives one solution and % warning that more solutions exist V = null(sym(A)) % basis for null space
A = 2 -1 -1 -1 2 -1 -1 -1 2 U = 2 -1 -1 0 1.5 -1.5 0 0 0 b = 1 -3 2 Warning: Matrix is singular to working precision. x = NaN NaN NaN Warning: The system is rank-deficient. Solution is not unique. xpart = -1/3 -5/3 0 V = 1 1 1
Plotting planes for linear system (i)
Here the three planes have an intersection which is a line, corresponding to the solutions xpart+t*V we found before.
T1 = diag( b(1)./A(1,:) ) % three points satisfying eq.1 T2 = diag( b(2)./A(2,:) ) % three points satisfying eq.2 T3 = diag( b(3)./A(3,:) ) % three points satisfying eq.3 fillpoints(stretch(T1,9),'y'); hold on % plot plane for eq.1 fillpoints(stretch(T2,3),'g') % plot plane for eq.2 fillpoints(stretch(T3,5),'c'); % plot plane for eq.3 plotpoints(xpart-3*V,xpart+4*V,'r','LineWidth',3); % plot line of solutions hold off nice3d; alpha(0.8); view(24,10); title('red line is solution set')
T1 = 0.5 0 0 0 -1 0 0 0 -1 T2 = 3 0 0 0 -1.5 0 0 0 3 T3 = -2 0 0 0 -2 0 0 0 1

Solving linear system (ii)
Gaussian elimination gives a matrix U with nonzero diagonal elements, hence there is a unique solution.
A = [2 -1 -1; -1 4 -1; -1 -1 2] [L,U,p] = lu(A,'vector'); U % matrix U from Gaussian elimination b = [1;-3;2] x = A\b % unique solution
A = 2 -1 -1 -1 4 -1 -1 -1 2 U = 2 -1 -1 0 3.5 -1.5 0 0 0.85714 b = 1 -3 2 x = 1.3333 1.1869e-16 1.6667
Plotting planes for linear system (ii)
The three planes intersect in a single point x which is the solution of the linear system we found before.
T1 = diag( b(1)./A(1,:) ) % three points satisfying eq.1 T2 = diag( b(2)./A(2,:) ) % three points satisfying eq.2 T3 = diag( b(3)./A(3,:) ) % three points satisfying eq.3 fillpoints(stretch(T1,9),'y'); hold on % plot plane for eq.1 fillpoints(stretch(T2,3),'g') % plot plane for eq.2 fillpoints(stretch(T3,5),'c'); % plot plane for eq.3 plotpoints(x,'o'); label(x,'x') % plot solution point hold off nice3d; alpha(0.75); view(-135,20); title('Three planes intersect at solution point x')
T1 = 0.5 0 0 0 -1 0 0 0 -1 T2 = 3 0 0 0 -0.75 0 0 0 3 T3 = -2 0 0 0 -2 0 0 0 1
