8.4 Exercise 3 3.1) Enter the matrix A2= 0 -1 -8 -10 Find a basis for null(A2), you may use the ones from the previous section or compute it again. Call these vectors N1 and N2. Recall that any linear combination of the basis vectors will be in the subspace. That means kl * N1 + 2 * NV2 should be in the null-space of A2 no matter what kl and k2 are. so A2(kl * N1 + k2 * N2) = 0. We can demonstrate this with the following commands: for i = 1:10, kl = round(10*rand(1)); k2 = -round(10*rand(1)); V(:,i) = kl*N1 + k2*N2; AV(:,i) = A2*V(:,i); end Explanation: The rand function gives numbers between zero and 1 so multiplying by 10 and rounding off gives integers between zero and 10. Thus kl and k2 are randomly chosen numbers. Each column of the matrix V is a vector which is a linear combination of our basis vectors N1 and N2. The corresponding column in AV is the vector A2*(that column of V). Claim AV should be a matrix of zeros. Type: A2V= AV to check what the loop did. Wow! It works! All of the columns of V are in the null-space of A2! 3.2) Which of the following vectors are in the null-space of A2? ( you need to enter these vectors in MATLAB as matrices, column vectors). . v1= (5, -35, 81, 2) . v2- (38, 101, 17, -24) . v3= (-9, -26, -3, 5) . v1= (1, 0, 1, 1) Use MATLAB, then type the following and enter your answer in the following format: If you decided v1 is in the null-space of A2 type : % A2v1= is in null(A2) otherwise type: % A2v1= is NOT in null(A2) . ( note the answers are case sensitive.)