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- deal with the problem of solving Ax = b when det A = 0.22.Suppose that, for a given matrix A, there is a nonzero vector x such that Ax = 0. Show that there is also a nonzero vector y such that A*y = 0.If the null space of a 5 × 6 matrix A is 4-dimensional, what is the dimension of the row space of A?Find a basis for and the dimension of the null space of the given matrix. (problem 32 on picture)
- If a vector is in the range of a matrix A, then it is also in the null space of A. true of false?If the null space of a 5 × 6 matrix A is 4-dimensional, what is the dimension of the column space of A?In Problems 2 to 4, find out whether the given vectors are dependent or independent; if they are dependent, find a linearly independent subset. Write each of the given vectors as a linear combination of the independent vectors.2. (1, −2, 3), (1, 1, 1), (−2, 1, −4), (3, 0, 5) 3. (0, 1, 1), (−1, 5, 3), (1, 0, 2), (2, −15, 1)4. (3, 5, −1), (1, 4, 2), (−1, 0, 5), (6, 14, 5)
- This question is regarding vectors and matrices, what do I match these 4 parts with when the 5 choices to match with are: It equals 1 It is meaningless It equals 0 It equals 4 It equals the zero vectorIf A is a 7 X 9 matrix with a 2 dimensional null space, what is the rank of A? Could a 6 X 9 matrix have a two dimensional null space?I thought that a matrix spanned Rn if there was a pivot in every row. However, when a matrix is linearly dependent and the last row is made up of zeroes, it could still end up potentially spanning Rn. I'm a little bit confused about how that works because there isn't a pivot in every row in that situation. How does it still span when there's a free variable? For context, I'm looking at Practice Problem 2 in Section 1.3 of the Vector Equations notes in the textbook Linear Algebra and Its Applications, Fifth Edition by David C. Lay, Steven R. Lay, and Judi J. McDonald. If a picture should be needed, here it is.
- Below is the answers to problem 1 and 2, please help with problem 3. Problem 1 Use the svd() function in MATLAB to compute , the rank-1 approximation of . Clearly state what is, rounded to 4 decimal places. Also, compute the root-mean square error (RMSE) between and . Solution: %code %Define matrix A A = [1, 2, 3; 3, 3, 4; 5, 6, 7]; %Compute SVD of A [U, S, V] = svd(A); %Rank-1 approx A1 = U(:,1) * S(1,1) * V(:,1)'; RMSE = sqrt(mean((A(:) - A1(:)).^2)); %Display A1 rounded to 4 decimal places disp(round(A1, 4)); 1.7039 2.0313 2.4935 2.7243 3.2477 3.9867 4.9087 5.8517 7.1832 %Display RMSE disp(RMSE); 0.3257 Problem 2 Use the svd() function in MATLAB to compute , the rank-2 approximation of . Clearly state what is, rounded to 4 decimal places. Also, compute the root-mean square error (RMSE) between and . Which approximation is better, or ? Explain. Solution: %code A = [2, 4, 7; 3, 3, 5; 1, 6, 6]; % Compute SVD of A [U, S, V] = svd(A); % Rank-2…Suppose the columns of A are not independent. How could you find a matrix B so that P = B ( BT B )-1 BT does give the projection onto the column space of A? (The usual formula will fail when AT A is not invertible.)