Question

UNIC HaHdheld ull, 130 points is a "perfect" score. Good luck! Part I. True False. Circle T if the statement is always true, and F if the statement is at least sometimes false. [10 pts] 1) T F RREF(A) is unique. 2) T F The solution set to "Ax b" is a vector space. 3) T F Every square matrix is diagonalizable. 4) T F Adding a column to a column of annxn A does not change det(A) 5) T F A least-squares solution u to "Ax-b" is a vector u that minimizes lIAx-bl 6) T F If A has only real eigenvalues, then A is symmetric. 7) T F I If m x n A has rank n, then "Ax b" has exactly one solution. det(AB) = det(BA). F T if m x n A has rank m, then Col(A) = Rm. 9) F T If A is symmetric, then A is orthogonally diagonalizable. F 10) T

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Answer 1

Mathematics Sabyasachi May 21, 2019 Solution . . 1. TRUE. RREF(A) İs unique (Standard result) 2. FALSE. The solution set of Arb is not a vector space Reason: (i) Let x be a solution and c 1 be a constant, then A(cz) = cAx cb b. (ii) Let z and y be solutions, then A(ay) Ar + Ay b+b- 2bf b 3, FALSE. An n × n matrix A is diagonalizable if and only if it has n linearly independent eigenvectors. Consider A= as a real matrix. The characteristic equation of A is: λ2 + 1-0 λ-ì, _ i. Which has no eigenvalue and det(A λ|)= det 0 hence no eigenvector. 4. TRUE. Adding a column to a column of A does not change det A (Standard result) 5. TRUE. A least square solution to Ar - b minimizes ||Ax -b (Standard result) 6. FALSE. The matrix A- has eigenvalues 1,0. But A is not symmetric 7. TRUE. The solution of A2 = b is u +υ, where u is a solution of Ax = 0 and u is the particular solution of Ax = b. Since rank(A) = n, nullity(A) n-n = 0, So, Ax = 0 has only one solution and hence Axbhas only one solution 0v 8, TRUE. det(AB) = det(BA) (Standard result) 9. TRUE. We know if A is an mx n matrix, then col(A) is a subspace of R". Since rank of A is m, so col(A) R 10. TRUE. If A is symmetric then A is orthogonally diagonalizable (Standard result)