University:
Massachusetts Institute of Technology
Course:
18.06SC | Linear Algebra
Academic year:

2023
Views:

414

Pages:

2
Author:

Baylee Morrison

Exercises on solving Ax = b and row reduced form R Problem 8.1: (3.4 #13.(a,b,d) Introduction to Linear Algebra: Strang) Ex plain why these are all false: a) The complete solution is any linear combination of x p and xn . b) The system Ax = b has at most one particular solution. c) If A is invertible there is no solution xn in the nullspace. Solution: a) The coefﬁcient of x p must be one. b) If xn ∈ N( A) is in the nullspace of A and x p is one particular solution, then x p + xn is also a particular solution. c) There’s always xn = 0. Problem 8.2: (3.4 #28.) Let � � � � 1 2 3 5 U= and c = . 0 0 4 8 Use Gauss-Jordan elimination to reduce the matrices [U 0] and [U c] to [ R 0] and [ R d]. Solve Rx = 0 and Rx = d. Check your work by plugging your values into the equations Ux = 0 and Ux = c. First we transform [U 0] into [ R 0]: � � � � � � 1 2 3 0 1 2 3 0 1 2 0 0 [U 0 ] = −→ −→ = [ R 0]. 0 0 4 0 0 0 1 0 0 0 1 0 Solution: We now solve Rx = 0 via back substitution: ⎡ ⎤ ⎡ ⎤ � � x1 � � � � 2 1 2 0 ⎣ 0 x1 + 2x2 = 0 x2 ⎦ = −→ −→ x = ⎣ −1 ⎦ , 0 0 1 0 x3 = 0 x3 0 where we used the free variable x2 = −1. (cx is a solution for all c.) We check that this is a correct solution by plugging it into Ux = 0: ⎡ ⎤ � � � � 2 1 2 3 ⎣ 0 ⎦ −1 = � 0 0 4 0 0 Next, we transform [U c] into [ R d]: � � � � � � 1 2 3 5 1 2 3 5 1 2 0 −1 −→ −→ [U c ] = = [ R d ]. 0 0 4 8 0 0 1 2 0 0 1 2 We now solve Rx = d via back substitution: ⎡ ⎤ ⎡ ⎤ � � x1 � � � � −3 1 2 0 ⎣ −1 x1 + 2x2 = −1 x2 ⎦ = −→ −→ x = ⎣ 1 ⎦ , 0 0 1 2 x3 = 2 2 x3 where we used the free variable x2 = 1. Finally, we check that this is the correct solution by plugging it into the equation Ux = c: ⎡ ⎤ � � −3 � � 1 2 3 ⎣ 5 1 ⎦= � 0 0 4 8 2 Problem 8.3: (3.4 #36.) Suppose Ax = b and Cx = b have the same (complete) solutions for every b. Is it true that A = C? Solution: Yes. In order to check that A = C as matrices, it is enough to check that Ay = Cy for all vectors y of the correct size (or just for the standard basis vectors, since multiplication by them “picks out the columns”). So let y be any vector of the correct size, and set b = Ay. Then y is certainly a solution to Ax = b, and so by our hypothesis must also be a solution to Cx = b; in other words, Cy = b = Ay.