Mark each statement True or Falls, and justify youranswer. (a) If A is 3×5 matri

Question

Mark each statement True or Falls, and justify youranswer. (a) If A is 3×5 matrix and T is a transformation definedby T(X) = AX, then the domain of T is R3 . (b) A transformation T is linear if and only if T(0) = 0 (c) If there is a b in Rn such that the equation AX = b isinconsistent, then the transformation T : X ! AX is not one-to-one. (d) An example of the linear combination vectors v1 and v2 isthe vector 5 v1 . (e) Asking whether the linear system corresponding to anaugmented matrix [a1 a2 a3 b] has solution amounts to asking whether bis in span{a1, a2, a3} . (f) If S is a linearly dependent set, the each vector is alinear combination of the other vectors in S . Mark each statement True or Falls, and justify youranswer. (a) If A is 3×5 matrix and T is a transformation definedby T(X) = AX, then the domain of T is R3 . (b) A transformation T is linear if and only if T(0) = 0 (c) If there is a b in Rn such that the equation AX = b isinconsistent, then the transformation T : X ! AX is not one-to-one. (d) An example of the linear combination vectors v1 and v2 isthe vector 5 v1 . (e) Asking whether the linear system corresponding to anaugmented matrix [a1 a2 a3 b] has solution amounts to asking whether bis in span{a1, a2, a3} . (f) If S is a linearly dependent set, the each vector is alinear combination of the other vectors in S .

Explanation / Answer

a) True. You can identify the domain of atransformation by seeing how many rows it's associated matrixhas. In this case, it is 3. b) True. If I remember right there's a theoremthat says a transformation T is only linear if the zero vector ismapped to zero, but I do remember proving on a test that thetransformation T(x)= Ax + e1 is not linear because itdoesn't map the zero vector to zero. c) True. If the equation Ax=b is inconsistent then thereis not a pivot that corresponds to each unknown. That is,there would exist a free variable in the associated matrix. Because of this, T could not be 1-1. d) True. 5v1 is a linear combination ofv1 and v2 with scalars 5 for v1and 0 for v2. e) True. If b is in the span of the columns of a then itcan be written as a linear combination of those columns. Therefore there exists atleast one solution. f) True. If S is a linearly dependent set theneach vector can be written as a linear combination of the othervectors in the set.

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