following questions:

1 - 4 -5 If T is defined by T(x) = Ax, find a vector x whose image under T is b, and determine whether x is unique. Let A = and b = - 3 5 1 - 10 (. . . Find a single vector x whose image under T is b. X= Is the vector x found in the previous step unique? O A. No, because there is a free variable in the system of equations. O B. Yes, because there is a free variable in the system of equations. O C. No, because there are no free variables in the system of equations. O D. Yes, because there are no free variables in the system of equations.-14 14 18 12 1 1 0 2 - 5 Let b = , and let A be the matrix . ls b in the range of the linear transformation XHAX? Why or why not? - 3 0 1 4 5 7 - 2 0 - 4 O ls b in the range of the linear transformation? Why or why not? 0 A. No, b is not in the range of the linear transformation because the system represented by the appropriate augmented matrix is inconsistent. O B. Yes, b is in the range of the linear transformation because the system represented by the appropriate augmented matrix is consistent. O C. Yes, b is in the range of the linear transformation because the system represented by the appropriate augmented matrix is inconsistent. O D. No, b is not in the range of the linear transformation because the system represented by the appropriate augmented matrix is consistent. by Let A = and b = Show that the equation Ax = b does not have a solution for some choices of b, and describe the set of all b for which Ax = b does have a solution. 9 b 2 . . How can it be shown that the equation Ax = b does not have a solution for some choices of b? O A. Row reduce the matrix A to demonstrate that A has a pivot position in every row. O B. Find a vector x for which Ax = b is the identity vector. O C. Row reduce the matrix A to demonstrate that A does not have a pivot position in every row. O D. Find a vector b for which the solution to Ax = b is the identity vector. O E. Row reduce the augmented matrix | A b to demonstrate that A b has a pivot position in every row. Describe the set of all b for which Ax = b does have a solution. The set of all b for which Ax = b does have a solution is the set of solutions to the equation 0 = b, + b2. (Type an integer or a decimal.)