-1 4 Let A= and b= by Show that the equation Ax = b does not have a solution for some choices of b, and describe the set of allb for which Ax =b does have a solution. 3 - 12

College Algebra (MindTap Course List)
12th Edition
ISBN:9781305652231
Author:R. David Gustafson, Jeff Hughes
Publisher:R. David Gustafson, Jeff Hughes
Chapter6: Linear Systems
Section6.2: Guassian Elimination And Matrix Methods
Problem 78E
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-1
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.
3 - 12
How can it be shown that the equation Ax = b does not have a solution for some choices of b?
A. Row reduce the matrix A to demonstrate that A does not have a pivot position in every row.
O B. Find a vector b for which the solution to Ax = b is the identity vector.
OC. Row reduce the augmented matrix [ A b]to demonstrate that [ A b] has a pivot position in every row.
OD. Row reduce the matrix A to demonstrate that A has a pivot position in every row.
OE. Find a vector x for which Ax = bis the identity vector.
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, + bz.
(Type an integer or a decimal.)
Transcribed Image Text:-1 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. 3 - 12 How can it be shown that the equation Ax = b does not have a solution for some choices of b? A. Row reduce the matrix A to demonstrate that A does not have a pivot position in every row. O B. Find a vector b for which the solution to Ax = b is the identity vector. OC. Row reduce the augmented matrix [ A b]to demonstrate that [ A b] has a pivot position in every row. OD. Row reduce the matrix A to demonstrate that A has a pivot position in every row. OE. Find a vector x for which Ax = bis the identity vector. 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, + bz. (Type an integer or a decimal.)
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