Let A = 1 - 4 2 -2 - 1 4 0 0 2 b₁ and b = b₂ Show that the equation Ax=b does not have a solution for all possible b b3 and describe the set of all b for which Ax=b does have a solution.

College Algebra (MindTap Course List)
12th Edition
ISBN:9781305652231
Author:R. David Gustafson, Jeff Hughes
Publisher:R. David Gustafson, Jeff Hughes
Chapter6: Linear Systems
Section6.3: Matrix Algebra
Problem 71E
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1 -2 - 1
4-8
- 4
2 0 2
and describe the set of all b for which Ax=b does have a solution.
Let A =
b₁
0 and b = b₂ Show that the equation Ax=b does not have a solution for all possible b
b3
How can it be shown that the equation Ax=b does not have a solution for all possible b? Choose the correct answer
below.
O A. Row reduce the augmented matrix [ A b] to demonstrate that [ A b] has a pivot position in every row.
B. Row reduce the matrix A to demonstrate that A does not have a pivot position in every row.
C. Row reduce the matrix A to demonstrate that A has a pivot position in every row.
D. Find a vector x for which Ax=b is the zero vector.
O E. Find a vector b for which the solution to Ax=b is the zero vector.
Describe the set of all b for which Ax=b does have a solution.
0 = 2b₁ +2b₂ + b3
(Type an expression using b₁,b₂, and b3 as the variables and 1 as the coefficient of b3.)
Transcribed Image Text:1 -2 - 1 4-8 - 4 2 0 2 and describe the set of all b for which Ax=b does have a solution. Let A = b₁ 0 and b = b₂ Show that the equation Ax=b does not have a solution for all possible b b3 How can it be shown that the equation Ax=b does not have a solution for all possible b? Choose the correct answer below. O A. Row reduce the augmented matrix [ A b] to demonstrate that [ A b] has a pivot position in every row. B. Row reduce the matrix A to demonstrate that A does not have a pivot position in every row. C. Row reduce the matrix A to demonstrate that A has a pivot position in every row. D. Find a vector x for which Ax=b is the zero vector. O E. Find a vector b for which the solution to Ax=b is the zero vector. Describe the set of all b for which Ax=b does have a solution. 0 = 2b₁ +2b₂ + b3 (Type an expression using b₁,b₂, and b3 as the variables and 1 as the coefficient of b3.)
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