1 2 and b = Let A= 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 solution, -3 -6 b2 How can it be shown that the equation Ax = b does not have a solution for some choices of b? O A. Find a vector b for which the solution to Ax = b is the identity vector. O B. Row reduce the matrix A to demonstrate that A does not have a pivot position in every row. O C. Row reduce the augmented matrix Ab to demonstrate that Ab has a pivot position in every row. O D. Row reduce the matrix A to demonstrate that A has a pivot position in every row. O E. Find a vector x for which Ax = b is 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, + b2.

1 2 and b = Let A= 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 solution, -3 -6 b2 How can it be shown that the equation Ax = b does not have a solution for some choices of b? O A. Find a vector b for which the solution to Ax = b is the identity vector. O B. Row reduce the matrix A to demonstrate that A does not have a pivot position in every row. O C. Row reduce the augmented matrix Ab to demonstrate that Ab has a pivot position in every row. O D. Row reduce the matrix A to demonstrate that A has a pivot position in every row. O E. Find a vector x for which Ax = b is 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, + b2.

Advanced Engineering Mathematics

10th Edition

ISBN:9780470458365

Author:Erwin Kreyszig

Publisher:Erwin Kreyszig

Chapter2: Second-order Linear Odes

Section: Chapter Questions

Problem 1RQ

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Concept explainers

Linear Algebra

In mathematics, problems can be solved by the arithmetic operators like addition, subtraction, multiplication, and division. In algebra, we represent the numbers by any letter called a variable. If these variables are of degree 1, then it is studied under linear algebra. The first-degree equations involving algebraic expressions are called linear equations. That is, if you represent it by drawing a graph you will get a straight line.

Question

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.
b2
-3 -6
How can it be shown that the equation Ax = b does not have a solution for some choices of b?
O A. Find a vector b for which the solution to Ax = b is the identity vector.
O B. Row reduce the matrix A to demonstrate that A does not have a pivot position in every row.
O C. Row reduce the augmented matrix
Ab to demonstrate that Ab
has a pivot position in every row.
O D. Row reduce the matrix A to demonstrate that A has a pivot position in every row.
O E. Find a vector x for which Ax = b is 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, + b2-

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Let A = 3 - 3 12 - 12 does have a solution. and b = b2 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 How can it be shown that the equation Ax = b does not have a solution for some choices of b? A. Find a vector x for which Ax=b is the identity vector. B. Row reduce the augmented matrix A b to demonstrate that [a b] has a pivot position in every row. C. Find a vector b for which the solution to Ax=b is the identity vector. D. Row reduce the matrix A to demonstrate that A does not have a pivot position in every row. E. Row reduce the matrix A to demonstrate that A 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 = (Type an integer or a decimal.) b₁ + b₂.
Can each vector in R* be written as a linear combination of the columns of the matrix A? Do the columns of A span R*? 3 1 7 - 12 - 1 - 1 A = - 4 - 2 2 0 - 10 20 0 2 - 1 Can each vector in R be written as a linear combination of the columns of the matrix A? Select the correct choice below and fill in the answer box to complete your choice, (Type an integer or decimal for each matrix element.) O A. No, because the reduced echelon form of A is O B. Yes, because the reduced echelon form of A is Help me solve this View an example Get more help - Clear all Check answer 82 9+ G 99+ W ENG DELL DII F3 prt sc F1 F4 FS beme end F7 F9 insert dele F10 F11 F12 %23 %2$ & 2 3 4 5 6 8
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The vectors -3 -3 -9 w 3= 12 20 + k 15 are linearly independent if and only if k +
solve question with complete explanation asap and get multiple upvotes
Let A = -6-30 -8 30] We want to determine if the columns of matrix A and are linearly independent. To do that we row reduce A. times the first row to the second. To do this we add We conclude that A. The columns of A are linearly independent. B. The columns of A are linearly dependent. C. We cannot tell if the columns of A are linearly independent or not.