Exercise 6 Is the set of vectors V = {i² + 2t – 1,3t + 5, -t² + t + 6} linearly dependent?
Exercise 6 Is the set of vectors V = {i² + 2t – 1,3t + 5, -t² + t + 6} linearly dependent?
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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Help with exercise 6 please?
![Exercise 6
Is the set of vectors
V = {r² + 2t – 1,3t + 5, –t² + t + 6}
linearly dependent?
Example 6
What about this set of vectors?
O = {t² + 2t + 1, t? - t +3,3t +1}
Is it linearly dependent? Let's check. First, set it as a linear combination of the vectors set equal to the zero vector, .
a (t? + 2t + 1) +b(t? - t+3) + c(3t + 1) =
(a + b)t? + (2a – b+3c)t + (a + 3b +c)
= 0
Again, make sure you see how this works algebraically. This yields the following system of equations, augmented matrix and resulting
RREF matrix.
a +6 = 0
2a - b+ 3c =
a + 36 +c =
1
0: 0
1 0 0: 0
RREF,
2
-1 3: 0
0 10: 0
1
3
1
0 0 1
Remember from previous tutorials that this means that this system only has one solution. That solution is (a, b, c) = (0,0,0). This
is the trivial solution! Because this is the only solution, we do NOT say that the set of vectors 0 is linearly dependent. In fact,
was say that the set of vectors O is linearly independent. Going back to the top of the tutorial, this also means that none of
the vectors in 0 can be written as a linearly combination of the others. That makes them independent, yes? They cannot be
written as a linear combination of the others. Or, they are not defined by the other vectors. They are their own vector. Alone.
Independent!
What you should notice about all of these Examples and Exercises is that when you do the row-reduction procedure the right
column stays full of zeros. It's always just zeros. For that reason you can almost entirely ignore it. Then, the matter of linear
dependence boils down to what the A part of the matrix reduces to. Just check the solutions. Is it just all zeros? Linearly
Independent. Is it anything other than all zeros? Li nearly Dependent.](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F2c762e73-cba5-4fbf-869f-39b9ab2042ca%2Fcfbff989-64f3-4545-b97b-74ef820e72d9%2Fv9e0mdt_processed.png&w=3840&q=75)
Transcribed Image Text:Exercise 6
Is the set of vectors
V = {r² + 2t – 1,3t + 5, –t² + t + 6}
linearly dependent?
Example 6
What about this set of vectors?
O = {t² + 2t + 1, t? - t +3,3t +1}
Is it linearly dependent? Let's check. First, set it as a linear combination of the vectors set equal to the zero vector, .
a (t? + 2t + 1) +b(t? - t+3) + c(3t + 1) =
(a + b)t? + (2a – b+3c)t + (a + 3b +c)
= 0
Again, make sure you see how this works algebraically. This yields the following system of equations, augmented matrix and resulting
RREF matrix.
a +6 = 0
2a - b+ 3c =
a + 36 +c =
1
0: 0
1 0 0: 0
RREF,
2
-1 3: 0
0 10: 0
1
3
1
0 0 1
Remember from previous tutorials that this means that this system only has one solution. That solution is (a, b, c) = (0,0,0). This
is the trivial solution! Because this is the only solution, we do NOT say that the set of vectors 0 is linearly dependent. In fact,
was say that the set of vectors O is linearly independent. Going back to the top of the tutorial, this also means that none of
the vectors in 0 can be written as a linearly combination of the others. That makes them independent, yes? They cannot be
written as a linear combination of the others. Or, they are not defined by the other vectors. They are their own vector. Alone.
Independent!
What you should notice about all of these Examples and Exercises is that when you do the row-reduction procedure the right
column stays full of zeros. It's always just zeros. For that reason you can almost entirely ignore it. Then, the matter of linear
dependence boils down to what the A part of the matrix reduces to. Just check the solutions. Is it just all zeros? Linearly
Independent. Is it anything other than all zeros? Li nearly Dependent.
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