Alg1231W4T2 - Linear dependence
Thursday, 16 August 2018
10:53 PM
Created with Microsoft OneNote 2016.
![Untitled picture.png Machine generated alternative text:
45
0
-15
- -16
-10
20
15
—5
10
15
Using the app above and experimenting with the sliders, find integers r and s such that
in each of the cases below:
Untitled picture.png Machine generated alternative text:
12
ii) u
—2
and v
3
2
and v
-3
1
and w
—3
-5
and w
—1
—4
3
11
-8
In this case we need
O and s
. In this case we need
3
O and s
Now, find scalars r, s and t (not all zero) such that
if u
2
and v
—1
4
and w
3
32
14
In this case we could use
Untitled picture.png Machine generated alternative text:
You have been asked to find out if the set B
VI ¯
{VI, v2, v3} of vectors is linearly independent, where
1
—1
2
1
1
1
—1
—4
and if not, find a linear relation between them. Recall that this really amounts to seeing if you can find solutions to the system
1
—1
2
1
1
—1
—4
which are not all zero. Of course y z 0
is a solution, but it doesn't count
Fortunately, you have mastered the delicate art of row reduction from MATH1131 , so you can readily deduce that a solution in fact does exist, namely
Note: enter you answer using the syntax [1,2 , 3] .
This means the set B {VI , v2, v3} is a linearly dependent set
An example of a linear relation between these vectors is
13
Ink Drawings






](Alg1231W4T2%20-%20Linear%20dependence_files/image005.png)
![Untitled picture.png Machine generated alternative text:
You have been asked to find out if the set B
VI ¯
{VI, v2, v3} of vectors is linearly independent, where
1
—1
2
1
1
1
—1
—4
and if not, find a linear relation between them. Recall that this really amounts to seeing if you can find solutions to the system
1
—1
2
1
1
—1
—4
which are not all zero. Of course y z 0
is a solution, but it doesn't count
Fortunately, you have mastered the delicate art of row reduction from MATH1131 , so you can readily deduce that a solution in fact does exist, namely
Note: enter you answer using the syntax [1,2 , 3] .
This means the set B {VI , v2, v3} is a linearly dependent set
An example of a linear relation between these vectors is
13
Untitled picture.png Machine generated alternative text:
The set C
, where
is a linearly dependent set. We can show this by using row reduction to find a non-zero solution to the system
For example
And, in this case, all non-zero solutions are just scalar multiples of this one.
Interestingly [select all that apply]:
we can write VI as a linear combination of v2, v3 and v 4
we can write v2 as a linear combination of VI , and v 4
O we can write v3 as a linear combination of VI , v2 and v 4
we can write v 4 as a linear combination of VI, v2 and v3.


](Alg1231W4T2%20-%20Linear%20dependence_files/image007.png)
![Untitled picture.png Machine generated alternative text:
When considering the span of a set of vectors, a vector in the set which can be expressed as a linear combination of the others in the set can be
considered redundant, in the sense that removing it will not affect the span.
From the our working above we can conclude that [select all that apply]:
span{v2, v3, v 4}
span{V1 v3'
span{V1 v2' "'4 }
span{V1 v2' v3}•
Untitled picture.png Machine generated alternative text:
cosh (x)
12 —10
-2
—8
—6
-4
-6
4
-2
—6
2
4
6
a = 1.52
8
10
Forget about face recognition, James. The latest SPECTRE internet password requires you to combine functions in the infinite dimensional vector space of
, coshx, and sinhæ together with
functions to obtain a "0-key" to crack into their evil web-portal. The pass key app here shows the functions e ,e
the linear key combination
f@) = aec + be +ccoshx +dsinhx.
Your mission, if you choose to accept it, is to find some combination of coefficients that will:
i) ensure that the function
aec + be¯x + ccoshx + dsinhx
is the zero function (or at least very close to it), and
ii) at least one of a, b, c, d is exactly equal to 1.](Alg1231W4T2%20-%20Linear%20dependence_files/image009.png)
