00:02
Consider the vectors u equal 3, 2, 1 and v equal negative 1, 3, negative 2.
00:10
In part a we will find a non -zero vector b in the span of u and v, and in part b we find a vector d not in the span of u and v.
00:22
Remember the span of two vectors in this case, two vectors, the span in general of a set of vectors is the vector space of all linear combinations of those vectors.
00:36
So in part a just to find vector b in the span of u and v, which is a non -zero vector, we only need to find a linear combination of u and v that gives us a vector different from zero and that would be a vector b, a non -zero vector in the span of u and v.
01:02
So for example let's take, or not take but we can take b equal to u plus v, that is using scalars 1 and 1 for each vector u and v, the coefficients to the linear combination.
01:32
And so we can see this is 3, 2, 1 plus negative 1, 3, negative 2.
01:44
And that is 3 minus 1 is 2, 2 plus 3 is 5, and 1 plus negative 2 is negative 1.
01:57
So we can take this vector b, that is b equal to 5, negative 1.
02:12
Then b is different from zero vector and b belongs to the span of vectors u and v simply because b is a linear combination of u and v using the scalars 1 and 1.
02:30
And that's it, that's where a is.
02:33
In part b we want to find a vector d that is not in the span of u and v and that's not that easy as for a in the sense that we get to verify something about the properties of the span of u and v.
02:50
So let's say we have alpha and beta real scalars and a, b, and c any vector in r3.
03:20
Or maybe i can say here a vector instead of any, alpha and beta are a real scalar and a, b, c is a vector in r3 such that we have that a, b, and c, the vector in r3 is alpha times u plus beta times c.
03:55
That is we have a linear combination of u and v equal to a vector in r3.
04:02
That's all we have.
04:05
And that is alpha times 3 to 1 plus beta times negative 1, 3, negative 2.
04:19
That is, i'm going to write first the right hand side and i'm going to put it as the left hand side.
04:26
So it's 3 alpha minus beta.
04:30
I'm doing the linear combination here.
04:32
The second component is 2 alpha plus 3 beta and the last component is alpha minus 2 beta.
04:43
And that equal to a vector a, b, c in r3.
04:51
So this means the equality of two vectors, the corresponding coordinates are equal so 3 alpha minus beta is a, 2 alpha plus 3 beta is b, and alpha minus 2 beta equals c.
05:17
So i'm going to take the first and third equations.
05:28
So 3 alpha minus beta is a, that's the first equation here.
05:40
And the third equation i'm going to multiply both sides by 3.
05:44
So we get 3 alpha minus 6 beta equals 3c.
05:52
That is 3 times the third equation here.
06:01
And this one is just the first equation as it is.
06:05
First equation.
06:10
That is we are considering just those two equations with this modification of the third equation, multiplying both sides by 3.
06:18
And then we subtract the, let me put it here, and so subtracting let's say the first minus the second equation.
06:31
The first means this one, the second this one.
06:34
So 3 alpha cancel out, negative beta minus negative 6 beta is the same as 6 beta minus beta is 5 beta, equal a minus 3c.
06:51
From here, beta is equal to a minus 3c over 5.
07:01
We have that...