00:01
We are going to find the values of age for which the vectors are linearly dependent in each exercise.
00:08
We have four exercise where we get to find the value or values of age.
00:13
For which the vectors, the given vectors, are linearly dependent.
00:19
So let's start with 11.
00:22
So the statement here is that the vectors are linearly dependent if we can find linear combinations of the given vectors, which is equal to zero with coefficient not all equal to zero.
00:40
In that case, we say the vectors are linearly dependent.
00:43
If a linear combination of the vector equal to zero implies that all the coefficient has to be equal to zero, then the vectors are said to be linearly independent.
00:55
So we got to establish a linear combination of the vectors like this, b times the second vector, plus c times its third vector equal to zero the zero vector.
01:21
In this case, we are leading with vectors in r3, that is with three components.
01:29
So this implies, if we do all calculations here, that is each scalar times the vector and add the three vectors, we will get the following.
01:44
We will get a times 1, that is a plus 3 times 3.
01:47
V minus c is the first component of the resultant vector.
01:53
Then negative a minus 5b plus 5c is the second component.
02:00
The last component will be for a plus 7b plus c or hc better.
02:11
It's the same but we put the coefficient first and then the unknown variable or coefficient equal to 0 and then all the entries of these vector here get b0 so this implies that we got to fulfill three equations three linear equations a plus 3b minus c equals 0 minus a minus 5b plus 5c equals 0 and 4a plus 7b plus h and 4a plus 7b plus h c equals zero and this is a linear system of three equations with three unknowns a b and c and we can write that system of equations in matrix form as the matrix is the matrix of constant coefficients is 1 3 negative 1 negative 1 negative 5 5 and 4 7 h times the unknown vector a, b, c, that's equal to 0, 0 .0.
03:31
And we can see that this matrix, which we call big a, is just a matrix form by putting in each column the given vectors.
03:42
The first column, 1, negative 1, 4 is the first vector here.
03:47
3, negative 5, 7, second column is the second vector, and the last vector is the third column, negative 1 ,5, each.
03:54
So we can construct the matrix directly by looking at the vectors.
04:00
In the next exercise, we will do that way.
04:04
But here we have a system, and then this is an homogeneous system, and it will have only the zero solution that is the coefficient of say b and c have to be zero if this metric is invariable, and this matrix has an inverse if the determinant is different from zero.
04:23
So we want precisely just the values of age for which this determinant or the determinant of the matrix is zero.
04:32
In that case, the system will have certainly solutions different from the zero or all zero scalars abc.
04:46
So if we calculate the determinant of this matrix, we find that doing all the calculations here, we get 12 minus 2h.
05:07
That's normal because h is in this is a component of this mattress.
05:13
It's the element 3 -3.
05:15
And then we do all calculations.
05:16
We get this.
05:17
So the determinant of a is zero, which is the case we want.
05:22
That is the case where the system can have no zero solutions.
05:29
It's equivalent to 12 minus 2h equal to 0, which is equivalent to h equals 6.
05:36
And that's the only solution of this equation determinant of a equals 0.
05:42
So the only value for which the three vectors are linearly dependent is h equals 0.
05:53
So the vectors are linearly dependent if and only if age equals 6.
06:10
Now we can do 12.
06:15
And we do the same.
06:17
In this case, we're going to construct the system directly.
06:20
System is 2, negative 4 .1 is the first column, the first vector.
06:30
Second vector, negative 6, 7, negative 3.
06:40
And the third vector is 8h4.
06:48
That times the coefficients, which we understand are the scalar that are multiplying each vector, equal to 0.
07:00
And in this case, i'm going to calculate the determinant of this matrix.
07:07
I'm going to call it big a again.
07:09
The determinant of big a is two times the determinant of 7h, negative 3, 4, plus 6 determinant of negative 4h14, plus 8 times the determinant of negative 4h14, 4 ,000, plus 8 times the determinant of, negative 4, 7, 1, negative 3.
07:45
And so the determinant of a is 2 times 12 plus 6h plus 6 times negative 24, sorry, negative 16, 4 times 4 times 4, negative 16, minus each plus 8 times 12 minus 7.
08:23
So, the term of a is two times, sorry, this is two in the parentheses, and we get 24 plus 12 h, minus 96 minus 6 age.
08:52
Okay, i made a mistake, sorry.
08:55
I think i, something wrong here is three.
09:00
Boss is 12.
09:07
Okay.
09:09
So i'm going to rewind it a bit.
09:13
I made a mistake here.
09:17
It's correct this.
09:18
So this is correct because it's 2 times 7 -8 -neged -neged 3 -4.
09:26
Then we have 6, which in sign, negative 4 -h14, and 8 times negative 4 -7 -1 -9 -3.
09:36
That's correct.
09:37
And now this part was incorrect.
09:41
It's 2 times, and now it's 7 times 4 is 28 minus, minus, which become plus because we have a negative 3, 3h.
09:53
Okay, that was completely wrong before.
09:56
That's the correct number.
09:59
Plus 6 times negative 16 minus age, plus 8 times 12.
10:08
Here is positive because it's negative 4 times negative 3 minus 7 times 1 is 7.
10:15
Now that's the correct result.
10:18
And we have two times or better.
10:22
Two times 28 is 56 plus 6h minus 96 minus 6h minus 6h plus 96 or better.
10:42
No, it's okay.
10:44
96 minus 56...