00:01
Hi, are given here.
00:02
Omega not equal to 1 is a cube root of unity.
00:05
Let's be some of the equations are given in x, y, and z.
00:08
We find the value of x, and z in the relation for the option d.
00:12
So let's solve this now.
00:13
We'll use the parameters rule.
00:15
In the previous videos, we have discussed the primers rule.
00:18
The same we're going to use here.
00:20
So we have x equals in parameter school, delta 1 upon delta, and y equals z equals.
00:30
We have now the first kind of delta here, but it's even as 1 -1 -1.
00:35
Coefficient of x and one omega omega square the pervasive of y then one omega square omega the perplexer of v now we know that uh one plus omega plus omega square that equal zero of this relation we know already so finding out delta what we can do here we can use a c1 so apply here c1 102 we have c1 plus c2 plus c3 in that case we'll get here we'll give one plus omega plus omega square zero zero one omega omega square one omega square one omega square omega plus we have not we take here three here then we have this matter of this that is omega squared and negative we have that's what is that is that will give omega to the power four now we know that omega cube is one so omega to the power four is omega cube times omega that will be omiga so gives three times omega square negative omega next we'll find out here delta one so delta one we have the constant that is given as the column one that by the constant that is a bc next for the colon two as it is one omega omega square one omega square omega so this coming out to be delta one now we just here we just add here apply here r1 it tends to r1 plus r2 plus r3 so we get here one plus omega plus omega that is 0 so we get here a plus b plus c, then 1 plus omega plus omega, 0.
02:49
Similarly here, 0, r2 as it is, and r3 as it is.
03:03
For this we have got here, next bending along r1, we get a plus b plus c times we get here omega square, and negative omega to bar 4 is omega.
03:15
We have seen previously.
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So we got this delta 1.
03:21
Next we have for delta 2, we just have the first column as it is.
03:29
Second column replaced by a constant.
03:31
That column is one omega square.
03:34
Here we have omega.
03:39
So next, what we can do here and this, here is 1 -1, and here we have 1 here.
03:52
So we apply here, we have r1, 102.
04:02
So what you can do here is say r1 plus omega square r2 plus omega r2.
04:11
What do we get here? we can get if we have this is 1 plus omega square plus omega, that is 0.
04:19
Then we have a plus b omega square plus b omega.
04:24
It will give 1 plus omega to the power 4 plus omega square.
04:31
Now omega to the power 4 is omega.
04:34
Again coming out to be 0 so it's coming out to be 0 next r to an r2 a square that is another square that is 1c omega the next part is now we can expand along r1 so we get a plus b omega square plus b omega times we have b omega negative c omega square plus we have got here i'm sorry it's coming out to be negative a plus b omega square plus c omega and then we have one times omega negative one times omega square it's coming out to be delta two similarly we'll solve then delta three but it's coming out to be a plus v omega plus c omega square times we have omega square negative omega this will get now i find out xy and z.
05:46
So x delta 1 upon delta.
05:49
So we have delta that's coming as three times omega square, negative omega.
05:59
Then we had here delta 1.
06:04
That was this...