00:02
Hello students, let's solve this problem.
00:03
The problem says a be a 4 by 5 matrix and u be the reduced row echelian form of a.
00:11
And u is given here, that is 1 -0 -0 -0 -0 -0 -1 -0 -0 -2 -3 -0 -0 -0 -0 -0 -0 -0 -0 -0 -0 -1 -0 -0 -1 -2 -5 -0.
00:41
And a1, if a1, that is a 1, that is a 1, that is given 2 .1 minus 3 minus 2 tanspose and a2 is equal to minus 1 to 3 1 transpose.
01:04
Given that x 0 is the solution of ax equal to b where b equal to 05 -34 tennspose and x0 is equal to 32020.
01:28
Transpose determine the matrix a.
01:33
So means we have to find that matrix of a.
01:40
So, a is any matrix and ub, it's reduce echallion form, so that you can write down that n, a is equal to nu, is the null space are same, but the column space of a is spanned by the column of a corresponding its basis variable, the pivot element of u, and u is the pivot element.
02:16
So pivot elements are same.
02:21
So the column space of a is spanned by the first and the second and the fourth column, of a.
02:29
So first two columns are given is a1 is 2 .1 minus 3 minus 2 tennspos.
02:42
A2 minus 1 to 31 tennispose.
02:51
Given that 0 534 tennspose belongs to column space, this one is a belongs to the column space.
03:11
So, alpha of a1 plus beta of a2 plus gamma of a4, that is equal to 0534 transpose you can say.
03:29
So also given that ax0 that is equal to 0 .53 .4.
03:39
Transpose, where x0 is 320 .2 .0.
03:50
So that is the thing for this.
03:58
So that you get 3a1 plus of 2a2 plus 0 a3 plus 2a3 plus 2a4, 0a5, that is equal to 0 .534 tennspos.
04:19
So after calculating you get 3a1, 2a2 plus 2a4 that is equal to 05334 tennspose.
04:35
That is 2 of a4 equal to 0534 tennspos minus 3a1, 2a2.
04:48
2 because a1, a2 already given.
04:53
So 2a4 is equal to transpose means 0 5, 3 4 minus 3.
05:01
This one is written as 2 1 minus 3 minus 2, minus 1, 2, 3, 1.
05:12
So this calculation we have to do so that we get minus minus minus 4, minus 2, 6 and 8, this one.
05:26
So, a4 is equal to what we have to get...