For every we deline the cperturbed problem to be the linear...

Name: for every we deline the cperturbed problem to be the linear programming problem obtained by replacing b with be given basis matrix b show that the corresponding basic solution xee the e pert 03136
Uploaded: 2022-10-11T11:00:47-08:00
Duration: 6 min 18 s
Channel: Adi S
Description: for every we deline the cperturbed problem to be the linear programming problem obtained by replacing b with be given basis matrix b show that the corresponding basic solution xee the e pert 03136

For every we define the c-perturbed problem to be the linear programming problem obtained by replacing b with b(e). Given basis matrix B, show that the corresponding basic solution Xe(e) of the e-perturbed problem is equal to B^(-1)[b | I]. (6) Show that there exist some 5 such that all basic solutions to the e-perturbed problem are nondegenerate, for 0 < e < 5.

Transcript

00:01 Here in this question it is given that a is equals to 1 2 negative 1 then we have 1 1 1 then we have 6 negative 1 0 also we have the matrix c as 1 2 negative 1 then we have 1 1 1 1 then we have 7 1 1 7 1 negative 1 then we have 7 1 negative 1 also here we are given as e c is equals to a here we need to find matrix e so here for evaluating it since we are given that e c is equals to a so from here we have that e is equals to c inverse a so here we have that e is equals to c inverse a so here c inverse will be equals to 1 by determinant of c, adjoint of c.

01:29 So now here we have matrix c as 1 2 negative 1.

01:42 Here we have 1 1 1, 1, here we have 7 1 negative 1.

01:50 So here the determinant of c will be equals to 20 as here here so here we have 1, here we have negative 1, 1, negative 1, then we have negative 2, here we have negative 1, negative 7, then we have negative 1, here we have 1 negative 7.

02:25 On solving this we get the determinant as 20.

02:28 Now for evaluating the ad joint, we first evaluate the minor of c.

02:38 That will be equals to negative 2 negative 8 negative c here we have negative 1 c negative 30 then we have 3 negative 1 now here next we evaluate the core factors of c that will be equals to negative to 8 negative c then we have 1 6 13 then we have 3 negative 2 negative 1 so here for evaluating the ad joint of c it will be equals to transpose of co -factor of c so here it will be equals to negative to 8 negative 6 then we have 1 6 30 then we have 3 negative 2 negative 1.

04:24 So here we have evaluated the adjoint of c so now e that will be equals to c inverse a so here we have 1 by 20 here we have have a joint as negative 2, 1, 3...

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