This document discusses describing the set S = {x E R^n : ai^Tx = alphai,i = 1,... ,m} for different values of n and m.
It explains that for n = m, the set S is the solution to the system of linear equations Ax = b, where A is the matrix with rows ai and b is the vector with elements alphai.
When m < n, there may be an infinite number of solutions or no solution, depending on the consistency of the underdetermined system of equations.
When m > n, the set S could be empty, finite, or infinite depending on the linear dependence and consistency of the overdetermined system.
Seal of Good Local Governance (SGLG) 2024Final.pptx
Linear Optimization Linear Optimization Problem 3. For i = 1,. .. , .pdf
1. Linear Optimization Linear Optimization Problem 3. For i = 1,. .. , m, let ai = (ai1,. . . , ain)T in
R^n and alphai in R. a) For n = 1,2,3 and n > 3 with m = n describe the set S = {x E R^n : ai^Tx
= alphai,i = 1,... ,m} b) How does this description change when m n? d) Express the set S in
matrix notation.
Solution
I = 1,2,3,….M A1= A2= A3= AM= A11 A21 A31 …………….. AM1
A12 A22 A32 …………….. AM2 A13 A23 A33 …………….. AM3 ……………..
…………….. …………….. …………….. …………….. A1N A2N A3N ……………..
AMN USING ' TO INDICATE TRANSPOSE ..THAT IS TRANSPOSE OF Ai = Ai'
USING B INSTEAD OF ALPHA FOR EASE IN TYPING ..SO [ALPHA]I = Bi S = [ X IS
AN ELEMENT OF R^N :Ai' *X = Bi ] a)……. IN GENERAL IN THIS CASE WE
SHALL HAVE N LINEAR EQNS.IN N VARIABLES ..X1,X2,…XN ..SEE BELOW …
DEPENDING ON THE LINEAR DEPENDENC & CONSISTENCY CHARACTER OF THE
EQNS. WE MAY GET S AS A NULL SET OR FINITE SET OR AN INFINITE SET
FOR N= 1 = M …... WE GET THE SET BY THE EQN... … A1= A11 * X1 =
B1 X1 = B1 /[A11] ..SO … S = [ X1] = [B1/A11] FOR N= 2 = M …...
WE GET THE SET BY THE EQN... … A = X= B= A1= A2= A11 A21 * X1
= B1 A12 A22 X2 B2 SO WE GET ….. S =[ X ] = [X1,X2] …GIVEN BY ….
X= A INVERSE * B FOR N= 3 = M …... WE GET THE SET BY THE EQN... …
A = X= B= A1= A2= A3= A11 A21 A31 X1 B1 A12 A22 A32 * X2 =
B2 A13 A23 A33 X3 B3 SO WE GET ….. S =[ X ] = [X1,X2,X3] …GIVEN BY ….
X= A INVERSE * B IN GENERAL FOR N=M WE GET THE SET BY EQN…
A= X = B = A1= A2= A3= AM= A11 A21 A31 …………….. AN1 X1 B1 A12
A22 A32 …………….. AN2 * X2 = B2 A13 A23 A33 …………….. AN3 X3 B3
…………….. …………….. …………….. …………….. …………….. ……………..
…………….. A1N A2N A3N …………….. ANN XN = BN SO WE GET ….. S =[ X ]
= [X1,X2,X3…..XN] …GIVEN BY …. X= A INVERSE * B b)…… M < N
IN GENERAL IN THIS CASE WE SHALL HAVE M LINEAR EQNS.IN N VARIABLES
..X1,X2,…XN .. SINCE THE NUMBER OF EQNS. ARE LESS THAN THE NUMBER
OF VARIABLES , WE MAY GET S AS A NULL SET OR AN INFINITE SET
DEPENDING ON THE CONSISTENCY OF THE EQNS c)….. M > N IN
GENERAL IN THIS CASE WE SHALL HAVE M LINEAR EQNS.IN N VARIABLES
..X1,X2,…XN .. SINCE THE NUMBER OF EQNS. ARE MORE THAN THE NUMBER
OF VARIABLES , DEPENDING ON THE LINEAR DEPENDENC & CONSISTENCY
2. CHARACTER OF THE EQNS. WE MAY GET S AS A NULL SET OR FINITE SET OR
AN INFINITE SET d)…. ALREADY SHOWN ABOVE ..
![Linear Optimization Linear Optimization Problem 3. For i = 1,. .. , m, let ai = (ai1,. . . , ain)T in
R^n and alphai in R. a) For n = 1,2,3 and n > 3 with m = n describe the set S = {x E R^n : ai^Tx
= alphai,i = 1,... ,m} b) How does this description change when m n? d) Express the set S in
matrix notation.
Solution
I = 1,2,3,….M A1= A2= A3= AM= A11 A21 A31 …………….. AM1
A12 A22 A32 …………….. AM2 A13 A23 A33 …………….. AM3 ……………..
…………….. …………….. …………….. …………….. A1N A2N A3N ……………..
AMN USING ' TO INDICATE TRANSPOSE ..THAT IS TRANSPOSE OF Ai = Ai'
USING B INSTEAD OF ALPHA FOR EASE IN TYPING ..SO [ALPHA]I = Bi S = [ X IS
AN ELEMENT OF R^N :Ai' *X = Bi ] a)……. IN GENERAL IN THIS CASE WE
SHALL HAVE N LINEAR EQNS.IN N VARIABLES ..X1,X2,…XN ..SEE BELOW …
DEPENDING ON THE LINEAR DEPENDENC & CONSISTENCY CHARACTER OF THE
EQNS. WE MAY GET S AS A NULL SET OR FINITE SET OR AN INFINITE SET
FOR N= 1 = M …... WE GET THE SET BY THE EQN... … A1= A11 * X1 =
B1 X1 = B1 /[A11] ..SO … S = [ X1] = [B1/A11] FOR N= 2 = M …...
WE GET THE SET BY THE EQN... … A = X= B= A1= A2= A11 A21 * X1
= B1 A12 A22 X2 B2 SO WE GET ….. S =[ X ] = [X1,X2] …GIVEN BY ….
X= A INVERSE * B FOR N= 3 = M …... WE GET THE SET BY THE EQN... …
A = X= B= A1= A2= A3= A11 A21 A31 X1 B1 A12 A22 A32 * X2 =
B2 A13 A23 A33 X3 B3 SO WE GET ….. S =[ X ] = [X1,X2,X3] …GIVEN BY ….
X= A INVERSE * B IN GENERAL FOR N=M WE GET THE SET BY EQN…
A= X = B = A1= A2= A3= AM= A11 A21 A31 …………….. AN1 X1 B1 A12
A22 A32 …………….. AN2 * X2 = B2 A13 A23 A33 …………….. AN3 X3 B3
…………….. …………….. …………….. …………….. …………….. ……………..
…………….. A1N A2N A3N …………….. ANN XN = BN SO WE GET ….. S =[ X ]
= [X1,X2,X3…..XN] …GIVEN BY …. X= A INVERSE * B b)…… M < N
IN GENERAL IN THIS CASE WE SHALL HAVE M LINEAR EQNS.IN N VARIABLES
..X1,X2,…XN .. SINCE THE NUMBER OF EQNS. ARE LESS THAN THE NUMBER
OF VARIABLES , WE MAY GET S AS A NULL SET OR AN INFINITE SET
DEPENDING ON THE CONSISTENCY OF THE EQNS c)….. M > N IN
GENERAL IN THIS CASE WE SHALL HAVE M LINEAR EQNS.IN N VARIABLES
..X1,X2,…XN .. SINCE THE NUMBER OF EQNS. ARE MORE THAN THE NUMBER
OF VARIABLES , DEPENDING ON THE LINEAR DEPENDENC & CONSISTENCY](https://image.slidesharecdn.com/linearoptimizationlinearoptimizationproblem3-230324034846-fcc71ad3/85/Linear-Optimization-Linear-Optimization-Problem-3-For-i-1-pdf-1-320.jpg)
