Points to be noted: 1. Ek system Ax = B, jisme M equations aur N variables hein. Rank(A) = Rank(A,B) = m m variables = basic variables N - m variables = non basic variables (N - m) variables ki values zero mankar, remaining m variables ki values nikalne se jo solution prapt hoga wah basic solution kahlaegi. Isme m basic and N-m non basic variables honge. Maximum number of basic solutions obtained = C( N,m) N objects mein se m ka selection Example x + 2y + z = 4 2x + y + 5z = 5 Rank of A = Rank of (A,B) = 2 Basic variables= 2 Non basic variables= 1 Number of basic solutions = C(3,2) = 3 Basic solutions (2,1,0) , (5, 0, -1), (0, 5/3, 2/3) Sabhi BS non-degenerate hein. 2. Linear independence of basic vectors; Basic variables ke vectors ka set linearly independent hone chahiye. unki matrix non- singular honi chahiye (determinant ≠ 0) 3. Non degenerate basic solutions: jisme sabhi basic variables non-zero ho. Degenerate basic solution: jisme kam se kam ek basic variables zero ho. Theorem: Basic solutions ke existence aur sabhi BS ke non-degenerate hone ki necessary aur sufficient condition hai ki augmented matrix(A,B) ke sabhi m columns ke sets linearly independent hone chahiye. Linearly dependent aur independent hona vectors ke sets ki property hai. Linear Algebra mein isse acche se samjha rakha hai.
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Points to be noted:
1. Ek system Ax = B, jisme M equations aur N variables hein.
Rank(A) = Rank(A,B) = m
m variables = basic variables
N - m variables = non basic variables
(N - m) variables ki values zero mankar, remaining m variables ki values nikalne se jo solution prapt hoga wah basic solution kahlaegi.
Isme m basic and N-m non basic variables honge.
Maximum number of basic solutions obtained = C( N,m)
N objects mein se m ka selection
Example
x + 2y + z = 4
2x + y + 5z = 5
Rank of A = Rank of (A,B) = 2
Basic variables= 2
Non basic variables= 1
Number of basic solutions = C(3,2) = 3
Basic solutions
(2,1,0) , (5, 0, -1), (0, 5/3, 2/3)
Sabhi BS non-degenerate hein.
2. Linear independence of basic vectors;
Basic variables ke vectors ka set linearly independent hone chahiye. unki matrix non- singular honi chahiye (determinant ≠ 0)
3.
Non degenerate basic solutions: jisme sabhi basic variables non-zero ho.
Degenerate basic solution: jisme kam se kam ek basic variables zero ho.
Theorem: Basic solutions ke existence aur sabhi BS ke non-degenerate hone ki necessary aur sufficient condition hai ki augmented matrix(A,B) ke sabhi m columns ke sets linearly independent hone chahiye.
Linearly dependent aur independent hona vectors ke sets ki property hai. Linear Algebra mein isse acche se samjha rakha hai.
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