Page 92 - DMTH502_LINEAR_ALGEBRA

Linear Algebra

Notes

AX Y
A 11 A 12  A 1n
A A  A
A 21 22 2n
   
A A  A
Where m 1 m 2 mn ...(2)
x y
1 1
x y
X 2 , Y 2
 
x n y n
In solving the linear system of equations (1) we sometimes use the technique of elimination. We
can illustrate this method on the following homogeneous equations:
2x – x + x = 0
1 2 3
x + 3x + 4x = 0
1 2 3
If we add (–2) times the second equation to the first equation,
we obtain
–7x – 7x = 0
2 3
or x = – x
2 3
If we add (3) times the first equation to the second equation
we obtain
7x + 7x = 0
1 3
or x = –x
1 3
So we conclude that if (x , x , x ) is the solution then x = x = –x . Thus the set of solutions consists
1 2 3 1 2 3
of all triples (a, a, –a).
For the general system (1), suppose we select m, scalars c , c ,...c , multiply the jth equation by c
1 2 m j
and then add, we obtain the equations

m
(C A + ... + C A )x + ... + (C A + C A + ... + C A )x =  C y
1 11 m m1 1 1 1n 2 2n m mn n j j
j 1
Such an equation is called a linear combination of the equations in (1). Evidently any solution of
the entire system of equations (1) will also be the solution of this new equation. This is the
fundamental idea of the elimination process. Thus if we have another system of linear equations

B x  B x 2    B x  Z 1 
1n n
12
11 1

     ...(3)

B x  B x 2  B x  Z K 
K
1 1
K
2
Kn n
in which each of the K equations is a linear combination of the equations (1), then every solution
of (1) is a solution of the new system (2).

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