Show that :-
a
b
c
a2
b2
c2
a3
b3
c3
= abc(b-c)(c-a)(a-b)
a
b
c
a2
b2
c2
a3
b3
c3
= abc(b-c)(c-a)(a-b)
L.H.S. =
a
b
c
a2
b2
c2
a3
b3
c3
Take a common along R1,
b common along R2 and
c common along R3
= abc
1
1
1
a
b
c
a2
b2
c2
Property of invariance :-
R2 ➤R2 - R1
R3 ➤R3 - R1
R2 ➤R2 - R1
R3 ➤R3 - R1
= abc
1
0
0
a
b-a
c-a
a2
b2-a2
c2-a2
= abc
1
0
0
a
b-a
c-a
a2
(b+a)(b-a)
(c+a)(c-a)
Take (b-a) common along R2,
(c-a) common along R3
= abc(b-a)(c-a)
1
0
0
a
1
1
a2
b+a
c+a
Property of invariance :-
R3 ➤R3 - R2
R3 ➤R3 - R2
= abc(b-a)(c-a)
1
0
0
a
1
0
a2
b+a
c-b
Triangle property of determinant :-
As all the elements below the diagnol elements are zero so its determinant is given by product of its diagonal elements.
As all the elements below the diagnol elements are zero so its determinant is given by product of its diagonal elements.
= abc(b-a)(c-a)(c-b) = abc(a-b)(b-c)(c-a) = R.H.S. Proved.
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