Prove the following by using properties of determinants :-
x
x+2y
x+y
x+y
x
x+2y
x+2y
x+y
x
= 9y2(x+y)
x
x+2y
x+y
x+y
x
x+2y
x+2y
x+y
x
= 9y2(x+y)
L.H.S. =
x
x+2y
x+y
x+y
x
x+2y
x+2y
x+y
x
Property of invariance :-
C1 ➤C1 - C3
C1 ➤C1 - C3
=
x-x-2y
x+2y-x-y
x+y-x
x+y
x
x+2y
x+2y
x+y
x
=
-2y
y
y
x+y
x
x+2y
x+2y
x+y
x
Taking y common along C1
= y
-2
1
1
x+y
x
x+2y
x+2y
x+y
x
Property of invariance :-
R3 ➤R3 - R2
R3 ➤R3 - R2
= y
-2
1
0
x+y
x
2y
x+2y
x+y
-y
Taking y common along R3
= y2
-2
1
0
x+y
x
2
x+2y
x+y
-1
Property of invariance :-
C2 ➤C2 + 2C3
C2 ➤C2 + 2C3
= y2
-2
1
0
x+y+2(x+2y)
x+2(x+y)
2+2(-1)
x+2y
x+y
-1
= y2
-2
1
0
x+y+2x+4y
x+2x+2y
2-2
x+2y
x+y
-1
= y2
-2
1
0
3x+5y
3x+2y
0
x+2y
x+y
-1
Property of invariance :-
R2 ➤R2 + (1/2)R1
R2 ➤R2 + (1/2)R1
= y2
-2
0
0
3x+5y
3x+2y +
1
2(3x+5y)
0
x+2y
x+y +
1
2(x+2y)
-1
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.
= y2 . (-2) . {3x+2y +
1
2(3x+5y)} . (-1)
1
2(3x+5y)} . (-1)
= (-2)(-1) . y2 . {3x+2y +
1
2(3x+5y)}
1
2(3x+5y)}
= 2y2 {
6x+4y+3x+5y
2}
6x+4y+3x+5y
2}
= y2 . (9x + 9y)
= 9y2(x+y) = R.H.S. Proved
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