Prove the following by using properties of determinants :-
(x + y + 2z)
z
z
x
(y + z + 2x)
x
y
y
(z + x + 2y)
= 2(x + y + z)³
(x + y + 2z)
z
z
x
(y + z + 2x)
x
y
y
(z + x + 2y)
= 2(x + y + z)³
L.H.S. =
(x + y + 2z)
z
z
x
(y + z + 2x)
x
y
y
(z + x + 2y)
Applying property of varience :
C1 ➤ C1 + C2 + C3
=
2x + 2y + 2z
2x + 2y + 2z
2x + 2y + 2z
x
(y + z + 2x)
x
y
y
(z + x + 2y)
=
2(x + y + z)
2(x + y + z)
2(x + y + z)
x
(y + z + 2x)
x
y
y
(z + x + 2y)
Applying scalar multiple property :
Taking 2(x + y + z) common from the column C1.
Taking 2(x + y + z) common from the column C1.
= 2(x + y + z)
1
1
1
x
(y + z + 2x)
x
y
y
(z + x + 2y)
R2 ➤ R2 - R1
and
R3 ➤ R3 - R1
and
R3 ➤ R3 - R1
= 2(x + y + z)
1
0
0
x
(y + z + x)
0
y
0
(z + x + y)
Triangle property :
All the elements below the main diagonal is zero. So, the determinant of the matrix is given by the product of the diagonals elements.
= 2(x + y + z) × 1 × (x + y + z) × (x + y + z)
= 2(x + y + z)³ = R.H.S proved.
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