, then
Δ = 0
Δ = 2 abc
Δ = -abc
Δ = a2 + b2 + c2
then f (x) is divisible by
n2 + n
(n + 1)!
(n + 2)!
n! (n2 + n + 1)
The value of the determinant is
0
1
∞
ω
The value of is
abc
(a + b) (b + c) (c + a)
4 abc
None of these
, then x equals
1, 1, 0
0, -1, 1
1, -1, 3
0, 0, 3
[2, 3]
[3, 4]
[2, 4]
(2, 4)
a + b + c - 3 abc
3 (a + b) (b + c) (c + a)
(a - b) (b - c) (c - a)
(a - b) (b - c) (c - a) (a + b + c)
If the vertices of a triangle are (-2, -3) and (3, 2) (-1, -8) then area of the triangle.
40cm2
30cm2
15cm2
45cm2
20
-2
5
If each entry in any row, or each entry in any column of a determinant is 0, then the value of the determinant is equal to
3
-1
