Without expanding, show that the value of the following determinant is zero: a + b 2 a + b 3 a + b 2 a + b 3 a + b 4 a + b 4 a + b 5 a + b 6 a + b

We have to prove that the value of determinant is 0 . ∆ = a + b 2 a + b 3 a + b 2 a + b 3 a + b 4 a + b 4 a + b 5 a + b 6 a + b = a a a 2 a 2 a 2 a 4 a + b 5 a + b 6 a + b Applying R 1 → R 2 - R 1 and R 2 → R 3 - R 2 = 2 a a a a a a 4 a + b 5 a + b 6 a + b = 0

Without expanding, show that the value of the following determinant is zero: a + b 2 a + b 3 a + b 2 a + b 3 a + b 4 a + b 4 a + b 5 a + b 6 a + b

We have to prove that the value of determinant is 0 . ∆ = a + b 2 a + b 3 a + b 2 a + b 3 a + b 4 a + b 4 a + b 5 a + b 6 a + b = a a a 2 a 2 a 2 a 4 a + b 5 a + b 6 a + b Applying R 1 → R 2 - R 1 and R 2 → R 3 - R 2 = 2 a a a a a a 4 a + b 5 a + b 6 a + b = 0

Without expanding, show that the value of the following determinant is zero: 1 2 2 2 3 2 4 2 2 2 3 2 4 2 5 2 3 2 4 2 5 2 6 2 4 2 5 2 6 2 7 2

Consider 1 2 2 2 3 2 4 2 2 2 3 2 4 2 5 2 3 2 4 2 5 2 6 2 4 2 5 2 6 2 7 2 = 1 4 9 16 4 9 16 25 9 16 25 36 16 25 36 49 = 1 4 9 16 4 9 16 25 9 16 25 36 7 9 11 13 [ Applying R 4 → R 4 - R 3 ] = 1 4 9 16 4 9 16 25 5 7 9 11 7 9 11 13 [ Applying R 3 → R 3 - R 2 ] = 1 4 9 16 3 5 7 9 5 7 9 11 2 2 2 2 [ Applying R 2 → R 2 - R 1 ] = 1 4 9 16 3 5 7 9 2 2 2 2 2 2 2 2 [ Applying R 4 → R 4 - R 3 and then R 3 → R 3 - R 2 ] = 0

Without expanding, show that the value of the following determinant is zero: 2 x + 2 - x 2 2 x - 2 - x 2 1 3 x + 3 - x 2 3 x - 3 - x 2 1 4 x + 4 - x 2 4 x - 4 - x 2 1

Consider 2 x + 2 - x 2 2 x - 2 - x 2 1 3 x + 3 - x 2 3 x - 3 - x 2 1 4 x + 4 - x 2 4 x - 4 - x 2 1 = 2 2 x + 2 - 2 x + 2 2 2 x + 2 - 2 x - 2 1 3 2 x + 3 - 2 x + 2 3 2 x + 3 - 2 x - 2 1 4 2 x + 4 - 2 x + 2 4 2 x + 4 - 2 x - 2 1 = 4 2 2 x + 2 - 2 x - 2 1 4 3 2 x + 3 - 2 x - 2 1 4 4 2 x + 4 - 2 x - 2 1 [ Applying C 1 → C 1 - C 2 ] = 4 1 2 2 x + 2 - 2 x - 2 1 1 3 2 x + 3 - 2 x - 2 1 1 4 2 x + 4 - 2 x - 2 1 = 0

Without expanding, show that the value of the following determinant is zero: sin α cos α cos ( α + δ ) sin β cos β cos ( β + δ ) sin γ cos γ cos ( γ + δ )

Consider ∆ = sin α cos α cos ( α + δ ) sin β cos β cos ( β + δ ) sin γ cos γ cos ( γ + δ ) ∆ = sin α cos α cos α cos δ - sin α sin δ sin β cos β cos β cos δ - sin α sin δ sin γ cos γ cos γ cos δ - sin γ sin δ Applying C 3 → C 3 - cos δ C 2 + sin δ C 1 ∆ = sin α cos α 0 sin β cos β 0 sin γ cos γ 0 ⇒ sin α cos α cos ( α + δ ) sin β cos β cos ( β + δ ) sin γ cos γ cos ( γ + δ ) = 0