Question

If `[(3, -1, sin3x), (-7, 4, cos2x), (-11, 7, 2)]` is a singular matrix, then find all values of x, where x ∈ `[0, π/2]`.

Answer 1

⇒ Given matrix is singular:

`[(3, -1, sin3x), (-7, 4, cos2x), (-11, 7, 2)] = 0`

⇒ Expand along the first row:

= `3|(4, cos 2x), (7, 2)| - (-1) |(-7, cos 2x), (-11, 2)| + sin 3x |(-7, 4), (-11, 7)|`

= 3(8 − 7 cos 2x) + 1(−14 + 11 cos 2x) + sin 3x(−49 + 44)

= 24 − 21 cos 2x − 14 + 11 cos 2x − 5 sin 3x

= 10 − 10 cos 2x − 5 sin 3x

⇒ Since the matrix is singular,

10 − 10 cos 2x − 5 sin 3x = 0

Divide by 5:

2 − 2 cos 2x − sin 3x = 0

2(1 − cos 2x) = sin 3x

⇒ Using identity:

1 − cos 2x = 2 sin² x

4 sin² x = sin 3x

4 sin² x = 3 sin x − 4 sin³ x

4 sin³ x + 4 sin² x − 3 sin x = 0

sin x(4 sin² x + 4 sin x − 3) = 0

4 sin² x + 4 sin x − 3 = 0

⇒ Let sin ⁡x = t:

4t² + 4t − 3 = 0

4t² + 6t − 2t − 3 = 0

2t(2t + 3) − 1(2t + 3) = 0

(2t + 3)(2t − 1) = 0

`t = -3/2 or t = 1/2`

⇒ Since x ∈ `[0, π/2]`

sin x = 0 ⇒ x = 0

`sin x = 1/2 ⇒ x = π/6`

Therefore, `x = 0, π/6`