JEE Main 2023 · 29 January · evening · Single correct · +4 / −1
The set of all values of t∈ ℝ, for which the matrix
[ eᵗ, e⁻ᵗ(sin t - 2 cos t), e⁻ᵗ( - 2 sin t - cos t); eᵗ, e⁻ᵗ(2 sin t + cos t), e⁻ᵗ(sin t - 2 cos t); eᵗ, e⁻ᵗ cos t, e⁻ᵗ sin t ] is invertible, is :
- (a) { kπ ,k ∈ ℤ }
- (b) ℝ
- (c) { (2k + 1)π/2,k ∈ ℤ }
- (d) { kπ + π/4,k ∈ ℤ }
Answer and solution
Answer (b) ℝ
Solution
Maths sir · 4 steps
Pull out the exponentials, subtract rows, and the sines and cosines collapse into sin² t+cos² t.
- Factor out exponentials. A matrix is invertible exactly when its determinant is not zero. Take eᵗ out of column 1 and e⁻ᵗ out of columns 2 and 3, so the determinant becomes eᵗ × e⁻²ᵗ × D = e⁻ᵗD. Write s = sin t and c = cos t.
D = |1, s-2c, -2s-c; 1, 2s+c, s-2c; 1, c, s|
- Create zeros. Apply R₁ → R₁ - R₃ and R₂ → R₂ - R₃. Row 1 becomes (0, s-3c, -3s-c) and row 2 becomes (0, 2s, -2c), so column 1 is now (0, 0, 1).
- Expand along column 1. Only the entry 1 in row 3 survives, with sign +. So D = (s-3c)(-2c) - (-3s-c)(2s) = -2sc + 6c² + 6s² + 2sc = 6(s²+c²) = 6.
- So. The determinant is 6e⁻ᵗ, and e⁻ᵗ > 0 for every real t, so it is never zero. The matrix is invertible for all t ∈ ℝ, which is option (b).
The idea to keep. When a determinant is full of sin t and cos t, first factor common terms out of rows and columns, then create zeros. It often collapses to a constant through sin² t + cos² t = 1.
Answer from the NTA final answer key. Solution by PrepWiser.