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JEE Mathematics 2023 Question with Solution

The set of all values of t∈Rt \in \mathbb{R}, for which the matrix

[ete−t(sin⁡t−2cos⁡t)e−t(−2sin⁡t−cos⁡t)ete−t(2sin⁡t+cos⁡t)e−t(sin⁡t−2cos⁡t)ete−tcos⁡te−tsin⁡t]\begin{bmatrix} e^t & e^{-t}(\sin t - 2\cos t) & e^{-t}(-2\sin t - \cos t)\\ e^t & e^{-t}(2\sin t + \cos t) & e^{-t}(\sin t - 2\cos t)\\ e^t & e^{-t}\cos t & e^{-t}\sin t \end{bmatrix}

is invertible, is:

  • A

    {(2k+1)π2,k∈Z}\{(2k+1)\frac{\pi}{2}, k \in \mathbb{Z}\}

  • B

    {kπ+π4,k∈Z}\{k\pi + \frac{\pi}{4}, k \in \mathbb{Z}\}

  • C

    {kπ,k∈Z}\{k\pi, k \in \mathbb{Z}\}

  • D

    R\mathbb{R}

Answer

Correct answer:D

Step-by-step solution

Standard Method

Given: The matrix depends on t∈Rt \in \mathbb{R}.

Find: The set of values of tt for which the matrix is invertible.

A matrix is invertible when its determinant is non-zero.

From the solution working, factor the common terms from the columns:

det⁡(A)=et⋅e−t⋅e−tdet⁡[1sin⁡t−2cos⁡t−2sin⁡t+cos⁡t12sin⁡t+cos⁡tsin⁡t−2cos⁡t1cos⁡tsin⁡t]\det(A)=e^t\cdot e^{-t}\cdot e^{-t}\det\begin{bmatrix} 1 & \sin t-2\cos t & -2\sin t+\cos t\\ 1 & 2\sin t+\cos t & \sin t-2\cos t\\ 1 & \cos t & \sin t \end{bmatrix}

Apply the row operations shown in the solution:

R1→R1−R2,R2→R2−R3R_1 \to R_1-R_2, \qquad R_2 \to R_2-R_3

Then the determinant reduces to the determinant of

[0−sin⁡t−cos⁡t−3sin⁡t+cos⁡t02sin⁡t−2cos⁡t1cos⁡tsin⁡t]\begin{bmatrix} 0 & -\sin t-\cos t & -3\sin t+\cos t\\ 0 & 2\sin t & -2\cos t\\ 1 & \cos t & \sin t \end{bmatrix}

Now expand along the first column:

det⁡(A)=et⋅e−t⋅e−t[1⋅∣−sin⁡t−cos⁡t−3sin⁡t+cos⁡t2sin⁡t−2cos⁡t∣]\det(A)=e^t\cdot e^{-t}\cdot e^{-t}\left[1\cdot\begin{vmatrix}-\sin t-\cos t & -3\sin t+\cos t\\2\sin t & -2\cos t\end{vmatrix}\right]

Using the working provided,

det⁡(A)=e−t(2sin⁡tcos⁡t+6cos⁡2t+6sin⁡2t−2sin⁡tcos⁡t)\det(A)=e^{-t}\left(2\sin t\cos t+6\cos^2 t+6\sin^2 t-2\sin t\cos t\right)

So,

det⁡(A)=e−t⋅6\det(A)=e^{-t}\cdot 6

Since e−t≠0e^{-t}\neq 0 for every real tt, we get

det⁡(A)≠0for all t∈R\det(A)\neq 0 \quad \text{for all } t\in\mathbb{R}

Therefore, the matrix is invertible for all real values of tt. The correct option is D.

The solution's lists option A, but the solution working concludes the answer is R\mathbb{R}, so the solution has been treated as the authority.

Determinant Interpretation

Given: The matrix has a common factor structure involving ete^t and e−te^{-t}.

Find: Whether any real tt can make the determinant zero.

The exponential factor is

et⋅e−t⋅e−t=e−te^t\cdot e^{-t}\cdot e^{-t}=e^{-t}

which is never zero for any real tt. Therefore, invertibility depends only on the remaining trigonometric determinant.

The row operations shown do not change the zero/non-zero nature of the determinant. After simplification, the trigonometric part becomes

2sin⁡tcos⁡t+6cos⁡2t+6sin⁡2t−2sin⁡tcos⁡t=6(sin⁡2t+cos⁡2t)=62\sin t\cos t+6\cos^2 t+6\sin^2 t-2\sin t\cos t=6(\sin^2 t+\cos^2 t)=6

using

sin⁡2t+cos⁡2t=1\sin^2 t+\cos^2 t=1

Hence the determinant is always

6e−t6e^{-t}

which is never zero. So every real tt is allowed, and the answer set is R\mathbb{R}.

Common mistakes

  • Using the answer key key without checking the determinant working. Here the listed correct answer disagrees with the solution derivation; determinant-based reasoning shows the matrix is invertible for all real tt.

  • Forgetting that e−te^{-t} is never zero for any real tt. Exponential expressions do not vanish on R\mathbb{R}, so they cannot make the determinant zero.

  • Making an error in the row operations R1→R1−R2R_1\to R_1-R_2 and R2→R2−R3R_2\to R_2-R_3. Incorrect subtraction changes the trigonometric entries and can lead to a false restricted set of values.

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