Matrices & Determinants
Invertibility of matrix, determinant condition
nta_pyq_2023_jan
Grade None
Question:
The set of all values of $t \in \mathbb{R}$, for which the matrix $\begin{pmatrix} e^t & e^{-t}(\sin t - 2\cos t) & e^{-t}(-2\sin t - \cos t) \\ e^t & e^{-t}(2\sin t + \cos t) & e^{-t}(\sin t - 2\cos t) \\ e^t & e^{-t}\cos t & e^{-t}\sin t \end{pmatrix}$ is invertible, is
$\left\{(2k+1)\frac{\pi}{2}, k \in \mathbb{Z}\right\}$
$\left\{k\pi + \frac{\pi}{4}, k \in \mathbb{Z}\right\}$
$\{k\pi, k \in \mathbb{Z}\}$
$\mathbb{R}$
Step-by-Step Solution
Key Concept: Matrix is invertible iff its determinant is non-zero; factor out common terms and apply row operations
Factor: $e^t \cdot e^{-t} \cdot e^{-t}$ from columns. After row operations $R_1 \to R_1 - R_2$, $R_2 \to R_2 - R_3$, determinant simplifies to $e^{-t} \times (2\sin t \cos t + 6\cos^2 t + 6\sin^2 t - 2\sin t \cos t) = e^{-t} \times 6 \neq 0$ for all $t \in \mathbb{R}$. Answer: (4)
Correct Answer: $\mathbb{R}$