Question

If \(A=\left[ \begin{matrix} {{e}^{t}} & {{e}^{-t}}\text{cos }\!\!~\!\!\text{ }t & {{e}^{-t}}\text{sin }\!\!~\!\!\text{ }t \\ {{e}^{t}} & -{{e}^{-t}}\text{cos }\!\!~\!\!\text{ }t-{{e}^{-t}}\text{sin }\!\!~\!\!\text{ }t & -{{e}^{-t}}\text{sin }\!\!~\!\!\text{ }t+{{e}^{-t}}\text{cos }\!\!~\!\!\text{ }t \\ {{e}^{t}} & 2{{e}^{-t}}\text{sin }\!\!~\!\!\text{ }t & -2{{e}^{-t}}\text{cos }\!\!~\!\!\text{ }t \\ \end{matrix} \right]\) then A is:

1. Invertible for all t ∈ R
2. Invertible only if t = π
3. Not invertible for any t ∈ R
4. Invertible only if \(\text{t}=\frac{\pi }{2}\)

Answer 1

Correct Answer - Option 1 : Invertible for all t ∈ R

det (A) = |A|

\(=\left[ \begin{matrix} {{e}^{t}} & {{e}^{-t}}\text{cos }\!\!~\!\!\text{ }t & {{e}^{-t}}\text{sin }\!\!~\!\!\text{ }t \\ {{e}^{t}} & -{{e}^{-t}}\text{cos }\!\!~\!\!\text{ }t-{{e}^{-t}}\text{sin }\!\!~\!\!\text{ }t & -{{e}^{-t}}\text{sin }\!\!~\!\!\text{ }t+{{e}^{-t}}\text{cos }\!\!~\!\!\text{ }t \\ {{e}^{t}} & 2{{e}^{-t}}\text{sin }\!\!~\!\!\text{ }t & -2{{e}^{-t}}\text{cos }\!\!~\!\!\text{ }t \\ \end{matrix} \right]\) 

\(={{e}^{t}}\cdot {{e}^{-t}}\cdot {{e}^{-t}}\left| \begin{matrix} 1 & \text{cos }\!\!~\!\!\text{ }t & \text{sin }\!\!~\!\!\text{ }t \\ 1 & -cos~t-\text{sin }\!\!~\!\!\text{ }t & -\text{sin }\!\!~\!\!\text{ }t+\text{cos }\!\!~\!\!\text{ }t \\ 1 & 2\text{sin }\!\!~\!\!\text{ }t & -2\text{cos }\!\!~\!\!\text{ }t \\ \end{matrix} \right|\) 

\(={{e}^{-t}}\left| \begin{matrix} 0 & 2\text{cos }\!\!~\!\!\text{ }t+\text{sin }\!\!~\!\!\text{ }t & 2\text{sin }\!\!~\!\!\text{ }t-\text{cos }\!\!~\!\!\text{ }t \\ 0 & -\text{cos }\!\!~\!\!\text{ }t-3\text{sin }\!\!~\!\!\text{ }t & -\text{sin }\!\!~\!\!\text{ }t+3\text{cos }\!\!~\!\!\text{ }t \\ 1 & 2\text{sin }\!\!~\!\!\text{ }t & -2\text{cos }\!\!~\!\!\text{ }t \\ \end{matrix} \right|\) 

(R₁ → R₁ – R₂ and R₂ → R₂ + R₃)

\(={{e}^{-t}}\left| \begin{matrix} 0 & -5\text{sin }\!\!~\!\!\text{ }t & 5\text{cos }\!\!~\!\!\text{ }t \\ 0 & -\text{cos }\!\!~\!\!\text{ }t-3\text{sin }\!\!~\!\!\text{ }t & -\text{sin }\!\!~\!\!\text{ }t+3\text{cos }\!\!~\!\!\text{ }t \\ 1 & 2\text{sin }\!\!~\!\!\text{ }t & -2\text{cos }\!\!~\!\!\text{ }t \\ \end{matrix} \right|\) 

R₁ → R₁ + 2R₂

= e^-t[(–5 sin t)(–sin t + 3 cos t) – 5 cos t(–cos t – 3 sin t)

= e^-t(5 sin² t – 15 sin t cos t + 5 cos² t + 15 sin t cos t)

= e^-t(5 sin² t + 5 cos² t)

= 5e^-t ≠ 0, ∀ t ∈ R

∴ A is invertible for all t ∈ R.