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:

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:

TuteeHUB · Accepted Answer

Invertible only if t = π

TuteeHUB · Answer

Invertible for all t ∈ R

TuteeHUB · Answer

Not invertible for any t ∈ R

TuteeHUB · Answer

Invertible only if $\text{t}=\frac{\pi }{2}$

1.	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:
A.	Invertible for all t ∈ R
B.	Invertible only if t = π
C.	Not invertible for any t ∈ R
D.	Invertible only if \(\text{t}=\frac{\pi }{2}\)
Answer» B. Invertible only if t = π

Discussion

No Comment Found

Related MCQs

Reply to Comment