If A= e^t e^(-t)cos t e^(-t)sin⁡ t e^t -e^(-t)cos⁡t-e^(-t)sin⁡t -e^(-t)sin⁡t+e^(-t)cos⁡t e^t 2e^(-t)sin⁡t -2e^(-t)cos⁡t , then A is

Question

If A= e^t e^(-t)cos t e^(-t)sin⁡ t e^t -e^(-t)cos⁡t-e^(-t)sin⁡t -e^(-t)sin⁡t+e^(-t)cos⁡t e^t 2e^(-t)sin⁡t -2e^(-t)cos⁡t , then A is:

MARKS Content Team · Accepted Answer

Correct answer: (D) Invertible for all t∈R. Since given matrix A is invertible ⇒A≠0 A=e^(-t) 1 cos t sin t 1 -cos t-sin t -sin t+cos t 1 2 sin t -2 cos t R_2→R_2-R_1 R_3→R_3-R_1 =e^(-t) 1 cos t sin t 0 -2 cos t-sin t -2 sin t+cos t 0 2 sin t-cos t -2 cos t-sin t =e^(-t)2cost+sint^2+2sint-cost^2 =5e^(-t) A=5 e^(-t)≠0 ∀ t∈R Hence, given matrix is always invertible.

If $A = [\begin{matrix} e^{t} & e^{- t} c o s t & e^{- t} \sin t \\ e^{t} & - e^{- t} \cos t - e^{- t} \sin t & - e^{- t} \sin t + e^{- t} \cos t \\ e^{t} & {2 e}^{- t} \sin t & - 2 e^{- t} \cos t \end{matrix}],$ then $A$ is:

Solution

Relevant Questions

If A=ete-tcos te-tsin⁡ tet-e-tcos⁡t-e-tsin⁡t-e-tsin⁡t+e-tcos⁡tet2e-tsin⁡t-2e-tcos⁡t, then A is:

Solution

Relevant Questions

If $A = [\begin{matrix} e^{t} & e^{- t} c o s t & e^{- t} \sin t \\ e^{t} & - e^{- t} \cos t - e^{- t} \sin t & - e^{- t} \sin t + e^{- t} \cos t \\ e^{t} & {2 e}^{- t} \sin t & - 2 e^{- t} \cos t \end{matrix}],$ then $A$ is: