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  • Solve! - then A is : - Matrices and Determinants - JEE Main

 

\begin{bmatrix} 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 \end{bmatrix}

then A is :

  • Option 1)

     

    not invertible for any t\epsilon R

  • Option 2)

     

    invertible only if t=\pi/2

  • Option 3)

     

    nvertible only if t=\pi

  • Option 4)

     

    invertible for all t\epsilon R

Answers (1)

best_answer

  

Property of determinant -

If two rows ( or two columns ) in a determinant have corresponding elements that are equal , the value of determinant is equal to zero 

-

 

 

Value of determinants of order 3 -

-

 

  |A| = e^{-1}\begin{vmatrix} 1 &\cos(t) &\sin (t) \\ 1& -\cos (t) - \sin (t) & -\sin (t) + \cos (t)\\ 1& 2\sin (t) & -2\cos (t) \end{vmatrix} \\ = e^{-t}[5\cos^2 {t} 5 \sin^2t]\forall t\in \mathbb R \\ = 5e^{-t} \neq 0 \forall t\in \mathbb R


Option 1)

 

not invertible for any t\epsilon R

Option 2)

 

invertible only if t=\pi/2

Option 3)

 

nvertible only if t=\pi

Option 4)

 

invertible for all t\epsilon R
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