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Please,please help me - Let and If the minimum value of det(A) is 8, then a value of d is: - Matrices and Determinants - JEE Main

Let d\epsilon R, andd\epsilon R,\begin{bmatrix} -2 &4+d &(sin\Theta) -2 \\ 1 & (sin\Theta)+2 &d \\ 5&(2sin\Theta )-d &(-sin\Theta )+2+2d \end{bmatrix}, \Theta \epsilon [0,2\pi ].

If the minimum value of det(A) is 8, then a value of d is:

  • Option 1)

     

    -5

  • Option 2)

     

    -7

  • Option 3)

     

    2(\sqrt{2}+1)

  • Option 4)

    2(\sqrt{2}+2)

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Value of determinants of order 3 -

-

 

det(A) = \begin{vmatrix} -2 & 4+d &sin\theta -2 \\ 1 & sin\theta +2 &d \\ 5&2sin\theta -d & -sin\theta + 2 + 2d \end{vmatrix}

from the concept

Apply row operation R_{1} \rightarrow R_{1}+ R_{3} -2R_{2}

= \begin{vmatrix} 1 & 0 &0 \\ 1 & sin\theta +2 & d\\ 5 & 2sin\theta - d & -sin\theta + 2 + 2d \end{vmatrix} \\ = \left ( 2 + sin\theta \right )\left ( 2+2d-sin\theta \right ) -d (2sin\theta - d) \\ = 4+4d-2sin\theta + 2sin\theta + 2dsin\theta -sin^{2}\theta -2dsin\theta + d^{2} \\ =d^{2 } + 4d + 4 - sin^{2}\theta \\ =(d+2)^{2} - sin^{2}\theta

For a given d, minimum value of det(A) is

\Rightarrow d= 1 or d = -5


Option 1)

 

-5

Option 2)

 

-7

Option 3)

 

2(\sqrt{2}+1)

Option 4)

2(\sqrt{2}+2)

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