GATE 2015 EE - SET 1 - Question 3

Solution : Let A be a 2 x 2 matrix with the sum of diagonal elements as - 6.

Let \lambda _{1} and \lambda _{2} be the eigen values of A.

\therefore The sum of the diagonal elements of A = - 6 .

\Rightarrow   \lambda _{1} + \lambda _{2} = -6 -----------------(1)

Det of A = \left | A \right | = \lambda _{1} \lambda _{2}

Now we have to find the maximum value of  \lambda _{1} and \lambda _{2}

Let f = \lambda _{1} and \lambda _{2}

\therefore  \lambda _{1} (-6- \lambda _{1} ) (from (1))

\therefore f = - 6 \lambda _{1} - \lambda {_{1}}^{2}

\Rightarrow f' = -6 - 2 \lambda _{1}

For f to have  maximum , f ' = 0

\Rightarrow - 6 - 2 \lambda _{1} = 0

\lambda _{1}  = - 3

Now f '' = - 2 < 0

\therefore f has a maximum at \lambda _{1}  = - 3

From (1)  \lambda _{1}  +  \lambda _{2} = -6

\Rightarrow  - 3 +  \lambda _{2} = -6

\Rightarrow  - 3 +  \lambda _{2} = -6

\Rightarrow    \lambda _{2} = -3 .

The maximum value of the determinant of  A = \lambda _{1}  \lambda _{2}

= (- 3 ) x (-3) = 9.

Hence , the correct answer is 9.