# 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.