Important properties of Eigen values:
- Sum of all eigen values$=$Sum of leading diagonal(principle diagonal) elements=Trace of the matrix.
- Product of all Eigen values$=Det(A)= \mid A \mid$
- Any square diagonal(lower triangular or upper triangular) matrix eigen values are leading diagonal (principle diagonal)elements itself.
Let upper triangular matrix $A = \begin{bmatrix} a_{11} &a_{12} \\ 0 &a_{22} \end{bmatrix}_{2\times 2}$
Given that: Sum of the diagonal elements of a $2\times 2$ upper triangular matrix is $-6.$
$\implies a_{11} + a_{22} = -6$
Let one eigen value is $\lambda$, then other eigen value is $-6-\lambda.$
Determinant $ = \mid A \mid = a_{11}a_{22} = \lambda (-6-\lambda)$
Lets $ y = \lambda (-6-\lambda)$
$\implies y = -6\lambda -\lambda^{2}\rightarrow(1)$
Differentiate with respect to $'\lambda'$ both side
$\implies \dfrac{\mathrm{d} y}{\mathrm{d} \lambda} = -6-2\lambda = 0\implies \lambda=-3$
Differentiate with respect to $'\lambda'$ both side
$\implies\dfrac{\mathrm{d^{2}} y}{\mathrm{d} \lambda^{2}} = -2<0\:{\color{Magenta}{\textbf{(Maxima)}}}$
$\therefore$ for $\lambda = -3,$ we got Maxima.
Now, we put $\lambda = -3$ in equation $(1)$ and get the maximum value of determinant.
$\implies y = -6(-3)- (-3)^{2}$
$\implies y = 18-9 = 9$
$\therefore\: \mid A \mid = a_{11}a_{22} = 9$
So, the correct answer is $9.$