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The sum of the diagonal elements of a $ 2\times 2 $ upper triangular matrix is $-6.$  Then, the maximum possible value of the determinant of the matrix is ________.
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Important properties of Eigen values:

  1. Sum of all eigen values$=$Sum of leading diagonal(principle diagonal) elements=Trace of the matrix.
  2. Product of all Eigen values$=Det(A)= \mid A \mid$
  3. 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 $9.$

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Sum of eigen value $=$ rank of matrix $=$ $-6$ , and product of eigen value $=$ determinant of matrix.

So, we need to find a breakup of $-6$ that maximizes the determinant. Like $\left ( -6,0 \right )$,$\left ( -5,-1 \right )$$\left ( -2,-4 \right )$, $\left ( -3,-3 \right )$ ... rest combinations like $\left ( 7/-8/-9/-10/....,1/2/3/4/ \right )$ all gives -ve determinant .... so maximum product is obtained by $\left ( -3,-3 \right )$ and hence maximum determinant is $9$
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