Let A = [ a _ ij ] be a 3 × 3 matrix, where a _ ij = \ cases 1 & ,...

We have,

{a}_{ij} = \left\{\begin{cases}1 & , & ifi = j \\ - x & , & if\left|i - j\right| = 1 \\ 2x + 1 & , & otherwise\end{cases}\right)</p><p>Then,</p><p>A = \left[$\begin{cases}1 & - x & 2x + 1 \\ - x & 1 & - x \\ 2x + 1 & - x & 1\end{cases}$\right]$</p><p>$⇒ \left|A\right| = \left(1 - {x}^{2}\right) + x\left(- x + 2{x}^{2} + x\right) + \left(2x + 1\right)\left({x}^{2} - 2x - 1\right)$</p><p>$⇒ \left|A\right| = 1 - {x}^{2} + 2{x}^{3} + \left(2{x}^{3} - 4{x}^{2} - 2x\right) + \left({x}^{2} - 2x - 1\right)$</p><p>$⇒ \left|A\right| = f\left(x\right) = 4{x}^{3} - 4{x}^{2} - 4x$</p><p>$⇒ {f}^{'}\left(x\right) = 4\left(3{x}^{2} - 2x - 1\right)$</p><p>For maxima or minima,${f}^{'}\left(x\right) = 0$</p><p>$⇒ 3{x}^{2} - 2x - 1 = 0$</p><p>$⇒ x = 1, \dfrac{- 1}{3}$</p><p>Now,</p><p>${f}^{' '}\left(x\right) = 4\left(6x - 2\right)$</p><p>${f}^{' '}\left(1\right) = 4\left(6 \cdot 1 - 2\right) = 16 > 0$( point of minima)</p><p>${f}^{' '}\left(- \dfrac{1}{3}\right) = 4\left(6 \cdot \left(- \dfrac{1}{3}\right) - 2\right) = - 16 < 0$( point of maxima)</p><p>Hence, required sum is</p><p>$f\left(1\right) + f\left(- \dfrac{1}{3}\right)$</p><p>$= - 4 + \dfrac{20}{27} = - \dfrac{88}{27}