Let A = [a_{ij}] be a 3 times 3 matrix,wherea | vedclass

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(B) The matrix $A$ is given by:$A = \begin{bmatrix} 1 & -x & 2x+1 \ -x & 1 & -x \ 2x+1 & -x & 1 \end{bmatrix}$Calculating the determinant $|A|$:$|A|

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$-\frac{88}{27}$

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(B) The matrix $A$ is given by:$A = \begin{bmatrix} 1 & -x & 2x+1 \ -x & 1 & -x \ 2x+1 & -x & 1 \end{bmatrix}$Calculating the determinant $|A|$:$|A| = 1(1 - x^2) + x(-x + x(2x+1)) + (2x+1)(x^2 - (2x+1))$$|A| = 1 - x^2 - x^2 + 2x^3 + x^2 + (2x+1)(x^2 - 2x - 1)$$|A| = 2x^3 - x^2 + 1 + (2x^3 - 4x^2 - 2x + x^2 - 2x - 1)$$|A| = 4x^3 - 4x^2 - 4x = f(x)$To find the critical points,set $f'(x) = 0$:$f'(x) = 12x^2 - 8x - 4 = 4(3x^2 - 2x - 1) = 4(3x+1)(x-1) = 0$Thus,$x = 1$ and $x = -\frac{1}{3}$.Evaluating $f(x)$ at these points:$f(1) = 4(1)^3 - 4(1)^2 - 4(1) = 4 - 4 - 4 = -4$ (Minimum value)$f(-\frac{1}{3}) = 4(-\frac{1}{27}) - 4(\frac{1}{9}) - 4(-\frac{1}{3}) = -\frac{4}{27} - \frac{12}{27} + \frac{36}{27} = \frac{20}{27}$ (Maximum value)The sum of the maximum and minimum values is:$-\frac{88}{27} = -4 + \frac{20}{27} = \frac{-108 + 20}{27} = -\frac{88}{27}$

Let $A = [a_{ij}]$ be a $3 \times 3$ matrix,where
$a_{ij} = 1$,if $i = j$
$a_{ij} = -x$,if $|i - j| = 1$
$a_{ij} = 2x + 1$,otherwise
Let a function $f: R \rightarrow R$ be defined as $f(x) = \det(A)$. Then the sum of maximum and minimum values of $f$ on $R$ is equal to:

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Let $A = [a_{ij}]$ be a $3 \times 3$ matrix,where$a_{ij} = 1$,if $i = j$$a_{ij} = -x$,if $|i - j| = 1$$a_{ij} = 2x + 1$,otherwiseLet a function $f: R \rightarrow R$ be defined as $f(x) = \det(A)$. Then the sum of maximum and minimum values of $f$ on $R$ is equal to: