If f(x) = left| begin{array}{ccc} 1 & x & x + | vedclass

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(A) Given determinant is $f(x) = \left| \begin{array}{ccc} 1 & x & x + 1 \ 2x & x(x - 1) & (x + 1)x \ 3x(x - 1) & x(x - 1)(x - 2) & (x + 1)x(x - 1)

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(A) Given determinant is $f(x) = \left| \begin{array}{ccc} 1 & x & x + 1 \ 2x & x(x - 1) & (x + 1)x \ 3x(x - 1) & x(x - 1)(x - 2) & (x + 1)x(x - 1) \end{array} ight|$.Taking $x$ common from $C_2$ and $(x+1)$ common from $C_3$ is not directly possible,but we can factor out $x$ from the second column and $x(x+1)$ from the third row.Alternatively,observe the columns. Notice that $C_3 = C_2 + C_1$ is not immediately obvious,but let's simplify by taking common factors:$f(x) = x(x-1) \cdot x \cdot \left| \begin{array}{ccc} 1 & 1 & x+1 \ 2x & x-1 & x(x+1) \ 3x(x-1) & (x-2) & (x+1)x \end{array} ight|$.Actually,a simpler way is to notice that for any $x$,the rows are linearly dependent. Let $R_3 = R_3 - (x-2)R_2$. Performing row operations or expanding the determinant shows that the expression simplifies to $0$ for all $x$.Specifically,$f(x) = 0$ for all $x \in \mathbb{R}$.Therefore,$f(100) = 0$.

If $f(x) = \left| \begin{array}{ccc} 1 & x & x + 1 \\ 2x & x(x - 1) & (x + 1)x \\ 3x(x - 1) & x(x - 1)(x - 2) & (x + 1)x(x - 1) \end{array} \right|$,then $f(100)$ is equal to

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