If we are given a function in determinant form as $f\left( x \right)=\left| \begin{align}
  & \,\,\,\,\,\,\,\,\,1\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,x\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,x+1 \\
 & \,\,\,\,\,\,\,\,2x\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,x\left( x-1 \right)\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\left( x+1 \right)x \\
 & 3x\left( x-1 \right)\,\,\,\,\,\,\,\,\,\,\,\,x\left( x-1 \right)\left( x-2 \right)\,\,\,\,\,\,\,\,\,\,\,\,\,\left( x+1 \right)x\left( x-1 \right)\, \\
\end{align} \right|$, then f (100) is equal to
(a) 0
(b) 1
(c) – 100
(d) 100

Hint: While solving this problem, first we need to observe the matrix closely, now, write the general form of a $3\times 3$ matrix to help determine the position of the elements. Now, apply the required column reduction and simplify the matrix according to the column reduction condition (${{C}_{3}}\to {{C}_{3}}-{{C}_{2}}$). Now, you will get two columns as identical, which will give you the determinant of the matrix as zero. Finally, substitute the value of $x$ as 100 and find the answer of $f\left( 100 \right)$ from the obtained value after the column reduction.

Complete step-by-step solution:
Here, we have the matrix in the terms of $x$, $\left| \begin{align}
  & \,\,\,\,\,\,\,\,\,1\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,x\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,x+1 \\
 & \,\,\,\,\,\,\,\,2x\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,x\left( x-1 \right)\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\left( x+1 \right)x \\
 & 3x\left( x-1 \right)\,\,\,\,\,\,\,\,\,\,\,\,x\left( x-1 \right)\left( x-2 \right)\,\,\,\,\,\,\,\,\,\,\,\,\,\left( x+1 \right)x\left( x-1 \right)\, \\
\end{align} \right|$
First, let us write the general form of a $3\times 3$ matrix, we get
$\left| \begin{align}
  & {{a}_{11}}\,\,\,\,\,{{a}_{12}}\,\,\,\,\,{{a}_{13}} \\
 & {{a}_{21}}\,\,\,\,\,{{a}_{22}}\,\,\,\,\,{{a}_{23}} \\
 & {{a}_{31}}\,\,\,\,\,{{a}_{32}}\,\,\,\,\,{{a}_{33}} \\
\end{align} \right|$
Let us use a column reduction method, to make some changes in the third column.
So, we need to subtract elements of the second column from the corresponding third column.
That is, ${{a}_{13}}={{a}_{13}}-{{a}_{12}}$; ${{a}_{23}}={{a}_{23}}-{{a}_{22}}$; ${{a}_{33}}={{a}_{33}}-{{a}_{32}}$.
We get,
We need to use this column reduction,${{C}_{3}}\to {{C}_{3}}-{{C}_{2}}$
$f\left( x \right)=\left| \begin{align}
  & \,\,\,\,\,\,\,\,\,1\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,x\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,x+1-x \\
 & \,\,\,\,\,\,\,\,2x\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,x\left( x-1 \right)\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\left( x+1 \right)x-x\left( x-1 \right) \\
 & 3x\left( x-1 \right)\,\,\,\,\,\,\,\,\,\,\,\,x\left( x-1 \right)\left( x-2 \right)\,\,\,\,\,\,\,\,\,\,\,\,\,\left( x+1 \right)x\left( x-1 \right)\,-x\left( x-1 \right)\left( x-2 \right) \\
\end{align} \right|$
Let us simplify with the help of basic mathematical operations, we get$f\left( x \right)=\left| \begin{align}
  & \,\,\,\,\,\,\,\,\,1\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,x\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,1 \\
 & \,\,\,\,\,\,\,\,2x\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,x\left( x-1 \right)\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,{{x}^{2}}+x-{{x}^{2}}+x \\
 & 3x\left( x-1 \right)\,\,\,\,\,\,\,\,\,\,\,\,x\left( x-1 \right)\left( x-2 \right)\,\,\,\,\,\,\,\,\,\,\,\,\,x\left( x-1 \right)\left[ \left( x+1 \right)-\left( x-2 \right) \right] \\
\end{align} \right|$
Now, let us simplify it further to obtain the third column, we get
$f\left( x \right)=\left| \begin{align}
  & \,\,\,\,\,\,\,\,\,1\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,x\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,1 \\
 & \,\,\,\,\,\,\,\,2x\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,x\left( x-1 \right)\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,x+x \\
 & 3x\left( x-1 \right)\,\,\,\,\,\,\,\,\,\,\,\,x\left( x-1 \right)\left( x-2 \right)\,\,\,\,\,\,\,\,\,\,\,\,\,x\left( x-1 \right)\left[ x+1-x+2 \right] \\
\end{align} \right|$
          $\,=\left| \begin{align}
  & \,\,\,\,\,\,\,\,\,1\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,x\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,1 \\
 & \,\,\,\,\,\,\,2x\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,x\left( x-1 \right)\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,2x \\
 & 3x\left( x-1 \right)\,\,\,\,\,\,\,\,\,\,\,\,x\left( x-1 \right)\left( x-2 \right)\,\,\,\,\,\,\,\,\,\,\,\,\,\,x\left( x-1 \right)\left( 3 \right)
\end{align} \right|$
From the above matrix, we can see that column one and column three are identical to each other.
We know, when two rows or two columns are identical to each other, the determinant of that particular matrix is equal to zero.
Therefore, $f\left( x \right)=0$
Now, since the answer to the function of $x$ is a constant which is zero, then it means that no matter whatever we take the value of $x$, the answer will remain the same constant, which is zero.
Therefore, let us substitute the value of $x$ as 100, we get
$f\left( 100 \right)=0$
Hence, the value of $f\left( 100 \right)=0$.

Note: A matrix is determined by $m\times n$, which indicates the matrix having $m$ number of rows and $n$ number of columns. The above problem could be solved using another method where we could get all the elements in the third row as zero with the help of using the column reduction method, ${{C}_{3}}\to {{C}_{3}}-\left( {{C}_{1}}+{{C}_{2}} \right)$, also using the rule of determinants which says, if all the elements in any one of the rows or columns are zero, the determinant of that matrix is zero.