Question

If $f\left( x \right)=\left| \begin{matrix}
   1 & x & x+1 \\
   2x & x\left( x-1 \right) & x\left( x+1 \right) \\
   3x\left( x-1 \right) & x\left( x-1 \right)\left( x-2 \right) & x\left( x-1 \right)\left( x+1 \right) \\
\end{matrix} \right|$, then find the value of $f\left( 50 \right)+f\left( 51 \right)+.....+f\left( 99 \right)$.

Answer 1

Hint: We first use the row and column operations to simplify the determinant value. We take $x$ common from the second row and $x\left( x-1 \right)$ from the third row. Then we expand the determinant to find the final value.

Complete step by step solution:
We need to find the simplified value of $f\left( x \right)=\left| \begin{matrix}
   1 & x & x+1 \\
   2x & x\left( x-1 \right) & x\left( x+1 \right) \\
   3x\left( x-1 \right) & x\left( x-1 \right)\left( x-2 \right) & x\left( x-1 \right)\left( x+1 \right) \\
\end{matrix} \right|$.
We can apply row operations on the determinant value without changing the initial form.
There are certain operations which we can apply for the problems. We can switch two rows or columns which causes the determinant to switch sign. We can add a multiple of one row to another which causes the determinant to remain the same. We can multiply a row as a constant result in the determinant scaling by that constant.
First, we take $x$ common from the second row and $x\left( x-1 \right)$ from the third row.
We get $f\left( x \right)={{x}^{2}}\left( x-1 \right)\left| \begin{matrix}
   1 & x & \left( x+1 \right) \\
   2 & \left( x-1 \right) & \left( x+1 \right) \\
   3 & \left( x-2 \right) & \left( x+1 \right) \\
\end{matrix} \right|$
So, we take the form of ${{R}_{2}}^{'}={{R}_{2}}-{{R}_{1}}$.
We get $f\left( x \right)={{x}^{2}}\left( x-1 \right)\left| \begin{matrix}
   1 & x & \left( x+1 \right) \\
   2 & \left( x-1 \right) & \left( x+1 \right) \\
   3 & \left( x-2 \right) & \left( x+1 \right) \\
\end{matrix} \right|={{x}^{2}}\left( x-1 \right)\left| \begin{matrix}
   1 & x & \left( x+1 \right) \\
   1 & -1 & 0 \\
   3 & \left( x-2 \right) & \left( x+1 \right) \\
\end{matrix} \right|$.
Similarly, we take the form of ${{R}_{3}}^{'}={{R}_{3}}-{{R}_{1}}$.
We get $f\left( x \right)={{x}^{2}}\left( x-1 \right)\left| \begin{matrix}
   1 & x & \left( x+1 \right) \\
   1 & -1 & 0 \\
   3 & \left( x-2 \right) & \left( x+1 \right) \\
\end{matrix} \right|={{x}^{2}}\left( x-1 \right)\left| \begin{matrix}
   1 & x & \left( x+1 \right) \\
   1 & -1 & 0 \\
   2 & -2 & 0 \\
\end{matrix} \right|$.
Now we expand the determinant value through the third column.
So, $f\left( x \right)={{x}^{2}}\left( x-1 \right)\left| \begin{matrix}
   1 & x & \left( x+1 \right) \\
   1 & -1 & 0 \\
   2 & -2 & 0 \\
\end{matrix} \right|={{x}^{2}}\left( x-1 \right)\left( x+1 \right)\left[ -2+2 \right]=0$.
We get the value of $f\left( x \right)=0$. We can put the values of $a=50,51,....99$ to get $f\left( a \right)=0$ as the function is $x$ independent.
Therefore, $f\left( 50 \right)+f\left( 51 \right)+.....+f\left( 99 \right)=0$.

Note: The key point is that row operations don't change whether or not a determinant is 0; at most they change the determinant by a non-zero factor or change its sign. We can use row operations to reduce the matrix to reduced row-echelon form.