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Question: If \(f\left( x \right)=\left| \begin{matrix} 1 & x & x+1 \\\ 2x & x\left( x-1 \right) & x\...

If f(x)=∣1xx+1 2xx(x−1)x(x+1) 3x(x−1)x(x−1)(x−2)x(x−1)(x+1) ∣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(50)+f(51)+.....+f(99)f\left( 50 \right)+f\left( 51 \right)+.....+f\left( 99 \right).

Explanation

Solution

We first use the row and column operations to simplify the determinant value. We take xx common from the second row and x(x−1)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(x)=∣1xx+1 2xx(x−1)x(x+1) 3x(x−1)x(x−1)(x−2)x(x−1)(x+1) ∣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 xx common from the second row and x(x−1)x\left( x-1 \right) from the third row.
We get f(x)=x2(x−1)∣1x(x+1) 2(x−1)(x+1) 3(x−2)(x+1) ∣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 R2′=R2−R1{{R}_{2}}^{'}={{R}_{2}}-{{R}_{1}}.
We get f(x)=x2(x−1)∣1x(x+1) 2(x−1)(x+1) 3(x−2)(x+1) ∣=x2(x−1)∣1x(x+1) 1−10 3(x−2)(x+1) ∣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 R3′=R3−R1{{R}_{3}}^{'}={{R}_{3}}-{{R}_{1}}.
We get f(x)=x2(x−1)∣1x(x+1) 1−10 3(x−2)(x+1) ∣=x2(x−1)∣1x(x+1) 1−10 2−20 ∣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(x)=x2(x−1)∣1x(x+1) 1−10 2−20 ∣=x2(x−1)(x+1)[−2+2]=0f\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(x)=0f\left( x \right)=0. We can put the values of a=50,51,....99a=50,51,....99 to get f(a)=0f\left( a \right)=0 as the function is xx independent.
Therefore, f(50)+f(51)+.....+f(99)=0f\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.