Question

If we are given a determinant ${{D}_{r}}=\left| \begin{matrix} r & x & \dfrac{n\left( n+1 \right)}{2} \\\ 2r-1 & y & {{n}^{2}} \\\ 3r-2 & z & \dfrac{n\left( 3n-1 \right)}{2} \\\ \end{matrix} \right|$, then prove that $\sum\limits_{r=1}^{n}{{{D}_{r}}}=0$.

Answer 1

Perform row operations on the given determinant, ${{R}_{2}}\to {{R}_{2}}-2{{R}_{1}}$ and ${{R}_{3}}\to {{R}_{3}}-3{{R}_{1}}$ to simplify ${{D}_{r}}$. Here, ${{R}_{1}},{{R}_{2}}$ and ${{R}_{3}}$ denotes first, second and third row respectively. Now, perform a column operation, ${{C}_{1}}\to {{C}_{1}}-{{C}_{3}}$ and expand the given determinant. Take summation $\sum\limits_{r=1}^{n}{}$ on both sides and consider n, x, y, z as constants and simplify the summation of ${{D}_{r}}$. Use the formula for sum of ‘n’ terms of an A.P. given as: - ${{S}_{n}}=\dfrac{n}{2}\left( n+1 \right)$ and if k is a constant then $\sum\limits_{r=1}^{n}{k}=nk$, to get the answer.

Complete step-by-step solution
Here, we have been provided with a determinant expression given as: -
$\Rightarrow$ ${{D}_{r}}=\left| \begin{matrix} r & x & \dfrac{n\left( n+1 \right)}{2} \\\ 2r-1 & y & {{n}^{2}} \\\ 3r-2 & z & \dfrac{n\left( 3n-1 \right)}{2} \\\ \end{matrix} \right|$
We have to prove, $\sum\limits_{r=1}^{n}{{{D}_{r}}}=0$.
Now, in the expression of determinant ${{D}_{r}}$, we have 3 rows and 3 columns. Horizontal lines are rows and vertical lines are columns. We know that performing row and column operations does not change the value of determinant. Therefore, we have,
(i) Performing the operation, ${{R}_{2}}\to {{R}_{2}}-2{{R}_{1}}$ and ${{R}_{3}}\to {{R}_{3}}-3{{R}_{1}}$ that means terms of $~{{R}_{2}}$ is replaced with ${{R}_{2}}-2{{R}_{1}}$ and ${{R}_{3}}$ is replaced with ${{R}_{3}}-3{{R}_{1}}$, we get,

r & x & \dfrac{n\left( n+1 \right)}{2} \\\ \left( 2r-1 \right)-2r & y-2x & {{n}^{2}}-2\left[ \dfrac{n\left( n+1 \right)}{2} \right] \\\ \left( 3r-2 \right)-3r & z-3x & \dfrac{n\left( 3n-1 \right)}{2}-3\left[ \dfrac{n\left( n+1 \right)}{2} \right] \\\ \end{matrix} \right|$$ $$\Rightarrow {{D}_{r}}=\left| \begin{matrix} r & x & \dfrac{n\left( n+1 \right)}{2} \\\ -1 & y-2x & -\dfrac{2n}{2} \\\ -2 & z-3x & -\dfrac{4n}{2} \\\ \end{matrix} \right|$$ Taking (-1) common from second and third row and ‘n’ common from column 3, we get, $$\begin{aligned} & \Rightarrow {{D}_{r}}=\left( -1 \right)\times \left( -1 \right)\times \left( n \right)\left| \begin{matrix} r & x & \left( \dfrac{n+1}{2} \right) \\\ 1 & 2x-y & 1 \\\ 2 & 3x-z & 2 \\\ \end{matrix} \right| \\\ & \Rightarrow {{D}_{r}}=n\left| \begin{matrix} r & x & \left( \dfrac{n+1}{2} \right) \\\ 1 & 2x-y & 1 \\\ 2 & 3x-z & 2 \\\ \end{matrix} \right| \\\ \end{aligned}$$ (ii) Performing the operation, $${{C}_{1}}\to {{C}_{1}}-{{C}_{3}}$$, that means terms of $${{C}_{1}}$$ is replaced with $${{C}_{1}}-{{C}_{3}}$$, we get, $$\Rightarrow {{D}_{r}}=n\left| \begin{matrix} r-\left( \dfrac{n+1}{2} \right) & x & \left( \dfrac{n+1}{2} \right) \\\ 0 & 2x-y & 1 \\\ 0 & 3x-z & 2 \\\ \end{matrix} \right|$$ Now, expanding the determinant we get, $$\begin{aligned} & \Rightarrow {{D}_{r}}=n\left[ \left( r-\dfrac{n+1}{2} \right)\times \left\\{ \left( 2x-y \right)\times 2-\left( 3x-z \right)\times 1 \right\\}-x\times \left( 0\times 2-0\times 1 \right)+\left( \dfrac{n+1}{2} \right)\times \left( 0\times \left( 3x-z \right)-0\times \left( 2x-y \right) \right) \right] \\\ & \Rightarrow {{D}_{r}}=n\left[ \left( r-\dfrac{n+1}{2} \right)\left\\{ 4x-2y-3x+z \right\\} \right] \\\ & \Rightarrow {{D}_{r}}=n\left[ \left( r-\dfrac{n+1}{2} \right)\times \left\\{ x-2y+z \right\\} \right] \\\ & \Rightarrow {{D}_{r}}=n\times \left( x-2y+z \right)\left( r-\dfrac{n+1}{2} \right) \\\ \end{aligned}$$ Taking summation sign both sides, we get, $$\Rightarrow \sum\limits_{r=1}^{n}{{{D}_{r}}}=\sum\limits_{r=1}^{n}{n\times \left( x-2y+z \right)\left( r-\dfrac{n+1}{2} \right)}$$ Since, $${{D}_{r}}$$ is a function of r, therefore n, x, y and z can be considered as constant terms. So, taking these constant terms out of the summation sign, we get, $$\begin{aligned} & \Rightarrow \sum\limits_{r=1}^{n}{{{D}_{r}}}=n\left( x-2y+z \right)\times \sum\limits_{r=1}^{n}{\left( r-\dfrac{n+1}{2} \right)} \\\ & \Rightarrow \sum\limits_{r=1}^{n}{{{D}_{r}}}=n\left( x-2y+z \right)\times \left[ \sum\limits_{r=1}^{n}{r-\sum\limits_{r=1}^{n}{\left( \dfrac{n+1}{2} \right)}} \right] \\\ & \Rightarrow \sum\limits_{r=1}^{n}{{{D}_{r}}}=n\left( x-2y+z \right)\times \left[ \left( 1+2+3+...+n \right)-\left( \dfrac{n+1}{2} \right)\sum\limits_{r=1}^{n}{1} \right] \\\ & \Rightarrow \sum\limits_{r=1}^{n}{{{D}_{r}}}=n\left( x-2y+z \right)\times \left[ \left( 1+2+3+...+n \right)-\left( \dfrac{n+1}{2} \right)\times \left( 1+1+...ntimes \right) \right] \\\ \end{aligned}$$ Here, 1 + 2 + 3 + …. + n are n terms in A.P. with common difference equal to 1. So, applying the formula sum of ‘n’ terms of an A.P. given as, $${{S}_{n}}=\dfrac{n\left( n+1 \right)}{2}$$, we get, $$\Rightarrow \sum\limits_{r=1}^{n}{{{D}_{r}}}=n\left( x-2y+z \right)\times \left[ \dfrac{n\left( n+1 \right)}{2}-\left( \dfrac{n+1}{2} \right)\times n \right]$$ Cancelling the like terms, we get, $$\begin{aligned} & \Rightarrow \sum\limits_{r=1}^{n}{{{D}_{r}}}=n\left( x-2y+z \right)\times 0 \\\ & \Rightarrow \sum\limits_{r=1}^{n}{{{D}_{r}}}=0 \\\ \end{aligned}$$ Hence, proved. **Note:** One may note that we have not expanded the given determinant directly because we have to reduce the calculations. That is why row and column operations are performed. Also, you may note that we have considered n, x, y and z are constants. This is because $${{D}_{r}}$$ denotes this condition. You must remember properties of determinants to solve the above question.