Question

If ${{\Delta }_{k}}=\left| \begin{matrix} 1 & n & n \\\ 2k & {{n}^{2}}+n+1 & {{n}^{2}}+n \\\ 2k-1 & {{n}^{2}} & {{n}^{2}}+n+1 \\\ \end{matrix} \right|$ and $\sum\limits_{k=1}^{n}{{{\Delta }_{k}}}=56$ , then find the value of n.

Answer 1

Firstly, we have to take the summation of ${{\Delta }_{k}}$ from 1 to n. Then, we have to apply the properties $\sum\limits_{k=1}^{n}{1}=n$ , $\sum\limits_{k=1}^{n}{k}=\dfrac{n\left( n+1 \right)}{2}$ and $\sum\limits_{k=1}^{n}{ak}=a\sum\limits_{k=1}^{n}{k}$ . Then, we have to simplify and take the determinant. Substitute the given value of $\sum\limits_{k=1}^{n}{{{\Delta }_{k}}}$ and find the value of n.

Complete step by step answer:
We are given that ${{\Delta }_{k}}=\left| \begin{matrix} 1 & n & n \\\ 2k & {{n}^{2}}+n+1 & {{n}^{2}}+n \\\ 2k-1 & {{n}^{2}} & {{n}^{2}}+n+1 \\\ \end{matrix} \right|$ . Let us take the summation of ${{\Delta }_{k}}$ from 1 to n.

& \Rightarrow \sum\limits_{k=1}^{n}{{{\Delta }_{k}}}=\left| \begin{matrix} \sum\limits_{k=1}^{n}{1} & n & n \\\ \sum\limits_{k=1}^{n}{2k} & {{n}^{2}}+n+1 & {{n}^{2}}+n \\\ \sum\limits_{k=1}^{n}{2k-1} & {{n}^{2}} & {{n}^{2}}+n+1 \\\ \end{matrix} \right| \\\ & \Rightarrow \sum\limits_{k=1}^{n}{{{\Delta }_{k}}}=\left| \begin{matrix} \sum\limits_{k=1}^{n}{1} & n & n \\\ \sum\limits_{k=1}^{n}{2k} & {{n}^{2}}+n+1 & {{n}^{2}}+n \\\ \sum\limits_{k=1}^{n}{2k}-\sum\limits_{k=1}^{n}{1} & {{n}^{2}} & {{n}^{2}}+n+1 \\\ \end{matrix} \right| \\\ \end{aligned}$$ We know that $$\sum\limits_{k=1}^{n}{1}=n$$ , $$\sum\limits_{k=1}^{n}{k}=\dfrac{n\left( n+1 \right)}{2}$$ and $$\sum\limits_{k=1}^{n}{ak}=a\sum\limits_{k=1}^{n}{k}$$ . Therefore, the above determinant can be written as $$\Rightarrow \sum\limits_{k=1}^{n}{{{\Delta }_{k}}}=\left| \begin{matrix} n & n & n \\\ \dfrac{2n\left( n+1 \right)}{2} & {{n}^{2}}+n+1 & {{n}^{2}}+n \\\ \dfrac{2n\left( n+1 \right)}{2}-n & {{n}^{2}} & {{n}^{2}}+n+1 \\\ \end{matrix} \right|$$ Let us cancel the common terms in column 1. $$\begin{aligned} & \Rightarrow \sum\limits_{k=1}^{n}{{{\Delta }_{k}}}=\left| \begin{matrix} n & n & n \\\ \dfrac{\require{cancel}\cancel{2}n\left( n+1 \right)}{\require{cancel}\cancel{2}} & {{n}^{2}}+n+1 & {{n}^{2}}+n \\\ \dfrac{\require{cancel}\cancel{2}n\left( n+1 \right)}{\require{cancel}\cancel{2}}-n & {{n}^{2}} & {{n}^{2}}+n+1 \\\ \end{matrix} \right| \\\ & \Rightarrow \sum\limits_{k=1}^{n}{{{\Delta }_{k}}}=\left| \begin{matrix} n & n & n \\\ n\left( n+1 \right) & {{n}^{2}}+n+1 & {{n}^{2}}+n \\\ n\left( n+1 \right)-n & {{n}^{2}} & {{n}^{2}}+n+1 \\\ \end{matrix} \right| \\\ \end{aligned}$$ Now, we have to apply distributive property in column 1. $$\begin{aligned} & \Rightarrow \sum\limits_{k=1}^{n}{{{\Delta }_{k}}}=\left| \begin{matrix} n & n & n \\\ {{n}^{2}}+n & {{n}^{2}}+n+1 & {{n}^{2}}+n \\\ {{n}^{2}}+n-n & {{n}^{2}} & {{n}^{2}}+n+1 \\\ \end{matrix} \right| \\\ & \Rightarrow \sum\limits_{k=1}^{n}{{{\Delta }_{k}}}=\left| \begin{matrix} n & n & n \\\ {{n}^{2}}+n & {{n}^{2}}+n+1 & {{n}^{2}}+n \\\ {{n}^{2}} & {{n}^{2}} & {{n}^{2}}+n+1 \\\ \end{matrix} \right| \\\ \end{aligned}$$ Now, we have to apply the column transformation, that is, transform column 3 by subtracting column 1 from column 3. $$\begin{aligned} & \Rightarrow \sum\limits_{k=1}^{n}{{{\Delta }_{k}}}=\left| \begin{matrix} n & n & n-n \\\ {{n}^{2}}+n & {{n}^{2}}+n+1 & {{n}^{2}}+n-{{n}^{2}}+n \\\ {{n}^{2}} & {{n}^{2}} & {{n}^{2}}+n+1-{{n}^{2}} \\\ \end{matrix} \right|\text{ }{{C}_{3}}\to {{C}_{3}}-{{C}_{1}} \\\ & \Rightarrow \sum\limits_{k=1}^{n}{{{\Delta }_{k}}}=\left| \begin{matrix} n & n & 0 \\\ {{n}^{2}}+n & {{n}^{2}}+n+1 & 0 \\\ {{n}^{2}} & {{n}^{2}} & n+1 \\\ \end{matrix} \right| \\\ \end{aligned}$$ Now, we have to transform column 2 by subtracting column 1 from column 2. $$\begin{aligned} & \Rightarrow \sum\limits_{k=1}^{n}{{{\Delta }_{k}}}=\left| \begin{matrix} n & n-n & 0 \\\ {{n}^{2}}+n & {{n}^{2}}+n+1-{{n}^{2}}+n & 0 \\\ {{n}^{2}} & {{n}^{2}}-{{n}^{2}} & n+1 \\\ \end{matrix} \right|\text{ }{{C}_{2}}\to {{C}_{2}}-{{C}_{1}} \\\ & \Rightarrow \sum\limits_{k=1}^{n}{{{\Delta }_{k}}}=\left| \begin{matrix} n & 0 & 0 \\\ {{n}^{2}}+n & 1 & 0 \\\ {{n}^{2}} & 0 & n+1 \\\ \end{matrix} \right|...\left( i \right) \\\ \end{aligned}$$ Let us take the determinant. We know that $$\Delta =\left| \begin{matrix} {{a}_{1}} & {{a}_{2}} & {{a}_{3}} \\\ {{b}_{1}} & {{b}_{2}} & {{b}_{3}} \\\ {{c}_{1}} & {{c}_{2}} & {{c}_{3}} \\\ \end{matrix} \right|={{a}_{1}}\left( {{b}_{2}}{{c}_{3}}-{{b}_{3}}{{c}_{2}} \right)-{{a}_{2}}\left( {{b}_{1}}{{c}_{3}}-{{b}_{3}}{{c}_{1}} \right)+{{a}_{3}}\left( {{b}_{1}}{{c}_{2}}-{{b}_{2}}{{c}_{1}} \right)$$ Therefore, we can write the determinant of (i) as $$\begin{aligned} & \Rightarrow \sum\limits_{k=1}^{n}{{{\Delta }_{k}}}=n\left( n+1 \right)-0+0 \\\ & \Rightarrow \sum\limits_{k=1}^{n}{{{\Delta }_{k}}}=n\left( n+1 \right) \\\ \end{aligned}$$ We are given that $\sum\limits_{k=1}^{n}{{{\Delta }_{k}}}=56$ . $$\Rightarrow 56=n\left( n+1 \right)$$ Let us apply distributive property. $$\begin{aligned} & \Rightarrow 56={{n}^{2}}+n \\\ & \Rightarrow {{n}^{2}}+n-56=0 \\\ \end{aligned}$$ Let us factorize the above polynomial and find the value of n by splitting the middle term. $$\begin{aligned} & \Rightarrow {{n}^{2}}+8n-7n-56=0 \\\ & \Rightarrow n\left( n+8 \right)-7\left( n+8 \right)=0 \\\ & \Rightarrow \left( n-7 \right)\left( n+8 \right)=0 \\\ & \Rightarrow \left( n-7 \right)=0,\left( n+8 \right)=0 \\\ \end{aligned}$$ Let us consider $\left( n-7 \right)=0$ . $\Rightarrow n=7$ Now, let us consider $\left( n+8 \right)=0$ . $\Rightarrow n=-8$ The negative value is ignored. Hence, the value of n is 7. **Note:** Students must thoroughly learn the properties of the determinants and summations. They must know to perform row and column operations. Students must note that we have ignored the negative value of n since the summation is taken from 1 to n, that is, positive numbers.