Question

If $\alpha ,\beta \ne 0\text{ and }f\left( n \right)={{\alpha }^{n}}+{{\beta }^{n}}\text{ and }\left| \begin{matrix}
   3 & 1+f\left( 1 \right) & 1+f\left( 2 \right) \\
   1+f\left( 1 \right) & 1+f\left( 2 \right) & 1+f\left( 3 \right) \\
   1+f\left( 2 \right) & 1+f\left( 3 \right) & 1+f\left( 4 \right) \\
\end{matrix} \right|=K{{\left( 1-\alpha \right)}^{2}}{{\left( 1-\beta \right)}^{2}}{{\left( \alpha -\beta \right)}^{2}}$ then K is equal to?
\[\begin{align}
  & \text{A}.\text{ 1} \\
 & \text{B}.\text{ }-\text{1} \\
 & \text{C}.\text{ }\alpha \beta \\
 & \text{D}.\text{ }\dfrac{1}{\alpha \beta } \\
\end{align}\]

Answer 1

Hint: Here in the question, put all the values of f (n) in the given determinant expression according to their value of n. After doing this, we have to look over the symmetry of the determinant expression. Hence, we have to manipulate the terms. The normal method (or traditional method) of doing such questions leads to quite lengthy and time consuming approaches. Hence, we will apply the concept of determinant multiplication here. After that, solve each part separately and then multiply them in the final stage. According to the right hand side value given in the question, we have to manipulate our final solution, so that the value of K will be determined by comparison.

Complete step by step answer:
Now, we have given data i.e.
\[f\left( n \right)={{\alpha }^{n}}+{{\beta }^{n}}\text{ where }\alpha \text{,}\beta \ne \text{0}\]
\[\left| \begin{matrix}
   3 & 1+f\left( 1 \right) & 1+f\left( 2 \right) \\
   1+f\left( 1 \right) & 1+f\left( 2 \right) & 1+f\left( 3 \right) \\
   1+f\left( 2 \right) & 1+f\left( 3 \right) & 1+f\left( 4 \right) \\
\end{matrix} \right|\]
We can find value of f(1), f(2), f(3), f(4) by substituting n=1, 2, 3 and 4 respectively in \[f\left( n \right)={{\alpha }^{n}}+{{\beta }^{n}}\text{ where }\alpha \text{,}\beta \ne \text{0}\].
So, we will have f(1) = \[\alpha +\beta\] , f(2) = \[{{\alpha }^{2}}+{{\beta }^{2}}\] , f(3) = \[{{\alpha }^{3}}+{{\beta }^{3}}\] and f(4) = \[{{\alpha }^{4}}+{{\beta }^{4}}\]
Put the values of f(1), f(2), f(3), f(4) then we get:
\[\left| \begin{matrix}
   3 & 1+\alpha +\beta & 1+{{\alpha }^{2}}+{{\beta }^{2}} \\
   1+\alpha +\beta & 1+{{\alpha }^{2}}+{{\beta }^{2}} & 1+{{\alpha }^{3}}+{{\beta }^{3}} \\
   1+{{\alpha }^{2}}+{{\beta }^{2}} & 1+{{\alpha }^{3}}+{{\beta }^{3}} & 1+{{\alpha }^{4}}+{{\beta }^{4}} \\
\end{matrix} \right|\]
Now, see each term has 3 sub terms except 3 (the first value). Hence, manipulate 3 = 1+1+1
Therefore, we have
\[\left| \begin{matrix}
   1+1+1 & 1+\alpha +\beta & 1+{{\alpha }^{2}}+{{\beta }^{2}} \\
   1+\alpha +\beta & 1+{{\alpha }^{2}}+{{\beta }^{2}} & 1+{{\alpha }^{3}}+{{\beta }^{3}} \\
   1+{{\alpha }^{2}}+{{\beta }^{2}} & 1+{{\alpha }^{3}}+{{\beta }^{3}} & 1+{{\alpha }^{4}}+{{\beta }^{4}} \\
\end{matrix} \right|\text{ }\text{. }\text{. }\text{. }\text{. }\text{. }\text{. }\text{. }\text{. }\text{. }\text{. (i)}\]
See in above expression, there is some kind of symmetry is there. Hence, we will try to extract 2 separate determinants from above expression. But before doing that, we should know the concept of multiplication of two determinants.
\[\left| \begin{matrix}
   {{a}_{1}} & {{b}_{1}} \\
   {{a}_{2}} & {{b}_{2}} \\
\end{matrix} \right|\times \left| \begin{matrix}
   {{l}_{1}} & {{m}_{1}} \\
   {{l}_{2}} & {{m}_{2}} \\
\end{matrix} \right|=\left| \begin{matrix}
   {{a}_{1}}{{l}_{1}}+{{b}_{1}}{{l}_{2}} & {{a}_{1}}{{m}_{1}}+{{b}_{1}}{{m}_{2}} \\
   {{a}_{2}}{{l}_{1}}+{{b}_{2}}{{l}_{2}} & {{a}_{2}}{{m}_{1}}+{{b}_{2}}{{m}_{1}} \\
\end{matrix} \right|\]
We have used rows by columns concept (first row of determinant 1 multiplied with first column of determinant 2 and then with second column of determinant 2).
This same procedure followed for second row of determinant 1. Now, come to our question, expression (i) will be reduced to
\[\begin{align}
  & \left| \begin{matrix}
   1 & 1 & 1 \\
   1 & \alpha & \beta \\
   1 & {{\alpha }^{2}} & {{\beta }^{2}} \\
\end{matrix} \right|\times \left| \begin{matrix}
   1 & 1 & 1 \\
   1 & \alpha & {{\alpha }^{2}} \\
   1 & \beta & {{\beta }^{2}} \\
\end{matrix} \right|\text{ }\text{. }\text{. }\text{. }\text{. }\text{. }\text{. }\text{. }\text{. }\text{. }\text{. }\text{. }\text{. }\text{. }\text{. (ii)} \\
 & \begin{matrix}
   \downarrow \\
   \text{let det 1} \\
\end{matrix}\begin{matrix}
   \downarrow \\
   \text{ let determinant 2} \\
\end{matrix} \\
\end{align}\]
\[\text{Determinant 1}=\left| \begin{matrix}
   1 & 1 & 1 \\
   1 & \alpha & \beta \\
   1 & {{\alpha }^{2}} & {{\beta }^{2}} \\
\end{matrix} \right|\]
\[\text{Determinant 2}=\left| \begin{matrix}
   1 & 1 & 1 \\
   1 & \alpha & {{\alpha }^{2}} \\
   1 & \beta & {{\beta }^{2}} \\
\end{matrix} \right|\]
Because we know that, the value of a determinant remain unuttered if the rows and columns are interchanged.
Now, from equation (ii) we have:
\[\begin{align}
  & \left| \begin{matrix}
   1 & 1 & 1 \\
   1 & \alpha & {{\alpha }^{2}} \\
   1 & \beta & {{\beta }^{2}} \\
\end{matrix} \right|\times \left| \begin{matrix}
   1 & 1 & 1 \\
   1 & \alpha & {{\alpha }^{2}} \\
   1 & \beta & {{\beta }^{2}} \\
\end{matrix} \right| \\
 & \Rightarrow {{\left( \left| \begin{matrix}
   1 & 1 & 1 \\
   1 & \alpha & {{\alpha }^{2}} \\
   1 & \beta & {{\beta }^{2}} \\
\end{matrix} \right| \right)}^{2}} \\
 & \Rightarrow {{\left( \left| \begin{matrix}
   1 & 0 & 1 \\
   1 & \alpha -1 & {{\alpha }^{2}} \\
   1 & \beta -1 & {{\beta }^{2}} \\
\end{matrix} \right| \right)}^{2}}\text{ }\left( \text{apply }{{C}_{2}}\to {{C}_{2}}-{{C}_{1}} \right)\left( {{\text{C}}_{\text{2}}}=\text{column 2} \right) \\
\end{align}\]
Here, column-2 implies column-2 subtracted column-1. That means, we have to subtract each element of column-2 from their corresponding element in column-1
\[\begin{align}
  & \Rightarrow {{\left( \left| \begin{matrix}
   1 & 0 & 0 \\
   1 & \alpha -1 & {{\alpha }^{2}}-1 \\
   1 & \beta -1 & {{\beta }^{2}}-1 \\
\end{matrix} \right| \right)}^{2}}\text{ }\left( \text{apply }{{C}_{3}}\to {{C}_{3}}-{{C}_{1}} \right) \\
 & \Rightarrow {{\left( \left( \alpha -1 \right)\left( {{\beta }^{2}}-1 \right)-\left( \beta -1 \right)\left( {{\alpha }^{2}}-1 \right) \right)}^{2}} \\
\end{align}\]
Now here we can write, $\left( {{\beta }^{2}}-1 \right)=\left( \beta -1 \right)\left( \beta +1 \right)\text{ and }\left( {{\alpha }^{2}}-1 \right)=\left( \alpha -1 \right)\left( \alpha +1 \right)$
\[\Rightarrow {{\left( \left( \alpha -1 \right)\left( \beta -1 \right)\left( \beta +1 \right)-\left( \beta -1 \right)\left( \beta +1 \right)\left( \alpha +1 \right) \right)}^{2}}\]
Now taking $\left( \alpha -1 \right)\left( \beta -1 \right)$ we get:
\[\Rightarrow {{\left( \left( \alpha -1 \right)\left( \beta -1 \right)\left( \left( \beta +1 \right)-\left( \alpha +1 \right) \right) \right)}^{2}}\]
We can write $\left( \alpha -1 \right)\left( \beta -1 \right)=\left( 1-\alpha \right)\left( 1-\beta \right)$ so we get:
\[\begin{align}
  & \Rightarrow {{\left( \left( 1-\alpha \right)\left( 1-\beta \right)\left( -1 \right)\left( \alpha -\beta \right) \right)}^{2}} \\
 & \Rightarrow {{\left( -1 \right)}^{2}}{{\left( 1-\alpha \right)}^{2}}{{\left( 1-\beta \right)}^{2}}{{\left( \alpha -\beta \right)}^{2}} \\
 & \Rightarrow 1{{\left( 1-\alpha \right)}^{2}}{{\left( 1-\beta \right)}^{2}}{{\left( \alpha -\beta \right)}^{2}} \\
\end{align}\]

Hence, by comparing to RHS we get value of K = 1.

Note: Students are a little bit confused about which method they should apply here for determinant multiplication. See, in this particular question, we have applied the method of rows and columns. But, we can also apply rows by rows, columns by rows and columns by column. Answer will be the same. Also, we can solve the determinant 1 and determinant 2 separately and then multiply them. But that way can consume a student's time to manipulate the expression according to the question. Hence, we use here some properties of determinants and get the solution in a more efficient way.