Question

If \[\alpha \] , \[\beta \ne 0\] , and \[f\left( n \right) = {\alpha ^n} + {\beta ^n}\] and \[\left| {\begin{array}{*{20}{c}}
  3&{1 + f\left( 1 \right)}&{1 + f\left( 2 \right)} \\
  {1 + f\left( 1 \right)}&{1 + f\left( 2 \right)}&{1 + f\left( 3 \right)} \\
  {1 + f\left( 2 \right)}&{1 + f\left( 3 \right)}&{1 + f\left( 4 \right)}
\end{array}} \right| = K{\left( {1 - \alpha } \right)^2}{\left( {1 - \beta } \right)^2}{\left( {\alpha - \beta } \right)^2}\] , then K is equal to
A \[\alpha \beta \]
B \[\dfrac{1}{{\alpha \beta }}\]
C 1
D -1

Answer 1

Hint: In linear algebra, determinant is a special number that can be determined from a square matrix. To solve the given functions, we need to consider the LHS part, in which to find the value of K we get LHS = RHS and the value of K. Hence, using matrix elementary functions simplify the given functions in the determinant to get the value of K.

Complete step-by-step answer:
Let us write the given data,
 \[\left| {\begin{array}{*{20}{c}}
  3&{1 + f\left( 1 \right)}&{1 + f\left( 2 \right)} \\
  {1 + f\left( 1 \right)}&{1 + f\left( 2 \right)}&{1 + f\left( 3 \right)} \\
  {1 + f\left( 2 \right)}&{1 + f\left( 3 \right)}&{1 + f\left( 4 \right)}
\end{array}} \right| = K{\left( {1 - \alpha } \right)^2}{\left( {1 - \beta } \right)^2}{\left( {\alpha - \beta } \right)^2}\]
Let us consider the LHS terms, in which after simplifying we get the RHS terms i.e.,
 \[\left| {\begin{array}{*{20}{c}}
  3&{1 + f\left( 1 \right)}&{1 + f\left( 2 \right)} \\
  {1 + f\left( 1 \right)}&{1 + f\left( 2 \right)}&{1 + f\left( 3 \right)} \\
  {1 + f\left( 2 \right)}&{1 + f\left( 3 \right)}&{1 + f\left( 4 \right)}
\end{array}} \right|\]
We know that, \[f\left( n \right) = {\alpha ^n} + {\beta ^n}\] , hence
 \[f\left( 1 \right) = \alpha + \beta \]
 \[f\left( 2 \right) = {\alpha ^2} + {\beta ^2}\]
 \[f\left( 3 \right) = {\alpha ^3} + {\beta ^3}\]
 \[f\left( 4 \right) = {\alpha ^4} + {\beta ^4}\]
Now, substitute all the considered equations in the given determinant as:
  \[ = \left| {\begin{array}{*{20}{c}}
  3&{1 + \alpha + \beta }&{1 + {\alpha ^2} + {\beta ^2}} \\
  {1 + \alpha + \beta }&{1 + {\alpha ^2} + {\beta ^2}}&{1 + {\alpha ^3} + {\beta ^3}} \\
  {1 + {\alpha ^2} + {\beta ^2}}&{1 + {\alpha ^3} + {\beta ^3}}&{1 + {\alpha ^4} + {\beta ^4}}
\end{array}} \right|\]
Split the terms, by considering 2 determinants in product form with respect to rows and columns of \[{R_1}{R_2}{R_3}\] and \[{C_1}{C_2}{C_3}\] as:
 \[ = \left| {\begin{array}{*{20}{c}}
  1&1&1 \\
  1&\alpha &\beta \\
  1&{{\alpha ^2}}&{{\beta ^2}}
\end{array}} \right| \times \left| {\begin{array}{*{20}{c}}
  1&1&1 \\
  1&\alpha &{{\alpha ^2}} \\
  1&\beta &{{\beta ^2}}
\end{array}} \right|\] ……… 1
Let us assume equation 1 as \[A \times B\] .
Now, interchange \[{C_3} \leftrightarrow {R_3}\] of determinant A and transpose the B determinant i.e.,
 \[ = \left| {\begin{array}{*{20}{c}}
  1&1&1 \\
  1&\alpha &\beta \\
  1&{{\alpha ^2}}&{{\beta ^2}}
\end{array}} \right| \times \left| {\begin{array}{*{20}{c}}
  1&1&1 \\
  1&\alpha &\beta \\
  1&{{\alpha ^2}}&{{\beta ^2}}
\end{array}} \right|\]
As there are common terms involved, we get:
 \[ \Rightarrow {\left| {\begin{array}{*{20}{c}}
  1&1&1 \\
  1&\alpha &\beta \\
  1&{{\alpha ^2}}&{{\beta ^2}}
\end{array}} \right|^2}\]
Apply Elementary transformation of column matrix i.e., \[{C_2} \to {C_2} - {C_1}\] and \[{C_3} \to {C_3} - {C_1}\]
 \[ = {\left| {\begin{array}{*{20}{c}}
  1&0&0 \\
  1&{\alpha - 1}&{\beta - 1} \\
  1&{{\alpha ^2} - 1}&{{\beta ^2} - 1}
\end{array}} \right|^2}\]
Expanding the terms, as \[{R_2}{C_2}{C_3}\] and \[{R_3}{C_2}{C_3}\] implies to zero, hence we have \[{R_1}{C_1}\left\{ {\left( {{R_2}{C_2}} \right)\left( {{R_3}{C_3}} \right)\left( {{R_3}{C_2}} \right)\left( {{R_2}{C_3}} \right)} \right\}\] i.e.,
 \[ \Rightarrow {\left\{ {1\left| {\begin{array}{*{20}{c}}
  {\alpha - 1}&{\beta - 1} \\
  {{\alpha ^2} - 1}&{{\beta ^2} - 1}
\end{array}} \right|} \right\}^2}\]
Simplify the determinant of the matrix i.e.,
 \[\left\{ {{R_1}{C_1} \times {R_2}{C_2} - {R_2}{C_1} \times {R_1}{C_2}} \right\}\] :
 \[ \Rightarrow {\left\{ {\left( {\alpha - 1} \right)\left( {{\beta ^2} - 1} \right) - \left( {\beta - 1} \right)\left( {{\alpha ^2} - 1} \right)} \right\}^2}\]
As this is of the form, \[{a^2} - {b^2} = \left( {a + b} \right)\left( {a - b} \right)\] , hence we get:
 \[ \Rightarrow {\left\{ {\left( {\alpha - 1} \right)\left( {\beta - 1} \right)\left( {\beta + 1} \right) - \left( {\beta - 1} \right)\left( {\alpha - 1} \right)\left( {\alpha + 1} \right)} \right\}^2}\]
 \[ \Rightarrow {\left( {\alpha - 1} \right)^2}{\left( {\beta - 1} \right)^2}{\left\{ {\beta + 1 - \alpha - 1} \right\}^2}\]
As +1 and -1 implies to zero, hence we get:
 \[{\left( {\alpha - 1} \right)^2}{\left( {\beta - 1} \right)^2}{\left( {\beta - \alpha } \right)^2}\]
Hence,
 \[ \Rightarrow {\left( {\alpha - 1} \right)^2}{\left( {\beta - 1} \right)^2}{\left( {\beta - \alpha } \right)^2} = {\left( {1 - \alpha } \right)^2}{\left( {1 - \beta } \right)^2}{\left( {\alpha - \beta } \right)^2}\]
Therefore, \[K = 1\] .
Hence, option C is the right answer.
So, the correct answer is “Option C”.

Note: While solving the determinants we have applied elementary column matrix since interchanging of columns makes the functions of the matrix equal to the given function, as an elementary matrix which differs from the identity matrix by one single elementary row. When a system is written horizontally, we can obtain systems equivalent to it by performing elementary transformation of column matrix: multiplying a column by a non-zero constant; adding a multiple of one column to another column; interchanging columns.