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Q)

Without expanding evaluate the determinent $\begin{vmatrix} (a^x+a^{-x})^2 & (a^x-a^{-x})^2 & 1 \\ (a^y+a^{-y})^2 & (a^y-a^{-y})^2 & 1 \\ (a^z+a^{-z})^2 & (a^z-a^{-z})^2 & 1 \end{vmatrix}$ where $a>0$ ,and $x,y,z \in R$

$\begin{array}{1 1} 0 \\ 1 \\ -1 \\ \pm 1 \end{array} $

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A)
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  • (i)If two rows or columns are identical, then the value of the determinant is zero
  • (ii)Elementary transformations can be made by
  • (a) interchanging the rows or columns
  • (b) By adding or subtracting two or more rows or columns
Let $\Delta=\begin{vmatrix} (a^x+a^{-x})^2 & (a^x-a^{-x})^2 & 1 \\ (a^y+a^{-y})^2 & (a^y-a^{-y})^2 & 1 \\ (a^z+a^{-z})^2 & (a^z-a^{-z})^2 & 1 \end{vmatrix}$
Apply $C_1 \to C_1-C_2$
$(a+b)^2-(a-b)^2=4ab)$
$\Delta=\begin{vmatrix} 4 & (a^x-a^{-x})^2 & 1 \\ 4 & (a^y-a^{-y})^2 & 1 \\ 4 & (a^z-a^{-z})^2 & 1 \end{vmatrix}$
Take 4 as the common factor from $C_1$
$\Delta=4\begin{vmatrix} 1 & (a^x-a^{-x})^2 & 1 \\ 1 & (a^y-a^{-y})^2 & 1 \\ 1 & (a^z-a^{-z})^2 & 1 \end{vmatrix}$
Since two coloumns are identical, then the value of the determinant is 0
Therefore $\Delta =0$
Solution : option A is correct
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