Evaluate the determinant: left| begin{array}{ | vedclass

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(A) Let the given determinant be $\Delta$. Using the identity $(p+q)^2 - (p-q)^2 = 4pq$,we apply the column operation $C_1 	o C_1 - C_2$: $(a^x + a^{

Vedclass · Accepted Answer

$0$

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(A) Let the given determinant be $\Delta$. Using the identity $(p+q)^2 - (p-q)^2 = 4pq$,we apply the column operation $C_1 	o C_1 - C_2$: $(a^x + a^{-x})^2 - (a^x - a^{-x})^2 = 4(a^x)(a^{-x}) = 4(a^0) = 4$. Similarly,for the second and third rows,we get $4$ and $4$. Thus,the determinant becomes: $\Delta = \left| \begin{array}{ccc} 4 & (a^x - a^{-x})^2 & 1 \ 4 & (b^x - b^{-x})^2 & 1 \ 4 & (c^x - c^{-x})^2 & 1 \end{array} ight|$. Since column $C_1$ and column $C_3$ are proportional (specifically,$C_1 = 4C_3$),the value of the determinant is $0$.

Evaluate the determinant: $\left| \begin{array}{ccc} (a^x + a^{-x})^2 & (a^x - a^{-x})^2 & 1 \\ (b^x + b^{-x})^2 & (b^x - b^{-x})^2 & 1 \\ (c^x + c^{-x})^2 & (c^x - c^{-x})^2 & 1 \end{array} \right|$

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