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Question
If x, y, z ∈ R, then the value of determinant `|((2x^2 + 2^(-x))^2, (2^x - 2^(-x))^2, 1),((3^x + 3^(-x))^2, (3^x -3^(-x))^2, 1),((4^x + 4^(-x))^2, (4^x - 4^(-x))^2, 1)|` is equal to ______.
Solution
If x, y, z ∈ R, then the value of determinant `|((2x^2 + 2^(-x))^2, (2^x - 2^(-x))^2, 1),((3^x + 3^(-x))^2, (3^x -3^(-x))^2, 1),((4^x + 4^(-x))^2, (4^x - 4^(-x))^2, 1)|` is equal to 0.
Explanation:
We have, `|((2x^2 + 2^(-x))^2, (2^x - 2^(-x))^2, 1),((3^x + 3^(-x))^2, (3^x -3^(-x))^2, 1),((4^x + 4^(-x))^2, (4^x - 4^(-x))^2, 1)|`
C1 → C1 – C2
⇒ `|((2^x + 2^-x)^2 - (2^x - 2^-x)^2, (2^x - 2^-x)^2, 1),((3^x + ^-x)^2 - (3^x -3^-x)^2, (3^x 3^-x)^2, 1),((4^x + 4^-x)^2 - (4^x - 4^-x)^2, (4^x - 4^-x)^2, 1)|`
⇒ `|(4 *2^x * 2^-x, (2^x - 2^-x)^2, 1),(4 * 3^x * 3^x, (3^x- 3^-x)^2, 1),(4 * 4^x * 4^-x, (4^x - 4^x)^2, 1)|` .....[Applying (a + b)2 – (a – b)2 = 4ab]
⇒ `|(4, (2^x - 2^-x)^2, 1),(4, (3^x - 3^-x)^2, 1),(4, (4^x - 4^-x)^2, 1)|`
⇒ `4|(1, (2^x - 2^-x)^2, 1),(1, (3^x - 3^-x)^2, 1),(1, (4^x - 4^-x)^2, 1)|` ......(Taking 4 common from C1)
⇒ 4 · 0 = 0 ....(∵ C1 and C3 are identical columns)
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