Matrices & Determinants Question - None

Question

If a, b, c > 0 and x, y, z ∈ R, then the determinant

|\begin{matrix} {(a^{x} + a^{- x})}^{2} & {(a^{x} - a^{- x})}^{2} & 1 \\ {(b^{y} + b^{- y})}^{2} & {(b^{y} - b^{- y})}^{2} & 1 \\ {(c^{z} + c^{- z})}^{2} & {(c^{z} - c^{- z})}^{2} & 1 \end{matrix}|

is equal to -

Accepted Answer

Let the determinant be \Delta. Using the property (u+v)^2 - (u-v)^2 = 4uv, we apply the column operation C1 -> C1 - C2. The first column becomes: (a^x + a^-x)^2 - (a^x - a^-x)^2 = 4(a^x)(a^-x) = 4. Similarly, the other elements in the first column become 4. Since the first column is (4, 4, 4)^T, we can factor out 4. The determinant then has two identical columns (the first column becomes 1, 1, 1 and the third column is already 1, 1, 1). Thus, the value of the determinant is 0.

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