Matrices & Determinants
Determinant Properties
Grade Class 12
Question:
If a, b, c > 0 and x, y, z ∈ R, then the determinant <br> <img src="https://latex.codecogs.com/svg.latex?\begin{vmatrix} (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{vmatrix}"/> is equal to -
(A) a^x b^y c^z
(B) a^-x b^-y c^-z
(C) a^2x b^2y c^2z
(D) zero
Step-by-Step Solution
Key Concept: Use the identity (u+v)^2 - (u-v)^2 = 4uv. Here, (a^x + a^-x)^2 - (a^x - a^-x)^2 = 4(a^x)(a^-x) = 4. Applying C1 -> C1 - C2 makes the first column all 4s, and the third column is all 1s. Since C1 = 4 * C3, the determinant is 0.
Let the determinant be \Delta. Applying C1 \to C1 - C2, we get <br> <img src="https://latex.codecogs.com/svg.latex?\Delta = \begin{vmatrix} (a^x + a^{-x})^2 - (a^x - a^{-x})^2 & (a^x - a^{-x})^2 & 1 \\ (b^y + b^{-y})^2 - (b^y - b^{-y})^2 & (b^y - b^{-y})^2 & 1 \\ (c^z + c^{-z})^2 - (c^z - c^{-z})^2 & (c^z - c^{-z})^2 & 1 \end{vmatrix}"/> <br> Since (u+v)^2 - (u-v)^2 = 4uv, we have <br> <img src="https://latex.codecogs.com/svg.latex?\Delta = \begin{vmatrix} 4(a^x)(a^{-x}) & (a^x - a^{-x})^2 & 1 \\ 4(b^y)(b^{-y}) & (b^y - b^{-y})^2 & 1 \\ 4(c^z)(c^{-z}) & (c^z - c^{-z})^2 & 1 \end{vmatrix} = \begin{vmatrix} 4 & (a^x - a^{-x})^2 & 1 \\ 4 & (b^y - b^{-y})^2 & 1 \\ 4 & (c^z - c^{-z})^2 & 1 \end{vmatrix}"/> <br> Since C1 and C3 are proportional (C1 = 4 * C3), the value of the determinant is 0.
Correct Answer: D