Determinants
General
Grade 12
Question:
Prove that $$\begin{vmatrix} 2 & \alpha+\beta+\gamma+\delta & \alpha\beta+\gamma\delta \\ \alpha+\beta+\gamma+\delta & 2(\alpha+\beta)(\gamma+\delta) & \alpha\beta(\gamma+\delta)+\gamma\delta(\alpha+\beta) \\ \alpha\beta+\gamma\delta & \alpha\beta(\gamma+\delta)+\gamma\delta(\alpha+\beta) & 2\alpha\beta\gamma\delta \end{vmatrix} = 0$$
Step-by-Step Solution
Key Concept: General
The given determinant can be expressed as the product of two determinants: <br> $$\begin{vmatrix} 1 & 1 & 0 \\ \alpha+\beta & \gamma+\delta & 0 \\ \alpha\beta & \gamma\delta & 0 \end{vmatrix} \begin{vmatrix} 1 & \gamma+\delta & \gamma\delta \\ 1 & \alpha+\beta & \alpha\beta \\ 0 & 0 & 0 \end{vmatrix}$$ <br> Since the third column of the first determinant is zero and the third row of the second determinant is zero, both determinants are zero. <br> Therefore, the product is $0 \times 0 = 0$.
Correct Answer: A