Let D be the determinant given by $D=\begin{vmatrix}1&\cos (β-α)&\cos (γ-α)\\\cos (α-β)&1&\cos (γ-β)\\\cos (α - γ)&\cos (β-γ)&1\end{vmatrix}$, where $α,β$ and $γ$ are real numbers.
Statement-1: The value of D is zero.
Statement-2: The determinant D is expressible as the product of two determinants each equal to zero.
Answer & explanation
Correct answer: option 1
We have,
$D=\begin{vmatrix}1&\cos (β-α)&\cos (γ-α)\\\cos (α-β)&1&\cos (γ-β)\\\cos (α - γ)&\cos (β-γ)&1\end{vmatrix}$
$=\begin{vmatrix}\cos^2α+\sin^2α&\cos β\cos α+\sin β\sin α&\cos γ\cos α+\sin γ\sin α\\\cos α\cos β+\sin α\sin β&\cos^2β+\sin^2β&\cos γ\cos β+\sin γ\sin β\\\cos α \cos γ+\sin α \sin γ&\cos β\cos γ+\sin β\sin γ&\cos^2γ+\sin^2γ\end{vmatrix}$
$=\begin{vmatrix}\cos α&\sin α&0\\\cos β&\sin β&1\\\cos γ&\sin γ&1\end{vmatrix}\begin{vmatrix}\cos α&\sin α&0\\\cos β&\sin β&1\\\cos γ&\sin γ&1\end{vmatrix}=0$
So, both the statement are true and statement-2 is a correct explanation for statement-1.