Let $D_1 = \begin{vmatrix} a & b & a+b \\ c & d & c+d \\ a & b & a-b \end{vmatrix}$ and $D_2 = \begin{vmatrix} a & c & a+c \\ b & d & b+d \\ a & c & a+b+c \end{vmatrix}$ then the value of $\frac{D_1}{D_2}$ where $b \neq 0$ and $ad \neq bc$, is
Step-by-Step Solution
Key Concept: Perform row and column operations to simplify the determinants D1 and D2. For D1, R3 -> R3 - R1 gives |a b a+b; c d c+d; 0 0 -2b|. Expanding along R3 gives -2b(ad-bc). For D2, R3 -> R3 - R1 gives |a c a+c; b d b+d; 0 0 b|. Expanding along R3 gives b(ad-bc). Thus D1/D2 = -2b(ad-bc) / b(ad-bc) = -2.
For $D_1 = \begin{vmatrix} a & b & a+b \\ c & d & c+d \\ a & b & a-b \end{vmatrix}$. Applying $R_3 \to R_3 - R_1$, we get $D_1 = \begin{vmatrix} a & b & a+b \\ c & d & c+d \\ 0 & 0 & -2b \end{vmatrix} = -2b(ad-bc)$. For $D_2 = \begin{vmatrix} a & c & a+c \\ b & d & b+d \\ a & c & a+b+c \end{vmatrix}$. Applying $R_3 \to R_3 - R_1$, we get $D_2 = \begin{vmatrix} a & c & a+c \\ b & d & b+d \\ 0 & 0 & b \end{vmatrix} = b(ad-bc)$. Thus, $\frac{D_1}{D_2} = \frac{-2b(ad-bc)}{b(ad-bc)} = -2$.
Correct Answer: A