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: Use row and column operations to simplify the determinants. For D1, perform R3 -> R3 - R1. For D2, perform R3 -> R3 - R1.
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: 1