Question

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

• A. -2
• B. 0
• C. $-2 b$
• D. $2 b$

Answer 1

Answer: (A)

Using: $C_3 \rightarrow C_3-\left(C_1+C_2\right), D_1=\begin{vmatrix}a & b & a+b \\ c & d & c+d \\ a & b & a-b\end{vmatrix} \quad$ and $\quad D_2=\begin{vmatrix}a & c & a+c \\ b & d & b+d \\ a & c & a+b+c\end{vmatrix}$
$\therefore \quad \frac{ D _1}{ D _2}=\frac{-2 b ( ad - bc )}{ b ( ad - bc )}=-2$