Let D_1 = | , * 20 c a&b& a + b &d& c + d &b& a - b , | and D_2 =… - Vidyadip

MCQ

Let $D_1 =$ $\left| {\,\begin{array}{*{20}{c}}a&b&{a + b}\\c&d&{c + d}\\a&b&{a - b}\end{array}\,} \right|$ and $D_2 =$ $\left| {\,\begin{array}{*{20}{c}}a&c&{a + c}\\b&d&{b + d}\\a&c&{a + b + c}\end{array}\,} \right|$ then the value of $\frac{{{D_1}}}{{{D_2}}}$ where $b \ne 0$ and $ad \ne bc$, is

✓
$-2$
B
$0$
C
$- 2b$
D
$2b$

✓

Answer

Correct option: A.

$-2$

a
Using $\rightarrow C_3 \rightarrow C_3 - (C_1 + C_2),$ $D_1 =$ $\left|{\,\begin{array}{*{20}{c}}a&b&{a + b}\\c&d&{c + d}\\a&b&{a - b}\end{array}\,} \right|$ and $D_2 = $ $\left| {\,\begin{array}{*{20}{c}}a&c&{a + c}\\b&d&{b + d}\\ a&c&{a + b + c}\end{array}\,} \right|$

$\therefore$ $\frac{{{D_1}}}{{{D_2}}}$ $=$ $\frac{{ - 2b(ad - bc)}}{{b(ad - bc)}}$ $=$ $- 2$

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