Let D1=|a b a+b c d c+d a b a-b| and D2=|a c a+c b d b+d a c a+b+c| then the value of (D1/D2) where b ≠ 0 and ad ≠ bc, is

Question

Let D1=|a b a+b c d c+d a b a-b| and D2=|a c a+c b d b+d a c a+b+c| then the value of (D1/D2) where b ≠ 0 and ad ≠ bc, is (A) -2 (B) 0 (C) -2 b (D) 2 b

Tardigrade Admin · Accepted Answer

Using: C3 arrow C3-(C1+C2), D1=|a b a+b c d c+d a b a-b| and D2=|a c a+c b d b+d a c a+b+c| ∴ ( D 1/ D 2)=(-2 b ( ad - bc )/ b ( ad - bc ))=-2

Q. Let $D_{1} = ∣ ∣ a c a b d b a + b c + d a - b ∣ ∣$ and $D_{2} = ∣ ∣ a b a c d c a + c b + d a + b + c ∣ ∣$ then the value of $\frac{D _{1}}{D _{2}}$ where $b \neq = 0$ and $a d \neq = b c$ , is

-2

0

$- 2 b$

$2 b$

Using: $C_{3} \to C_{3} - (C_{1} + C_{2}), D_{1} = ∣ ∣ a c a b d b a + b c + d a - b ∣ ∣$ and $D_{2} = ∣ ∣ a b a c d c a + c b + d a + b + c ∣ ∣$
$∴ \frac{D _{1}}{D _{2}} = \frac{- 2 b ( a d - b c )}{b ( a d - b c )} = - 2$

Q. Let D1​=∣∣​aca​bdb​a+bc+da−b​∣∣​ and D2​=∣∣​aba​cdc​a+cb+da+b+c​∣∣​ then the value of D2​D1​​ where b=0 and ad=bc, is

-2

0

−2b

2b

Using: C3​→C3​−(C1​+C2​),D1​=∣∣​aca​bdb​a+bc+da−b​∣∣​ and D2​=∣∣​aba​cdc​a+cb+da+b+c​∣∣​ ∴D2​D1​​=b(ad−bc)−2b(ad−bc)​=−2

Q. Let $D_{1} = ∣ ∣ a c a b d b a + b c + d a - b ∣ ∣$ and $D_{2} = ∣ ∣ a b a c d c a + c b + d a + b + c ∣ ∣$ then the value of $\frac{D _{1}}{D _{2}}$ where $b \neq = 0$ and $a d \neq = b c$ , is

$- 2 b$

$2 b$

Using: $C_{3} \to C_{3} - (C_{1} + C_{2}), D_{1} = ∣ ∣ a c a b d b a + b c + d a - b ∣ ∣$ and $D_{2} = ∣ ∣ a b a c d c a + c b + d a + b + c ∣ ∣$
$∴ \frac{D _{1}}{D _{2}} = \frac{- 2 b ( a d - b c )}{b ( a d - b c )} = - 2$