Le t D1=|a&b&a+b c&b&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 is, where b ≠ 0 and a d ≠ b c .

Question

Le t D1=|a&b&a+b c&b&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 is, where b ≠ 0 and a d ≠ b c . (A)  (B)  (C)  (D)

Tardigrade Admin · Accepted Answer

Using: C3 arrow C3-(C1+C2) in D1 and D2, we have ∴ (D1/D2)=(-2 b(a d-b c)/b(a d-b c))=-2

Q. Le t $D_{1} = ∣ ∣ a c a b b b a + b c + d a - b ∣ ∣$ and $D_{2} = ∣ ∣ a b a c d c a + c b + d a + b + c ∣ ∣$ then the value of $D_{1} / D_{2}$ is, where $b \neq = 0$ and $a d \neq = b c$ ___.