Let D 1 - a b a-b c d c-d a b a-b - D 2 - a c a-c b d b-d a c a-b-c- then the value of - D 1 D 2 - where b- 0 and ad- bc- is -

Given nbsp; D _ 1 = a amp; b amp; a+b c amp; d amp; c+d a amp; b amp; a b D _ 1 = a amp; b amp; a c amp; ...

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Determinant

Let D 1 = a...

Question

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

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D_{1}

∣ ∣ ∣ ∣ \begin{matrix} a & b & a + b c & d & c + d a & b & a - b \end{matrix} ∣ ∣ ∣ ∣

D_{2}

∣ ∣ ∣ ∣ \begin{matrix} a & c & a + c b & d & b + d a & c & a + b + c \end{matrix} ∣ ∣ ∣ ∣

\frac{D_{1}}{D_{2}}

b \neq 0

a d \neq b c,

D_{1} = ∣ ∣ ∣ ∣ \begin{matrix} a & b & a + b c & d & c + d a & b & a - b \end{matrix} ∣ ∣ ∣ ∣

D_{2} = ∣ ∣ ∣ ∣ \begin{matrix} a & c & a + c b & d & b + d a & c & a + b + c \end{matrix} ∣ ∣ ∣ ∣

\frac{D_{1}}{D_{2}}

b \neq 0

a d \neq b c

D_{1} = ∣ ∣ ∣ ∣ \begin{matrix} x & a & b - 1 & 0 & x x & 2 & 1 \end{matrix} ∣ ∣ ∣ ∣

D_{2} = ∣ ∣ ∣ ∣ \begin{matrix} c x^{2} & 2 a & - b x & 2 & 1 - 1 & 0 & x \end{matrix} ∣ ∣ ∣ ∣ .

(x^{2} - 4 x - 7) (x^{2} - 2 x - 3) = 0

D_{1} + D_{2} = 0,

a + 4 b + c

P

\prod : 2 x - 3 y + 6 z - 4 = 0

△ P A B

A \equiv (2, 2, 1)

B \equiv (- 1, - 4, - 1)

14 sq. units

\prod

P

6 x + a y + b z + d_{1} = 0

6 x + a y + b z + d_{2} = 0, d_{1} > d_{2}

(\frac{d_{1} - d_{2} + a + b}{3})

D_{1} = ∣ ∣ ∣ ∣ \begin{matrix} x & a & b - 1 & 0 & x x & 2 & 1 \end{matrix} ∣ ∣ ∣ ∣

D_{2} = ∣ ∣ ∣ ∣ \begin{matrix} c x^{2} & 2 a & - b x & 2 & 1 - 1 & 0 & x \end{matrix} ∣ ∣ ∣ ∣ .

(x^{2} - 4 x - 7) (x^{2} - 2 x - 3) = 0

D_{1} + D_{2} = 0,

a + 4 b + c

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