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Algebra of Limits

let D1 = | ...

Question

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

A

- 2

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B

0

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C

- 2 b

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D

2 b

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Solution

The correct option is A $- 2$
$D_{1} ∣ ∣ ∣ ∣ \begin{matrix} a & b & a + b c & d & c + b a & b & a - b \end{matrix} ∣ ∣ ∣ ∣$
$= ∣ ∣ ∣ ∣ \begin{matrix} a & b & a + b c & d & c + d o & o & - 2 d \end{matrix} ∣ ∣ ∣ ∣ R_{3} \to R_{3} - R_{1}$
$= - 2 b (a d - b c)$
$D_{2} = ∣ ∣ ∣ ∣ \begin{matrix} a & c & a + c b & d & b + d a & c & a + b + c \end{matrix} ∣ ∣ ∣ ∣$
$= ∣ ∣ ∣ ∣ \begin{matrix} a & c & a + c b & d & b + d o & o & b \end{matrix} ∣ ∣ ∣ ∣ R_{3} \to R_{3} - R_{1}$
$= b (a a - b c)$
Now $\frac{D_{1}}{D_{2}} = \frac{- 2 b (a d - b c)}{b (a d - b c)}$
$= - 2$

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Similar questions

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}}

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} ∣ ∣ ∣ ∣

, then the value of

∣ ∣ ∣ \frac{D_{1}}{D_{2}} ∣ ∣ ∣

b \neq 0

a d \neq b c

, is ............

Q. Equal masses of two liquids having densities

d_{1}

d_{2}

are mixed together. The density of the resulting mixture is

Q. If the determinant

D =

∣ ∣ ∣ ∣ ∣ \begin{matrix} 1 & 1 & 1 α + β & α^{2} + β^{2} & 2 α β α + β & 2 α β & α^{2} + β^{2} \end{matrix} ∣ ∣ ∣ ∣ ∣

D_{1} =

∣ ∣ ∣ ∣ \begin{matrix} 1 & 0 & 0 0 & α & β 0 & β & α \end{matrix} ∣ ∣ ∣ ∣

, then find the determinant of

D_{2}

D_{2} = \frac{D}{D_{1}}

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