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Derivative of Some Standard Functions

If Δ1=x-y-z...

Question

If $Δ_{1} = ∣ ∣ ∣ ∣ \begin{matrix} x - y - z & 2 x & 2 x 2 y & y - z - x & 2 y 2 z & 2 z & z - x - y \end{matrix} ∣ ∣ ∣ ∣$ and $Δ_{2} = ∣ ∣ ∣ ∣ \begin{matrix} x + y + 2 z & x & y z & y + z + 2 x & y z & x & z + x + 2 y \end{matrix} ∣ ∣ ∣ ∣$ then

A

Δ_{1} = 2 Δ_{2}

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B

Δ_{2} = 2 Δ_{1}

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C

Δ_{1} = Δ_{2}

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D

none of these

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Solution

The correct option is A $Δ_{2} = 2 Δ_{1}$
Consider, $Δ_{1} = ∣ ∣ ∣ ∣ \begin{matrix} x - y - z & 2 x & 2 x 2 y & y - z - x & 2 y 2 z & 2 z & z - x - y \end{matrix} ∣ ∣ ∣ ∣$
$C_{1} \to C_{1} - C_{2}, C_{2} \to C_{2} - C_{3}$
$Δ_{1} = ∣ ∣ ∣ ∣ ∣ \begin{matrix} - (x + y + z) & 0 & 2 x x + y + z & - (x + y + z) & 2 y 0 & x + y + z & z - x - y \end{matrix} ∣ ∣ ∣ ∣ ∣$
$Δ_{1} = {(x + y + z)}^{2} ∣ ∣ ∣ ∣ \begin{matrix} - 1 & 0 & 2 x 1 & - 1 & 2 y 0 & 1 & z - x - y \end{matrix} ∣ ∣ ∣ ∣$
$R_{2} \to R_{2} + R_{3}$
$Δ_{1} = {(x + y + z)}^{2} ∣ ∣ ∣ ∣ \begin{matrix} - 1 & 0 & 2 x 1 & 0 & y + z - x 0 & 1 & z - x - y \end{matrix} ∣ ∣ ∣ ∣$
$\Rightarrow Δ_{1} = (x + y + z)^{3}$
Now, $Δ_{2} = ∣ ∣ ∣ ∣ \begin{matrix} x + y + 2 z & x & y z & y + z + 2 x & y z & x & z + x + 2 y \end{matrix} ∣ ∣ ∣ ∣$
$C_{1} \to C_{1} + C_{2} + C_{3}$
$Δ_{2} = ∣ ∣ ∣ ∣ ∣ \begin{matrix} 2 (x + y + z) & x & y 2 (x + y + z) & y + z + 2 x & y 2 (x + y + z) & x & z + x + 2 y \end{matrix} ∣ ∣ ∣ ∣ ∣$
$Δ_{2} = 2 (x + y + z) ∣ ∣ ∣ ∣ \begin{matrix} 1 & x & y 1 & y + z + 2 x & y 1 & x & z + x + 2 y \end{matrix} ∣ ∣ ∣ ∣$
$R_{1} \to R_{1} - R_{2,} R_{2} \to R_{2} - R_{3}$
$Δ_{2} = 2 (x + y + z) ∣ ∣ ∣ ∣ ∣ \begin{matrix} 0 & - (x + y + z) & 0 0 & x + y + z & - (x + y + z) 1 & x & z + x + 2 y \end{matrix} ∣ ∣ ∣ ∣ ∣$
$\Rightarrow Δ_{2} = 2 (x + y + z)^{3}$
$\Rightarrow Δ_{2} = 2 Δ_{1}$

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Similar questions

∆_{1} = |\begin{matrix} 1 & 1 & 1 \\ a & b & c \\ a^{2} & b^{2} & c^{2} \end{matrix}|, ∆_{2} = |\begin{matrix} 1 & b c & a \\ 1 & c a & b \\ 1 & a b & c \end{matrix}|, then

∆_{1} + ∆_{2} = 0

∆_{1} + 2 ∆_{2} = 0

∆_{1} = ∆_{2}

(d) none of these

Δ_{1} = ∣ ∣ ∣ ∣ \begin{matrix} x & sin θ & cos θ - sin θ & - x & 1 cos θ & 1 & x \end{matrix} ∣ ∣ ∣ ∣

Δ_{2} = ∣ ∣ ∣ ∣ \begin{matrix} x & sin 2 θ & cos 2 θ - sin 2 θ & - x & 1 cos 2 θ & 1 & x \end{matrix} ∣ ∣ ∣ ∣

x \neq 0

θ \in (0, \frac{π}{2})

If

$Δ_{1}$ , $∣ ∣ ∣ ∣ \begin{matrix} 1 & 1 & 1 a & b & c a^{2} & b^{2} & c^{2} \end{matrix} ∣ ∣ ∣ ∣$ , $Δ_{2}$ = $∣ ∣ ∣ ∣ \begin{matrix} 1 & b c & a 1 & c a & b 1 & a b & c \end{matrix} ∣ ∣ ∣ ∣$ then

Δ_{1} = ∣ ∣ ∣ ∣ \begin{matrix} 7 & x & 2 - 5 & x + 1 & 3 4 & x & 7 \end{matrix} ∣ ∣ ∣ ∣, Δ_{2} = ∣ ∣ ∣ ∣ \begin{matrix} x & 2 & 7 x + 1 & 3 & - 5 x & 7 & 4 \end{matrix} ∣ ∣ ∣ ∣

Δ_{1} - Δ_{2} = 0

Δ_{1} = ∣ ∣ ∣ ∣ ∣ \begin{matrix} 1 & cos α & cos β cos α & 1 & cos γ cos β & cos γ & 1 \end{matrix} ∣ ∣ ∣ ∣ ∣

Δ_{2} = ∣ ∣ ∣ ∣ ∣ \begin{matrix} 0 & cos α & cos β cos α & 0 & cos γ cos β & cos γ & 0 \end{matrix} ∣ ∣ ∣ ∣ ∣

Δ_{1} = Δ_{2}

{sin}^{2} α + {sin}^{2} β + {sin}^{2} γ

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