If x- y- z are in GP-then using properties of determinants-show that Delta-lbeginvmatrix p x-y - x - y 1p y-z - y - z 0 - p x-y - p y-zend vmatrix -0- where y-y-z and p is any real number-

Let Δ= px+y amp;x amp;y py+z amp; y amp; z 0 amp; px+y amp; py+z By C_1→ C_1 pC_1 C_3,Δ = 0 amp;x amp;y 0 amp; y amp; z ...

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Byju's Answer

Standard XII

Chemistry

Gibb's Energy and Nernst Equation

If x, y, z ar...

Question

If x,y,z are in GP,then using properties of determinants,show that $Δ = ∣ ∣ ∣ ∣ \begin{matrix} p x + y & x & y p y + z & y & z 0 & p x + y & p y + z \end{matrix} ∣ ∣ ∣ ∣$ =0,where $x \neq y \neq z$ and p is any real number.

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Solution

Let $Δ = ∣ ∣ ∣ ∣ \begin{matrix} p x + y & x & y p y + z & y & z 0 & p x + y & p y + z \end{matrix} ∣ ∣ ∣ ∣$
By $C_{1} \to C_{1} - p C_{1} - C_{3}, Δ = ∣ ∣ ∣ ∣ ∣ \begin{matrix} 0 & x & y 0 & y & z - p^{2} x - p y - p y - z & p x + y & p y + z \end{matrix} ∣ ∣ ∣ ∣ ∣$
Expanding along $C_{1}, Δ = (- p^{2} x - 2 p y - z) (x z - y^{2})$
$∵$ x,y,z are in GP,so $x z = y^{2} \Rightarrow x z - y^{2} = 0 ∴ Δ = 0$

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Similar questions

Q. If x,y,z are in GP, using properties of determinants, show that

∣ ∣ ∣ ∣ \begin{matrix} p x + y & x & y p y + z & y & z 0 & p x + y & p y + z \end{matrix} ∣ ∣ ∣ ∣ = 0

, where

x \neq y \neq z

and p is any real number.

Using properties of determinant, prove that:

∣ ∣ ∣ ∣ ∣ \begin{matrix} x & x^{2} & 1 + {p x}^{3} y & y^{2} & 1 + {p y}^{3} z & z^{2} & 1 + {p z}^{3} \end{matrix} ∣ ∣ ∣ ∣ ∣

= (1 + p x y z) (x - y) (y - z) (z - x)

Q. Using properties of determinants, prove that

ω ∣ ∣ ∣ ∣ ∣ \begin{matrix} x & y & z x^{2} & y^{2} & z^{2} y + z & z + x & x + y \end{matrix} ∣ ∣ ∣ ∣ ∣ = (x - y) (y - z) (z - x) (x + y + z)

Using the properties of determinants, show that:

∣ ∣ ∣ ∣ ∣ \begin{matrix} x & x^{2} & y z y & y^{2} & z x z & z^{2} & x y \end{matrix} ∣ ∣ ∣ ∣ ∣ = (x - y) (y - z) (z - x) (x y + y z + z x)

Using properties of determinants prove the following questions.

$∣ ∣ ∣ ∣ ∣ \begin{matrix} x & x^{2} & 1 + p x^{3} y & y^{2} & 1 + p y^{3} z & z^{2} & 1 + p z^{3} \end{matrix} ∣ ∣ ∣ ∣ ∣ = (1 + p x y z) (x - y) (y - z) (z - x)$ , where p is any scalar.

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If x,y,z are in GP,then using properties of determinants,show that Δ=∣∣ ∣∣px+yxypy+zyz0px+ypy+z∣∣ ∣∣=0,where x≠y≠z and p is any real number.

Let Δ=∣∣ ∣∣px+yxypy+zyz0px+ypy+z∣∣ ∣∣ By C1→C1−pC1−C3,Δ=∣∣ ∣ ∣∣0xy0yz−p2x−py−py−zpx+ypy+z∣∣ ∣ ∣∣ Expanding along C1,Δ=(−p2x−2py−z)(xz−y2) ∵ x,y,z are in GP,so xz=y2⇒xz−y2=0 ∴Δ=0

If x,y,z are in GP,then using properties of determinants,show that $Δ = ∣ ∣ ∣ ∣ \begin{matrix} p x + y & x & y p y + z & y & z 0 & p x + y & p y + z \end{matrix} ∣ ∣ ∣ ∣$ =0,where $x \neq y \neq z$ and p is any real number.