If D 1 - - - 1 x y z- 1 y z x- 1 z x y - and D 2 - - - 1 1 1- x y z- x 2 y 2 z 2 - then find relation between D1 and D2-

D_1= #160; 1 amp; #160; x amp; yz #160; 1 amp; #160; y amp; #160; zx #160; 1 amp; #160; z amp; #160; xy multiply 1st ...

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Byju's Answer

Standard XII

Mathematics

Properties of Determinants

If D 1 = |...

Question

If $D_{1} = ∣ ∣ ∣ ∣ \begin{matrix} 1 & x & y z 1 & y & z x 1 & z & x y \end{matrix} ∣ ∣ ∣ ∣$ and $D_{2} = ∣ ∣ ∣ ∣ ∣ \begin{matrix} 1 & 1 & 1 x & y & z x^{2} & y^{2} & z^{2} \end{matrix} ∣ ∣ ∣ ∣ ∣$ then find relation between $D_{1}$ and $D_{2}$ .

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Solution

$D_{1} = ∣ ∣ ∣ ∣ \begin{matrix} 1 & x & y z 1 & y & z x 1 & z & x y \end{matrix} ∣ ∣ ∣ ∣$
multiply 1st row by $x$ and

divide 2 nd row by $y$

by 3rd row by $z$

$= \frac{1}{x y z} ∣ ∣ ∣ ∣ ∣ \begin{matrix} x & x^{2} & x y z y & y^{2} & x y z z & z^{2} & x y z \end{matrix} ∣ ∣ ∣ ∣ ∣$

Take $x y z$ from 3rd column

$= \frac{x y z}{x y z} ∣ ∣ ∣ ∣ ∣ \begin{matrix} x & x^{2} & 1 y & y^{2} & 1 z & z^{2} & 1 \end{matrix} ∣ ∣ ∣ ∣ ∣$

[column are changed to rows but determinant will not change]

$= ∣ ∣ ∣ ∣ ∣ \begin{matrix} x & y & z x^{2} & y^{2} & z^{2} 1 & 1 & 1 \end{matrix} ∣ ∣ ∣ ∣ ∣$

Interchanged $C_{3}$ and $C_{1}$ and again $C_{2}$ and $C_{3}$

$= ∣ ∣ ∣ ∣ ∣ \begin{matrix} 1 & 1 & 1 x^{2} & y^{2} & z^{2} x & y & z \end{matrix} ∣ ∣ ∣ ∣ ∣$

$C_{3} \leftrightarrow C_{1}$

$= (- 1) (- 1) ∣ ∣ ∣ ∣ ∣ \begin{matrix} 1 & 1 & 1 x & y & z x^{2} & y^{2} & z^{2} \end{matrix} ∣ ∣ ∣ ∣ ∣ \Rightarrow D_{1} = D_{2}$

[When you interchange two column/rows determinant will change its sign]

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