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Question

If $$D _ { 1 } = \left| \begin{array} { r r r } { 1 } & { x } & { y z } \\ { 1 } & { y } & { z x } \\ { 1 } & { z } & { x y } \end{array} \right|$$ and $$D _ { 2 } = \left| \begin{array} { c c c } { 1 } & { 1 } & { 1 } \\ { x } & { y } & { z } \\ { x ^ { 2 } } & { y ^ { 2 } } & { z ^ { 2 } } \end{array} \right|$$ then find relation between $$D_1$$ and $$D_2$$.


Solution

$$D_1=\begin{vmatrix}  1&  x& yz \\  1&  y&  zx\\  1&  z&  xy\end{vmatrix}$$
multiply 1st row by $$x$$ and 

divide 2 nd row by $$y$$

by 3rd row by $$z$$

$$=\dfrac{1}{xyz}\begin{vmatrix}  x&  x^2&  xyz\\  y&  y^2& xyz \\  z&  z^2&  xyz\end{vmatrix}$$

Take $$xyz$$ from 3rd column

$$=\dfrac{xyz}{xyz}\begin{vmatrix}  x&  x^2& 1 \\ y & y^2 & 1 \\ z &  z^2&  1\end{vmatrix}$$

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

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

Interchanged $$C_3$$ and $$C_1$$ and again $$C_2$$ and $$C_3$$

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

$$C_3\leftrightarrow C_1$$

$$=(-1)(-1)\begin{vmatrix} 1 & 1 &  1\\  x&  y&  z\\  x^2&  y^2&  z^2\end{vmatrix}\Rightarrow D_1=D_2$$

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

Mathematics

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