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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