. Let $D1$ = det $\begin{pmatrix} a & b & c \\ x & y & z \\ p & q & r \\ \end{pmatrix}$ and $D2$ = det$\begin{pmatrix} -x & a & -p \\ y & -b & q \\ z & -c & r \\ \end{pmatrix} $
$(i)D_1=D_2\ \ \ \ \ \ $ $(ii)D_1=2D_2\ \ \ \ $ $(iii)D_1=-D_2\ \ \ \ $ $(iv)D_2=2D_1\ \ \ \ $
I am trying to find out the relation between given deteriminants so I am using some properties of determinants on $D_2$
$D2$ = $\begin{vmatrix} -x & a & -p \\ y & -b & q \\ z & -c & r \\ \end{vmatrix} \implies (-1)\begin{vmatrix} x & -a & p \\ y & -b & q \\ z & -c & r \\ \end{vmatrix} \text{Took $(-1)$ common from first row}$
$\implies (1)\begin{vmatrix} x & a & p \\ y & b & q \\ z & c & r \\ \end{vmatrix} \text{Took $(-1)$ common from $2nd$ column}$
$\implies(-1)\begin{vmatrix} a & x & p \\ b & y & q \\ c & z & r \\ \end{vmatrix} \text{interchanged first and 2nd column}$
$\implies(-1)D_1$
I don't have an answer available for this question so I want to know if I did it right? Can anyone tell me any other way of solving this? I Just wanna know any other method.