If $A=\left|\begin{array}{ccc}a_{1} & b_{1} & c_{1} \\ a_{2} & b_{2} & c_{2} \\ a_{3} & b_{3} & c_{3}\end{array}\right|$ and $B=\left|\begin{array}{ccc}c_{1} & c_{2} & c_{3} \\ a_{1} & a_{2} & a_{3} \\ b_{1} & b_{2} & b_{3}\end{array}\right|$,then

Let $A = [a_{ij}]$ and $B = [b_{ij}]$ be two $3 \times 3$ real matrices such that $b_{ij} = (3)^{(i+j-2)} a_{ji}$,where $i, j = 1, 2, 3$. If the determinant of $B$ is $81$,then the determinant of $A$ is:

Show that $\Delta=\left|\begin{array}{ccc} (y+z)^{2} & x y & z x \\ x y & (x+z)^{2} & y z \\ x z & y z & (x+y)^{2} \end{array}\right|=2 x y z(x+y+z)^{3}$

Consider the following statements :
$(a)$ If any two rows or columns of a determinant are identical,then the value of the determinant is zero.
$(b)$ If the corresponding rows and columns of a determinant are interchanged,then the value of the determinant does not change.
$(c)$ If any two rows (or columns) of a determinant are interchanged,then the value of the determinant changes in sign.
Which of these are correct?

Using properties of determinants,prove that:
$\left|\begin{array}{ccc}3 a & -a+b & -a+c \\ -b+a & 3 b & -b+c \\ -c+a & -c+b & 3 c\end{array}\right|=3(a+b+c)(a b+b c+c a)$

Let $D_1 = \begin{vmatrix} a & b & a+b \\ c & d & c+d \\ a & b & a-b \end{vmatrix}$ and $D_2 = \begin{vmatrix} a & c & a+c \\ b & d & b+d \\ a & c & a+b+c \end{vmatrix}$. Then the value of $\frac{D_1}{D_2}$,where $b \neq 0$ and $ad \neq bc$,is: