If Delta = begin{vmatrix} a & b & c x & y & | vedclass

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(D) By the property of determinants,if each element of a row (or column) of a determinant is multiplied by a constant $k$,then its value gets multipli

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$k^3\Delta$

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(D) By the property of determinants,if each element of a row (or column) of a determinant is multiplied by a constant $k$,then its value gets multiplied by $k$. Since there are $3$ rows in the given $3 	imes 3$ determinant,we can factor out $k$ from each of the three rows. $\begin{vmatrix} ka & kb & kc \ kx & ky & kz \ kp & kq & kr \end{vmatrix} = k \cdot k \cdot k \begin{vmatrix} a & b & c \ x & y & z \ p & q & r \end{vmatrix} = k^3 \Delta$.

If $\Delta = \begin{vmatrix} a & b & c \\ x & y & z \\ p & q & r \end{vmatrix}$,then $\begin{vmatrix} ka & kb & kc \\ kx & ky & kz \\ kp & kq & kr \end{vmatrix}$ =

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