Question:

Given the determinant: $$ \begin{vmatrix}a&b&c\\x&y&z\\p&q&r\end{vmatrix}=k $$ We want to evaluate the determinant: $$ D = \begin{vmatrix}6x&2y&2z\\-3p&-q&-r\\3a&b&c\end{vmatrix} $$ Step 1: Factor out common multipliers from each row. Factor 2 from the first row: $$ D = 2 \begin{vmatrix}3x&y&z\\-3p&-q&-r\\3a&b&c\end{vmatrix} $$ Factor -1 from the second row: $$ D = 2 \cdot (-1) \begin{vmatrix}3x&y&z\\3p&q&r\\3a&b&c\end{vmatrix} $$ $$ D = -2 \begin{vmatrix}3x&y&z\\3p&q&r\\3a&b&c\end{vmatrix} $$ Step 2: Factor out common multipliers from columns. Factor 3 from the first column: $$ D = -2 \cdot 3 \begin{vmatrix}x&y&z\\p&q&r\\a&b&c\end{vmatrix} $$ $$ D = -6 \begin{vmatrix}x&y&z\\p&q&r\\a&b&c\end{vmatrix} $$ Step 3: Relate the resulting determinant to the given determinant $k$. Let $D' = \begin{vmatrix}x&y&z\\p&q&r\\a&b&c\end{vmatrix}$. The given determinant is $k = \begin{vmatrix}a&b&c\\x&y&z\\p&q&r\end{vmatrix}$. To transform $k$ into $D'$, we perform row swaps: 1. Swap Row 1 and Row 2 of $k$: $$ \begin{vmatrix}x&y&z\\a&b&c\\p&q&r\end{vmatrix} = -k $$ 2. Swap Row 2 and Row 3 of the resulting determinant: $$ \begin{vmatrix}x&y&z\\p&q&r\\a&b&c\end{vmatrix} = -(-k) = k $$ Thus, $D' = k$. Step 4: Substitute the value of $D'$ back into the expression for $D$. $$ D = -6 \cdot k $$ $$ D = -6k $$