If \( \begin{vmatrix}2a & 2a & 3b+c\\3b & 2a+c & 3b\\2a+3b & c & c\end{vmatrix}=48 \),then \( abc \) is equal to
Answer & explanation
Correct answer: option 4
The correct answer is Option (4) → −2
The row operation \( R_1\rightarrow R_1+R_2-R_3 \) does not change the determinant and puts a zero in the first row:
\( \displaystyle R_1:\ (2a+3b-2a-3b,\ 2a+2a+c-c,\ 3b+c+3b-c)=(0,\ 4a,\ 6b) \)
\( \displaystyle \Delta=\begin{vmatrix}0 & 4a & 6b\\3b & 2a+c & 3b\\2a+3b & c & c\end{vmatrix} \)
Expanding along the first row:
\( \displaystyle \Delta=-4a\begin{vmatrix}3b & 3b\\2a+3b & c\end{vmatrix}+6b\begin{vmatrix}3b & 2a+c\\2a+3b & c\end{vmatrix} \)
\( \displaystyle =-4a\left[3bc-3b(2a+3b)\right]+6b\left[3bc-(2a+c)(2a+3b)\right] \)
\( \displaystyle =-4a(3bc-6ab-9b^2)+6b(3bc-4a^2-6ab-2ac-3bc) \)
\( \displaystyle =-12abc+24a^2b+36ab^2-24a^2b-36ab^2-12abc \)
\( \displaystyle =-24abc \)
So \( -24abc=48 \), which gives \( abc=-2 \).