Explanation:
(B) : Given,
\(\left|\begin{array}{ccc} 4 x & 6 x+2 & 8 x+1 \\ 6 x+2 & 9 x+3 & 12 x \\ 8 x+1 & 12 x & 16 x+2 \end{array}\right|=0\)
Operating \(\mathrm{C}_{3 \rightarrow} \mathrm{C}_{3}-2 \mathrm{C}_{1}\)
\(\left|\begin{array}{ccc} 4 x & 6 x+2 & 1 \\ 6 x+2 & 9 x+3 & -4 \\ 8 x+1 & 12 x & 0 \end{array}\right|=0\)
Now, taking 2 common from \(\mathrm{C}_{1}\), we get
\(\left|\begin{array}{ccc} 2 x & 6 x+2 & 1 \\ 3 x+1 & 9 x+3 & -4 \\ 4 x+\frac{1}{2} & 12 x & 0 \\ \operatorname{ng} C_{2} \xrightarrow[\rightarrow]{ } C_{2}-3 C_{1} \end{array}\right|=0\)
Operating \(\mathrm{C}_{2} \rightarrow \mathrm{C}_{2}-3 \mathrm{C}\)
\(\left|\begin{array}{ccc} 2 x & 2 & 1 \\ 3 x+1 & 0 & -4 \\ 4 x+\frac{1}{2} & \frac{-3}{2} & 0 \end{array}\right|=0\)
Multiply by 2 in column 2 , we get
\(\left|\begin{array}{ccc} 2 x & 4 & 1 \\ 3 x+1 & 0 & -4 \\ 4 x+\frac{1}{2} & -3 & 0 \end{array}\right|=0\)
Operating \(\mathrm{R}_{2} \xrightarrow{2} \mathrm{R}_{2}+4 \mathrm{R}_{1}\)
\(\left|\begin{array}{ccc} 2 x & 4 & 1 \\ 11 x+1 & 16 & 0 \\ 4 x+\frac{1}{2} & -3 & 0 \end{array}\right|=0\)
\(1(-33 x-3)-64 x-8\)
\(-33 x-3-64 x-8=0\)
\(-97 x-11=0\)
\(x=\frac{-11}{97}\)
\(\text { Now, } \quad 1(-33 x-3)-64 x-8=0\)