Question

If $a,b,c$ are non-zero and different from $1$, then the value of $\left|\begin{array}{ccc}{\log }_{a}1& {\log }_{a}b& {\log }_{a}c\\ {\log }_a\left(\frac{1}{b}\right)& {\log }_{b}1& {\log }_a\left(\frac{1}{c}\right)\\ {\log }_a\left(\frac{1}{c}\right)& {\log }_{a}c& {\log }_{c}1\end{array}\right|$ is

• A. $0$
• B. $1+{\log }_a(a+b+c)$
• C. ${\log }_a(ab+bc+ca)$
• D. $1$
• E. ${\log }_a(a+b+c)$

Answer 1

The correct option is A $0$

$\Delta =\left|\begin{array}{ccc}{\log }_{a}1& {\log }_{a}b& {\log }_{a}c\\ {\log }_a\frac{1}{b}& {\log }_{b}1& {\log }_a\frac{1}{c}\\ {\log }_a\frac{1}{c}& {\log }_{a}c& {\log }_{c}1\end{array}\right|\Rightarrow \Delta =\left|\begin{array}{ccc}0& {\log }_{a}b& {\log }_{a}c\\ -{\log }_{a}b& 0& -{\log }_{a}c\\ -{\log }_{a}c& {\log }_{a}c& 0\end{array}\right|\Rightarrow \Delta =0-{\log }_{a}b\{0-{({\log }_{a}c)}^2\}+{\log }_{a}c\{-{\log }_{a}b{\log }_{a}c-0\}\Rightarrow \Delta ={\log }_{a}b{({\log }_{a}c)}^2-{\log }_{a}b{({\log }_{a}c)}^2=0$

So, option A is correct.