Question

If a, b, c are non-zero and different from 1, then the value of \[\left| {\begin{array}{*{20}{c}}
  {{{\log }_a}1}&{{{\log }_a}b}&{{{\log }_a}c} \\
  {{{\log }_a}(\dfrac{1}{b})}&{{{\log }_b}1}&{{{\log }_a}(\dfrac{1}{c})} \\
  {{{\log }_a}(\dfrac{1}{c})}&{{{\log }_a}c}&{{{\log }_c}1}
\end{array}} \right|\] is
A) 0
B) \[1 + {\log _a}(a + b + c)\]
C) \[{\log _a}(a + b + c)\]
D) 1

Answer 1

Hint: Determinant means a matrix is an array of many numbers. We will assign the given determinant as delta. Then we will use the value i.e. log 1 = 0. Also, we will use the division rule i.e. \[{\log _a}(\dfrac{m}{n}) = {\log _a}m - {\log _a}n\]. Next, we will substitute all the values in the given determinant. Thus, solving the determinant we will get the final output.

Complete step by step solution:
Let the given determinant be delta ( \[\Delta \] ),
\[\therefore \Delta = \left| {\begin{array}{*{20}{c}}
  {{{\log }_a}1}&{{{\log }_a}b}&{{{\log }_a}c} \\
  {{{\log }_a}(\dfrac{1}{b})}&{{{\log }_b}1}&{{{\log }_a}(\dfrac{1}{c})} \\
  {{{\log }_a}(\dfrac{1}{c})}&{{{\log }_a}c}&{{{\log }_c}1}
\end{array}} \right|\]
As we know, log 1= 0 with any base.
\[\therefore {\log _a}1 = 0\] , \[{\log _b}1 = 0\] and \[{\log _c}1 = 0\]
Substituting the above value in the given determinant, we will get,
\[ \Rightarrow \Delta = \left| {\begin{array}{*{20}{c}}
  0&{{{\log }_a}b}&{{{\log }_a}c} \\
  {{{\log }_a}(\dfrac{1}{b})}&0&{{{\log }_a}(\dfrac{1}{c})} \\
  {{{\log }_a}(\dfrac{1}{c})}&{{{\log }_a}c}&0
\end{array}} \right|\]
As we know, \[{\log _a}(\dfrac{m}{n}) = {\log _a}m - {\log _a}n\].
Applying this division rule, we will get the values as below:

First,
\[{\log _a}(\dfrac{1}{b})\]
\[ = {\log _a}1 - {\log _a}b\]
\[ = - {\log _a}b\]

And,
\[{\log _a}(\dfrac{1}{c})\]
\[ = {\log _a}1 - {\log _a}c\]
\[ = - {\log _a}c\]

Now, substituting these above values, we will get,
\[ \Rightarrow \Delta = \left| {\begin{array}{*{20}{c}}
  0&{{{\log }_a}b}&{{{\log }_a}c} \\
  { - {{\log }_a}b}&0&{ - {{\log }_a}c} \\
  { - {{\log }_a}c}&{{{\log }_a}c}&0
\end{array}} \right|\]
On evaluating this above determinant, we will get,
\[ \Rightarrow \Delta = 0 - {\log _a}b\{ 0 - {({\log _a}c)^2}\} + {\log _a}c\{ {\log _a}c( - {\log _a}b) - 0( - {\log _a}c)\} \]
\[ \Rightarrow \Delta = - {\log _a}b\{ 0 - {({\log _a}c)^2}\} + {\log _a}c\{ - {\log _a}c({\log _a}b) - 0\} \]
Removing the brackets, we will get,
\[ \Rightarrow \Delta = 0 + {\log _a}b{({\log _a}c)^2} - {\log _a}b{({\log _a}c)^2} - 0\]
\[ \Rightarrow \Delta = {\log _a}b{({\log _a}c)^2} - {\log _a}b{({\log _a}c)^2}\]
\[ \Rightarrow \Delta = 0\]

Hence, the value of the given determinant \[\left| {\begin{array}{*{20}{c}}
  {{{\log }_a}1}&{{{\log }_a}b}&{{{\log }_a}c} \\
  {{{\log }_a}(\dfrac{1}{b})}&{{{\log }_b}1}&{{{\log }_a}(\dfrac{1}{c})} \\
  {{{\log }_a}(\dfrac{1}{c})}&{{{\log }_a}c}&{{{\log }_c}1}
\end{array}} \right| = 0\].

Note:
> The given matrix is in the form of a skew symmetric matrix. For any skew symmetric matrix $A$, $A=-A^T$, the determinant will be always zero.
> A logarithm of a number with a base is equal to another number. A logarithm is just the opposite function of exponentiation. A logarithm has various important properties that prove multiplication and division of logarithms can also be written in the form of logarithm of addition and subtraction respectively. Also, log 0 is undefined. Because, we never get the value 0, by raising any value to the power of anything else. The two most common types of logarithms are:-
1) Common Logarithm (or) Base 10 Logarithm
2) Natural Logarithm (or) Base e Logarithm