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

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 ( $\Delta$ ),

{{{\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