Question

Find the value of $\left| {\begin{array}{*{20}{c}}
  1&{{{\log }_b}a} \\
  {{{\log }_a}b}&1
\end{array}} \right| = \ldots $

Answer 1

Hint: For calculating the determinant of a matrix, first of all the given matrix should be a square matrix. If it is a square matrix and its order is $2 \times 2$, then its determinant is given by
$A = \left[ {\begin{array}{*{20}{c}}
  a&b \\
  c&d
\end{array}} \right]$
$\left| A \right| = ad - bc$
Further we will be using the formula \[{\log _b}a = \dfrac{{\log a}}{{\log b}}\] for getting the value of determinant.

Complete step-by-step solution:
In this question, we are given a matrix and we need to find its determinant value.
Now, the given matrix is a $2 \times 2$ square matrix.
The determinant is used for finding the inverse of the given matrix, used for telling things that are useful in systems of linear equations and also in calculus.
A determinant is denoted by $\left| A \right|$.
Calculating the value of determinant:
First of all, for calculating the value of a determinant, the given matrix must be a square matrix. That means it must have the same number of rows and columns. Then we just need to carry basic arithmetic operations.
For example: Let us take a $2 \times 2$ square matrix.
$A = \left[ {\begin{array}{*{20}{c}}
  a&b \\
  c&d
\end{array}} \right]$
Therefore, its determinant is given by
$\left| A \right| = ad - bc$
Here, our matrix is $A = \left[ {\begin{array}{*{20}{c}}
  1&{{{\log }_b}a} \\
  {{{\log }_a}b}&1
\end{array}} \right]$
So, here $a = 1,b = {\log _b}a,c = {\log _a}b,d = 1$
Hence, its determinant is given by
$
   \Rightarrow \left| A \right| = \left( 1 \right)\left( 1 \right) - \left( {{{\log }_b}a} \right)\left( {{{\log }_a}b} \right) \\
   \Rightarrow \left| A \right| = 1 - \left( {{{\log }_b}a} \right)\left( {{{\log }_a}b} \right) \\
 $
Now, we know that \[{\log _b}a = \dfrac{{\log a}}{{\log b}}\]. Therefore, above equation becomes
$ \Rightarrow \left| A \right| = 1 - \left( {\dfrac{{\log a}}{{\log b}}} \right)\left( {\dfrac{{\log b}}{{\log a}}} \right)$
Here, both $\log a$ and $\log b$ gets cancelled. Hence, we get
\[
   \Rightarrow \left| A \right| = 1 - 1 \\
   \Rightarrow \left| A \right| = 0 \\
 \]
Hence, the determinant value of $\left| {\begin{array}{*{20}{c}}
  1&{{{\log }_b}a} \\
  {{{\log }_a}b}&1
\end{array}} \right|$ is equal to 0.

Note: Calculating the determinant of a $3 \times 3$ matrix:
Example: Let us take a $3 \times 3$ matrix.
$A = \left[ {\begin{array}{*{20}{c}}
  a&b&c \\
  d&e&f \\
  g&h&i
\end{array}} \right]$
So, its determinant is given by
$\left| A \right| = a\left( {ei{\text{ }} - {\text{ }}fh} \right){\text{ }} - {\text{ }}b\left( {di{\text{ }} - {\text{ }}fg} \right){\text{ }} + {\text{ }}c\left( {dh{\text{ }} - {\text{ }}eg} \right)$