Question

If \[{a_1},{a_2},{a_3},{\rm{ }}....{a_n},....\] are in GP, then the value of the determinant \[\left| {\begin{array}{*{20}{l}}{\log {a_n}}&{\log {a_{n + 1}}}&{\log {a_{n + 2}}}\\{\log {a_{n + 3}}}&{\log {a_{n + 4}}}&{\log {a_{n + 5}}}\\{\log {a_{n + 6}}}&{\log {a_{n + 7}}}&{\log {a_{n + 8}}}\end{array}} \right|\], is
A) \[0\]
B) \[1\]
C) \[2\]
D) \[ - 2\]

Answer 1

Hint:
Here, we will use the basic concept of the Geometric progression (G.P.) as the series given is in GP form. In this question, we will apply the basic matrix operation to simplify the given matrix, and then by solving the matrix we will be able to get the value of the determinant of the matrix.

Complete step by step solution:
It is given that \[{a_1},{a_2},{a_3},{\rm{ }}....{a_n},....\]is in GP.
 Let, \[a\] be the first term of the G.P. and \[r\] is the common ratio.
We know that the G.P. series is in the form of\[a,ar,a{r^2},a{r^3},a{r^4},.........\]
Therefore, we can write\[{a_n} = a{r^n}\]or\[{a_{n + 1}} = a{r^{n + 1}}\].
So the matrix becomes
\[\left| {\begin{array}{*{20}{l}}{\log {a_n}}&{\log {a_{n + 1}}}&{\log {a_{n + 2}}}\\{\log {a_{n + 3}}}&{\log {a_{n + 4}}}&{\log {a_{n + 5}}}\\{\log {a_{n + 6}}}&{\log {a_{n + 7}}}&{\log {a_{n + 8}}}\end{array}} \right| = \left| {\begin{array}{*{20}{l}}{\log a{r^n}}&{\log a{r^{n + 1}}}&{\log a{r^{n + 2}}}\\{\log a{r^{n + 3}}}&{\log a{r^{n + 4}}}&{\log a{r^{n + 5}}}\\{\log a{r^{n + 6}}}&{\log a{r^{n + 7}}}&{\log a{r^{n + 8}}}\end{array}} \right|\]
Simplifying the above matrix, we get
\[ \Rightarrow \left| {\begin{array}{*{20}{l}}{\log {a_n}}&{\log {a_{n + 1}}}&{\log {a_{n + 2}}}\\{\log {a_{n + 3}}}&{\log {a_{n + 4}}}&{\log {a_{n + 5}}}\\{\log {a_{n + 6}}}&{\log {a_{n + 7}}}&{\log {a_{n + 8}}}\end{array}} \right| = \left| {\begin{array}{*{20}{l}}{\log {r^n}}&{\log {r^{n + 1}}}&{\log {r^{n + 2}}}\\{\log {r^{n + 3}}}&{\log {r^{n + 4}}}&{\log {r^{n + 5}}}\\{\log {r^{n + 6}}}&{\log {r^{n + 7}}}&{\log {r^{n + 8}}}\end{array}} \right|\]
Now we will apply the basic property of logarithmic function i.e. \[\log {a^b} = b\log a\]. Therefore, we get
\[ \Rightarrow \left| {\begin{array}{*{20}{l}}{\log {a_n}}&{\log {a_{n + 1}}}&{\log {a_{n + 2}}}\\{\log {a_{n + 3}}}&{\log {a_{n + 4}}}&{\log {a_{n + 5}}}\\{\log {a_{n + 6}}}&{\log {a_{n + 7}}}&{\log {a_{n + 8}}}\end{array}} \right| = \left| {\begin{array}{*{20}{l}}{n\log r}&{\left( {n + 1} \right)\log r}&{\left( {n + 2} \right)\log r}\\{\left( {n + 3} \right)\log r}&{\left( {n + 4} \right)\log r}&{\left( {n + 5} \right)\log r}\\{\left( {n + 6} \right)\log r}&{\left( {n + 7} \right)\log r}&{\left( {n + 8} \right)\log r}\end{array}} \right|\]
Now taking\[\log r\]common from all the terms of the matrix, we get
\[ \Rightarrow \left| {\begin{array}{*{20}{l}}{\log {a_n}}&{\log {a_{n + 1}}}&{\log {a_{n + 2}}}\\{\log {a_{n + 3}}}&{\log {a_{n + 4}}}&{\log {a_{n + 5}}}\\{\log {a_{n + 6}}}&{\log {a_{n + 7}}}&{\log {a_{n + 8}}}\end{array}} \right| = {\left( {\log r} \right)^3}\left| {\begin{array}{*{20}{l}}n&{\left( {n + 1} \right)}&{\left( {n + 2} \right)}\\{\left( {n + 3} \right)}&{\left( {n + 4} \right)}&{\left( {n + 5} \right)}\\{\left( {n + 6} \right)}&{\left( {n + 7} \right)}&{\left( {n + 8} \right)}\end{array}} \right|\]
Now we will perform determinant function i.e. \[{C_3} = {C_3} - {C_2}\]and\[{C_2} = {C_2} - {C_1}\]where C is the denoted for columns. Therefore, we get
\[ \Rightarrow \left| {\begin{array}{*{20}{l}}{\log {a_n}}&{\log {a_{n + 1}}}&{\log {a_{n + 2}}}\\{\log {a_{n + 3}}}&{\log {a_{n + 4}}}&{\log {a_{n + 5}}}\\{\log {a_{n + 6}}}&{\log {a_{n + 7}}}&{\log {a_{n + 8}}}\end{array}} \right| = {\left( {\log r} \right)^3}\left| {\begin{array}{*{20}{l}}n&1&1\\{\left( {n + 3} \right)}&1&1\\{\left( {n + 6} \right)}&1&1\end{array}} \right|\]
And according to the properties of the determinants if any two rows or columns of the determinants are equal then the value of the determinant is zero. Therefore, we get
\[ \Rightarrow \left| {\begin{array}{*{20}{l}}{\log {a_n}}&{\log {a_{n + 2}}}&{\log {a_{n + 4}}}\\{\log {a_{n + 6}}}&{\log {a_{n + 8}}}&{\log {a_{n + 10}}}\\{\log {a_{n + 12}}}&{\log {a_{n + 14}}}&{\log {a_{n + 16}}}\end{array}} \right| = {(\log r)^3} \times 0 = 0\]
Hence, 0 is the value of the determinant.

So, option A is the correct option.

Note:
Here we have found out the determinant of a GP series. Geometric progression (G.P.) is the sequence of numbers such that the common ratio between the consecutive numbers remains constant. We need to remember that determinants are calculated for square matrices only. Other than geometric progression we have arithmetic progression and harmonic progression also. A series in which two consecutive terms differ by the same common difference then the progression or series is called arithmetic progression. Harmonic progression is just the reciprocal of an arithmetic progression.