If $${a_1},{a_2},{a_3},.....$$ are positive numbers in G.P. then the value of \[\left| {\begin{array}{*{20}{c}}
{\log {a_n}}&{\log {a_{n + 1}}}&{\log {a_{n + 2}}}\\
{\log {a_{n + 1}}}&{\log {a_{n + 2}}}&{\log {a_{n + 3}}}\\
{\log {a_{n + 2}}}&{\log {a_{n + 3}}}&{\log {a_{n + 4}}}
\end{array}} \right|\]
A.
1
B.
4
C.
3
D.
0
Answer :
0
Solution :
If the G.P. be $$a,ar,a{r^2},.....$$ then $${a_n} = a{r^{n - 1}}$$
\[D = \left| {\begin{array}{*{20}{c}}
{\log a + \left( {n - 1} \right)\log r}&{\log a + n \log r}&{\log a + \left( {n + 1} \right)\log r}\\
{\log a + n\log r}&{\log a + \left( {n + 1} \right)\log r}&{\log a + \left( {n + 2} \right)\log r}\\
{\log a + \left( {n + 1} \right)\log r}&{\log a + \left( {n + 2} \right)\log r}&{\log a + \left( {n + 3} \right)\log r}
\end{array}} \right|\]
$${R_3} \to {R_3} - {R_2}$$ and $${R_2} \to {R_2} - {R_1}$$ gives,
\[ = \left| {\begin{array}{*{20}{c}}
{\log a + \left( {n - 1} \right)\log r}&{\log a + n \log r}&{\log a + \left( {n + 1} \right)\log r}\\
{\log r}&{\log r}&{\log r}\\
{\log r}&{\log r}&{\log r}
\end{array}} \right|\]
= 0, since $$R_2$$ and $$R_3$$ are identical.
Releted MCQ Question on Algebra >> Matrices and Determinants
Releted Question 1
Consider the set $$A$$ of all determinants of order 3 with entries 0 or 1 only. Let $$B$$ be the subset of $$A$$ consisting of all determinants with value 1. Let $$C$$ be the subset of $$A$$ consisting of all determinants with value $$- 1.$$ Then
A.
$$C$$ is empty
B.
$$B$$ has as many elements as $$C$$
C.
$$A = B \cup C$$
D.
$$B$$ has twice as many elements as elements as $$C$$
Let $$a, b, c$$ be the real numbers. Then following system of equations in $$x, y$$ and $$z$$
$$\frac{{{x^2}}}{{{a^2}}} + \frac{{{y^2}}}{{{b^2}}} - \frac{{{z^2}}}{{{c^2}}} = 1,$$ $$\frac{{{x^2}}}{{{a^2}}} - \frac{{{y^2}}}{{{b^2}}} + \frac{{{z^2}}}{{{c^2}}} = 1,$$ $$ - \frac{{{x^2}}}{{{a^2}}} + \frac{{{y^2}}}{{{b^2}}} + \frac{{{z^2}}}{{{c^2}}} = 1$$ has