If $$a > 0, b > 0, c > 0$$ are respectively the $$p^{th}, q^{th}, r^{th}$$ terms of G.P., then the value of the determinant \[\left| {\begin{array}{*{20}{c}}
{\log a}&p&1\\
{\log b}&q&1\\
{\log c}&r&1
\end{array}} \right|\] is
A.
$$0$$
B.
$$1$$
C.
$$ - 1$$
D.
None of these
Answer :
$$0$$
Solution :
Let $$A$$ be the $$1^{st}$$ term and $$R$$ the common ratio of G.P., then ;
$$\eqalign{
& a = {T_p} = A{R^{p - 1}} \cr
& \therefore \log a = \log A + \left( {p - 1} \right)\log R \cr} $$
Similarly, $$\log b = \log A + \left( {q - 1} \right)\log R$$
and $$\log c = \log A + \left( {r - 1} \right)\log R$$
\[\therefore \Delta = \left| {\begin{array}{*{20}{c}}
{\log A + \left( {p - 1} \right)\log R}&p&1\\
{\log A + \left( {q - 1} \right)\log R}&q&1\\
{\log A + \left( {r - 1} \right)\log R}&r&1
\end{array}} \right|\]
Split into two determinants and in the first take $$\log A$$ common and in the second take $$\log R$$ common
\[\Delta = \log A\left| {\begin{array}{*{20}{c}}
1&p&1\\
1&q&1\\
1&r&1
\end{array}} \right| + \log R\left| {\begin{array}{*{20}{c}}
{p - 1}&p&1\\
{q - 1}&q&1\\
{r - 1}&r&1
\end{array}} \right|\]
Apply $${C_1} \to {C_1} - {C_2} + {C_3}$$ in the second
\[\Delta = 0 + \log R\left| {\begin{array}{*{20}{c}}
0&p&1\\
0&q&1\\
0&r&1
\end{array}} \right| = 0\]
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