Question

If \(p,q,r\) are all positive and are the \(p^{th}\), \(q^{th}\) and \(r^{th}\) terms of a geometric progression respectively, then the value of the determinant
\[ \left|\begin{matrix} \log x & p & 1 \\ \log y & q & 1 \\ \log z & r & 1 \end{matrix}\right| \] equals

• A. \(\log xyz\)
• B. \((p-1)(q-1)(r-1)\)
• C. \(pqr\)
• D. \(0\)

Hint:

If any column of a determinant is a linear combination of the others, determinant becomes zero.

Answer 1

The Correct Option is D

Step 1: Use GP condition.
If \(x,y,z\) are \(p^{th}, q^{th}, r^{th}\) terms of a GP, then:
\[ \log x,\log y,\log z \] also form an arithmetic progression with respect to indices.
So:
\[ \log x = A + (p-1)d \]
\[ \log y = A + (q-1)d \]
\[ \log z = A + (r-1)d \]
Step 2: Express each log as linear function of index.
\[ \log x = dp + (A-d) \]
\[ \log y = dq + (A-d) \]
\[ \log z = dr + (A-d) \]
Step 3: Row dependence.
Thus first column is a linear combination of second and third columns:
\[ \log x = d(p) + (A-d)(1) \]
\[ \log y = d(q) + (A-d)(1) \]
\[ \log z = d(r) + (A-d)(1) \]
So:
\[ C_1 = d\,C_2 + (A-d)\,C_3 \]
Since one column is dependent on others, determinant is zero.
Final Answer:
\[ \boxed{0} \]