Understanding the Problem: Determinant of GP Logarithms
The question asks for the value of a specific 3x3 determinant. The entries of this determinant are the natural logarithms (ln) of the terms of a geometric progression (GP). We are given that \(a_1, a_2, a_3, \dots, a_9\) are terms in a GP. We need to evaluate the determinant:
A geometric progression is a sequence of non-zero numbers where each term after the first is found by multiplying the previous one by a fixed, non-zero number called the common ratio. If the first term is \(a\) and the common ratio is \(r\), the terms of a GP are \(a, ar, ar^2, ar^3, \dots, ar^{n-1}, \dots\).
In this problem, the terms are \(a_1, a_2, \dots, a_9\). Let the first term be \(A\) and the common ratio be \(R\). Then, the terms can be written as:
\(a_1 = A\)
\(a_2 = AR\)
\(a_3 = AR^2\)
\(a_4 = AR^3\)
\(a_5 = AR^4\)
\(a_6 = AR^5\)
\(a_7 = AR^6\)
\(a_8 = AR^7\)
\(a_9 = AR^8\)
In general, the \(i\)-th term of the GP is \(a_i = A \cdot R^{i-1}\).
Properties of Natural Logarithms (ln)
The natural logarithm, denoted by \(\ln\), is the logarithm to the base \(e\). Key properties we will use are:
Product rule: \(\ln(xy) = \ln x + \ln y\)
Power rule: \(\ln(x^y) = y \ln x\)
Applying Properties to the Determinant Elements
The elements of the determinant are \(\ln a_i\). Using the GP property \(a_i = A \cdot R^{i-1}\) and the logarithm properties, we can write each element:
\(\ln a_i = \ln (A \cdot R^{i-1})\)
Using the product rule for logarithms:
\(\ln a_i = \ln A + \ln (R^{i-1})\)
Using the power rule for logarithms:
\(\ln a_i = \ln A + (i-1) \ln R\)
Let \(L = \ln A\) and \(K = \ln R\). Then, \(\ln a_i = L + (i-1)K\).
Now let's write out the elements of the determinant using this form:
\(\ln a_1 = L + (1-1)K = L\)
\(\ln a_2 = L + (2-1)K = L + K\)
\(\ln a_3 = L + (3-1)K = L + 2K\)
\(\ln a_4 = L + (4-1)K = L + 3K\)
\(\ln a_5 = L + (5-1)K = L + 4K\)
\(\ln a_6 = L + (6-1)K = L + 5K\)
\(\ln a_7 = L + (7-1)K = L + 6K\)
\(\ln a_8 = L + (8-1)K = L + 7K\)
\(\ln a_9 = L + (9-1)K = L + 8K\)
Constructing the Determinant with Simplified Terms
Substituting these expressions into the determinant, we get:
We can use elementary column operations to simplify the determinant without changing its value. A key property is that if we subtract a multiple of one column from another column, the determinant's value remains unchanged. Also, if two columns of a determinant are identical, the determinant's value is zero.
Performing Column Operations
Let \(C_1, C_2, C_3\) be the three columns of the determinant. We will perform the following operations:
Operation 1: \(C_2 \leftarrow C_2 - C_1\) (Replace Column 2 with Column 2 minus Column 1)
Operation 2: \(C_3 \leftarrow C_3 - C_2\) (Replace Column 3 with Column 3 minus the *original* Column 2)
Let's see what the new columns become:
New \(C_2\):
Row 1: \((L + K) - L = K\)
Row 2: \((L + 4K) - (L + 3K) = K\)
Row 3: \((L + 7K) - (L + 6K) = K\)
So, the new \(C_2\) is \(\begin{pmatrix} K \\ K \\ K \end{pmatrix}\).
New \(C_3\):
Row 1: \((L + 2K) - (L + K) = K\)
Row 2: \((L + 5K) - (L + 4K) = K\)
Row 3: \((L + 8K) - (L + 7K) = K\)
So, the new \(C_3\) is \(\begin{pmatrix} K \\ K \\ K \end{pmatrix}\).
In the resulting determinant, the second column (\(C_2\)) and the third column (\(C_3\)) are identical. A fundamental property of determinants states that if any two rows or any two columns of a determinant are identical or proportional, the value of the determinant is zero.
Since \(C_2 = C_3\), the value of the determinant is 0.
Alternatively, one could perform operations \(C_2 \leftarrow C_2 - C_1\) and \(C_3 \leftarrow C_3 - C_1\).
New \(C_2 = \begin{pmatrix} K \\ K \\ K \end{pmatrix}\)
New \(C_3\):
Row 1: \((L + 2K) - L = 2K\)
Row 2: \((L + 5K) - (L + 3K) = 2K\)
Row 3: \((L + 8K) - (L + 6K) = 2K\)
New \(C_3 = \begin{pmatrix} 2K \\ 2K \\ 2K \end{pmatrix}\).
In this determinant, \(C_3 = 2 \cdot C_2\). Since the third column is proportional to the second column, the determinant's value is also 0.
Revision Table: Key Concepts
Concept
Definition/Property
Application
Geometric Progression (GP)
Sequence \(a_i = AR^{i-1}\)
Express terms \(a_i\) using \(a_1\) and common ratio \(R\).
Natural Logarithm (ln)
Logarithm base \(e\); \(\ln(xy) = \ln x + \ln y\), \(\ln(x^y) = y \ln x\)
Transform \(\ln a_i\) into linear expressions of \(\ln a_1\) and \(\ln R\).
Determinant Properties
Column operations preserve value; determinant is 0 if columns are identical or proportional.
Simplify the determinant expression to reveal proportional columns.
Additional Information: Related Math Concepts
Understanding the connection between sequences and determinants can be useful. In this problem, the logarithms of the GP terms form an arithmetic progression (AP).
If \(a_1, a_2, \dots, a_n\) are in GP, then \(\ln a_1, \ln a_2, \dots, \ln a_n\) are in AP. Why?
This is the form of an AP with first term \(\ln A\) and common difference \(\ln R\).
So, the determinant's rows are formed by consecutive terms of this AP:
Row 1: Terms 1, 2, 3 of the AP
Row 2: Terms 4, 5, 6 of the AP
Row 3: Terms 7, 8, 9 of the AP
A determinant whose rows (or columns) are in arithmetic progression has a value of 0. This provides another way to understand why the determinant is zero in this case.
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