Problem 48 - Algebra for JEE ADVANCED

Question

\begin{aligned} &\text { If } a_{1}, a_{2}, \ldots, a_{s^{\prime}}, \ldots \text { form a G.P. and } a_{i}>0, \text { for all } i \geq 1 \text {, }\\\ &\text { then }\left|\begin{array}{lll} \log a_{n} & \log a_{n+1} & \log a_{n+2} \\ \log a_{m+3} & \log a_{n+4} & \log a_{m+5} \\ \log a_{n+6} & \log a_{n+7} & \log a_{n+8} \end{array}\right| \text { is equal to } \end{aligned}a. 0 b. 1 c. 2 d. 3 (AIEEE, 2005)

Step-by-Step Solution

Answer

The determinant is 0.

Step 1: Understanding the Problem

We need to evaluate the determinant of a 3x3 logarithmic matrix, where the elements are terms of a geometric progression (G.P.). Since the sequence is in G.P., all terms can be expressed in terms of the first term and common ratio.

Step 2: Express Terms of G.P. using Common Ratio

If the sequence \(a_1, a_2, \ldots\) is a G.P. with common ratio \(r\), then each term \(a_n = a_1 r^{n-1}\). Thus, \(a_i = a_1 r^{i-1}\) for all terms.

Step 3: Express Logarithms in the Matrix

The elements in the determinant include \(\log a_i\). By applying the logarithmic property of products, \(\log a_i = \log a_1 + (i-1) \log r\). Hence each element in the matrix can be expressed using this property.

Step 4: Setup the Determinant with Logarithms

Substitute \(\log a_i = \log a_1 + (i-1) \log r\) in the determinant:\[\begin{vmatrix} \log a_1 + (n-1) \log r & \log a_1 + n \log r & \log a_1 + (n+1) \log r \ \log a_1 + (m+2) \log r & \log a_1 + (n+3) \log r & \log a_1 + (m+4) \log r \ \log a_1 + (n+5) \log r & \log a_1 + (n+6) \log r & \log a_1 + (n+7) \log r\end{vmatrix}\]

Step 5: Simplify Matrix Rows by Factorizing Common Terms

We notice a pattern: each row is structured by adding consecutive increments of \(\log r\). Factor \(\log r\) out of the difference between adjacent columns to simplify determinants.Factor out the common term along each row to see that each row becomes linearly dependent due to the factorization. The given determinant simplifies to zero.

Step 6: Conclusion from Linear Dependency

Since the rows are linearly dependent (each row can be expressed as a linear combination of another), the determinant of such a matrix is always zero. Thus, the determinant is 0.

Key Concepts

Logarithmic Properties

When dealing with logarithms, there are essential properties that can make our computations easier and much more manageable. These properties allow us to transform and manipulate logarithmic expressions in a way that reveals simpler forms or insights.

**Three Key Properties of Logarithms:**

• The Product Rule: \(\log_b(xy) = \log_b(x) + \log_b(y)\). This rule tells us that the logarithm of a product is the sum of the logarithms of each factor.
• The Quotient Rule: \(\log_b\left(\frac{x}{y}\right) = \log_b(x) - \log_b(y)\). It states that the logarithm of a quotient is the difference between the logarithm of the numerator and the denominator.
• The Power Rule: \(\log_b(x^n) = n \cdot \log_b(x)\). This indicates that the logarithm of an exponent is the exponent multiplied by the logarithm of the base.

In the provided exercise, the Power Rule is especially helpful.

By recognizing that each term in our determinant can be expressed as a sum of logarithms, the simplification becomes straightforward. We can rewrite terms like \(\log a_i = \log a_1 + (i-1) \log r\), revealing repeated increments of \(\log r\), which helps in factoring out common terms.

Geometric Progression

A geometric progression (G.P.) is a sequence where each term is obtained by multiplying the previous term by a constant, known as the common ratio. Understanding how a sequence progresses helps in translating terms and simplifying complex expressions.

**Characteristics of a G.P.:**

• First term: Let's denote it as \(a_1\).
• Common ratio: Denoted by \(r\), which is the factor by which we multiply each term to get the next one.
• Formula for the \(n\)-th term: \(a_n = a_1 r^{n-1}\). This formula allows us to calculate any term in the sequence.

In this exercise, understanding that each \(a_i = a_1 r^{i-1}\) allows us to see the logarithms of the terms in the matrix as part of a sequence.

This insight leads to realizing that the logarithmic expressions follow a predictable pattern, simplifying to a form that can be rearranged or factored.

Matrix Simplification

Matrix simplification is a crucial tactic in computing determinants, especially when dealing with special sequences or operations like logarithms. The determinant, which is a scalar value derived from a matrix, can reveal a lot about the system that the matrix represents.

**Simplifying Determinants of Matrices:**
To evaluate determinants, we often look for patterns or symmetries that reduce the complexity of computation. In logarithmic matrices derived from G.P., we notice that rows or columns might become linearly dependent.

• **Factoring Common Terms:** By expressing each matrix element as a sum of the base and an increment, common terms like \(\log a_1\) can be factored out. This helps to reveal linear dependency.
• **Identifying Linear Dependency:** When rows or columns in a matrix can be expressed as certain multiples of others, they are linearly dependent. This simplification often allows us to deduce that the determinant is zero, as in the exercise provided, where the matrix's rows are linear combinations of each other.

Ultimately, matrix simplification reduces a problem to its essence, making solving for determinants faster and less error-prone. It transforms a complex, layered problem into an observation of basic principles, like linearly dependent vectors leading to a zero determinant.