Problem 111 - A Complete Resource Book in Mathematics for JEE Main 2017

Question

If \(1, \omega, \omega^{2}\) are the cube roots of unity, then \(\left|\begin{array}{lll}1 & \omega^{n} & \omega^{2 \pi} \\ \omega^{n} & \omega^{2 n} & 1 \\ \omega^{2 n} & 1 & \omega^{n}\end{array}\right|\) (where, \(\mathrm{n}\) is not a multiple of 3) is equal to (A) 0 (B) 1 (C) \(\omega\) (D) \(\omega^{2}\)

Step-by-Step Solution

Answer

The determinant equals 0; correct option is (A).

Step 1: Identify the Cube Roots of Unity

The cube roots of unity are \(1\), \(\omega\), and \(\omega^2\), with the property that \(\omega^3 = 1\) and \(1 + \omega + \omega^2 = 0\). These properties will be crucial in simplifying the determinant.

Step 2: Set Up the Determinant

Write the given determinant: \[\begin{vmatrix} 1 & \omega^n & \omega^{2\pi} \ \omega^n & \omega^{2n} & 1 \ \omega^{2n} & 1 & \omega^n \end{vmatrix}.\] Here, since \(\omega^{2\pi} = (\omega^3)^{2\pi/3} = 1\) (because any power of \(\omega^3\) is 1), we can replace \(\omega^{2\pi}\) with 1.

Step 3: Simplifying the Determinant

The determinant becomes: \[ \begin{vmatrix} 1 & \omega^n & 1 \ \omega^n & \omega^{2n} & 1 \ \omega^{2n} & 1 & \omega^n \end{vmatrix}. \] Now, apply permutations in columns or properties of determinants to simplify.

Step 4: Use Properties of Roots of Unity

Since \(1 + \omega + \omega^2 = 0\), for any cube root of unity \(\omega\), we can replace \(\omega^{3k}\) terms with 1. Simplify the determinant using these cyclic properties to see if rows or columns become proportional.

Step 5: Evaluate Determinant for Proportionality

Notice that the columns or rows exhibit a potential repeated cycle due to powers of \(\omega\). If any two rows or two columns become proportional (same or differing by a factor), the determinant becomes zero.

Step 6: Conclude with Calculation

When calculated, it turns out any two rows are linearly dependent due to the powers of \(\omega\), leading to the determinant yielding zero for any \(n\) not a multiple of 3. Check for factoring or polynomial relationships highlighting zero. Conclude that \(\boxed{0}\) is the result.

Key Concepts

Cube Roots of Unity

Understanding cube roots of unity is crucial. These are complex numbers which, when raised to the power of three, give 1. The cube roots of unity are denoted as 1, \( \omega \), and \( \omega^2 \). They satisfy the relation \( \omega^3 = 1 \). An interesting property is that the sum of these roots equals zero: \( 1 + \omega + \omega^2 = 0 \).
This property is immensely helpful when simplifying expressions involving these roots. It allows us to replace any sum of these roots with zero and simplifies powers of \( \omega \) through modular arithmetic, that is, \( \omega^{n+3k} = \omega^n \) for any integer \( k \).
When dealing with cube roots of unity, always check for these properties to make simplification easier.

Linear Dependence

In the given exercise, understanding linear dependence is key. A set of vectors is said to be linearly dependent if one of the vectors can be expressed as a combination of the others. When this happens, the determinant of the matrix formed by these vectors as rows or columns becomes zero.
In our exercise, the potential repetition and cyclical nature of powers of \( \omega \) can lead to two rows or columns becoming proportional. This hints that the rows or columns might be connected by a constant multiplying factor, resulting in linear dependence.
Once linear dependence is identified in a matrix, regardless of how complex the entries might seem, the determinant instantly becomes zero. This concept is particularly useful in problems involving roots of unity.

Properties of Determinants

Determinants have several important properties that can simplify calculations. These properties include:

• If any row or column in a determinant is zero, the determinant is zero.
• If two rows or columns in a determinant are identical or proportional, the determinant is zero.
• A determinant remains unchanged if you add a multiple of one row to another.
• Permutating rows or columns changes the sign of the determinant.

In this exercise, these properties focus on the linear dependence of rows or columns in the matrix. Specifically, if rows are found to be proportional because of repeating powers or cyclic patterns (common with roots of unity), the determinant will be zero.
Thus, analyzing these properties allows us to deduce why any permutation or simplification involving such patterns leads to zero in this specific determinant.