Problem 59 - Algebra for JEE ADVANCED

Question

\(\Delta_{1}=\left|\begin{array}{ccc}y^{5} z^{6}\left(z^{3}-y^{3}\right) & x^{4} z^{6}\left(x^{3}-z^{3}\right) & x^{4} y^{5}\left(y^{3}-x^{3}\right) \\\ y^{2} z^{3}\left(y^{6}-z^{6}\right) & x z^{3}\left(z^{6}-x^{6}\right) & x y^{2}\left(x^{6}-y^{6}\right) \\ y^{2} z^{3}\left(z^{3}-y^{3}\right) & x z^{3}\left(x^{3}-z^{3}\right) & x y^{2}\left(y^{3}-x^{3}\right)\end{array}\right|\) and \(\Delta_{2}=\left|\begin{array}{ccc}x & y^{2} & z^{3} \\ x^{4} & y^{5} & z^{6} \\ x^{7} & y^{8} & z^{9}\end{array}\right|\). Then \(\Delta_{1} \Delta_{2}\) is equal to a. \(\Delta_{3}^{3}\) b. \(\Delta_{2}^{2}\) c. \(\Delta_{2}^{4}\) d. None of these

Step-by-Step Solution

Answer

(d) None of these.

Step 1: Simplify each element of the first matrix

Each element in the matrix used to find \(\Delta_{1}\) has a pattern with polynomials and powers. We need to recognize a common sequence: Replace the term \((a^3 - b^3)\) with the factorized form \((a-b)(a^2 + ab + b^2)\) so that \(y^5 z^6(y^3 - z^3) = y^5 z^6 (y-z)(y^2 + yz+z^2)\), and similarly apply to other terms too.

Step 2: Notice Zero rows Condition for Determinant

Observe that the second and third rows of the first matrix \(\Delta_{1}\) are proportional:1. \(y^2 z^3(y^6 - z^6) = y^2 z^3(y^3 - z^3)(y^3 + z^3)\)2. \(y^2 z^3(z^3 - y^3) = -y^2 z^3(y^3 - z^3)\)This observation might lead to parallel row vectors, implying that \(\Delta_{1}\) might be zero.

Step 3: Finding \(\Delta_{1}\)

Since the second and third row in \(\Delta_{1}\) are parallel due to being proportional, it implies two rows of the matrix are linearly dependent. Hence, the determinant \(\Delta_{1} = 0\).

Step 4: Finding \(\Delta_{2}\)

The determinant \(\Delta_{2}\) needs computation:Using properties of determinants and the pattern in the matrix, determine \(\Delta_{2}\) as a Vandermonde determinant (difference of powers across rows resembling a Vandermonde structure).Thus,\[\Delta_{2} = (x-y)(y-z)(z-x) \cdot (x^2) \cdot(x^4) = (x-y)(y-z)(z-x)x^6\]

Step 5: Calculate \(\Delta_{1} \Delta_{2}\)

We have found that \(\Delta_{1} = 0\). Multiply \(\Delta_{1}\) by \(\Delta_{2}\):\[\Delta_{1} \Delta_{2} = 0 \cdot (x-y)(y-z)(z-x)x^6 = 0\]

Step 6: Determine Which Option Matches

Compare \(\Delta_{1} \Delta_{2} = 0\) with the provided options, i.e., \(\Delta_{2}^{2}, \Delta_{2}^{3}, \) and \(\Delta_{2}^{4}\), which are all non-zero and not identical to zero.Thus, the correct option is (d) None of these.

Key Concepts

Matrix simplification

Matrix simplification is a technique to make matrices easier to handle, especially when calculating determinants. This involves reducing complex expressions to simpler forms.
A key part of simplifying involves recognizing patterns and using algebraic identities.

• For example, when you see terms like \(a^3 - b^3\), you can factor them using the difference of cubes formula: \(a^3 - b^3 = (a-b)(a^2 + ab + b^2)\). This can simplify computations significantly.
• In some cases, simplification involves identifying proportional rows or columns. If two rows (or columns) are multiples of each other, the determinant is zero due to linear dependence.

Applying these techniques can turn a complicated matrix into something much simpler, making it easier to calculate the determinant and understand the overall structure.

Vandermonde determinant

The Vandermonde determinant arises from a particular kind of matrix, distinguished by its pattern of elements raised to successive powers. A standard Vandermonde matrix has the form:

• The first column consists of first power elements, like \(x, y, z, \ldots\)
• Each subsequent column contains powers of these elements, e.g., \(x^2, y^2, z^2, \ldots\) in the second column, \(x^3, y^3, z^3, \ldots\) in the third, and so on.
• The determinant of such a matrix is particularly interesting because it reveals relationships between the numbers, like \( (x-y)(y-z)(z-x)\) for three variables.

This type of determinant often helps in solving problems by breaking down complex polynomial relationships into easier linear factors.

Linear dependence

Linear dependence occurs in the context of matrices when two or more rows (or columns) are proportional, meaning one can be expressed as a multiple of another. This concept is crucial when calculating determinants, as:

• If even two rows (or columns) of a square matrix are linearly dependent, the determinant is zero.
• This makes linear dependence a key shortcut in recognizing when a determinant is zero without performing full calculations.

Recognizing linear dependence can save significant time and effort, especially in larger matrices, since it immediately indicates a lack of non-trivial "solutions" or independence among the vectors represented by the matrix rows or columns.

Polynomial factorization

Polynomial factorization is breaking down a polynomial into simpler polynomial factors that multiply to give the original polynomial. It's a technique used in many mathematical areas, including matrix algebra.

• Common forms include difference of squares, perfect square trinomials, and the aforementioned difference of cubes.
• For example, recognizing \( a^3 - b^3 \) allows us to write it as \( (a-b)(a^2 + ab + b^2) \), simplifying calculations and revealing underlying structures more clearly.
• In matrix settings, factorization can simplify the problem and help with direct calculation of determinants by reducing complexity.

Understanding polynomial factorization is key to dealing with matrices that contain polynomial elements, as it facilitates the simplification process and highlights mathematical relationships.