Problem 79 - A Complete Resource Book in Mathematics for JEE Main

Question

If \(\left|\begin{array}{lll}a & a^{2} & 1+a^{3} \\ b & b^{2} & 1+b^{3} \\ c & c^{2} & 1+c^{3}\end{array}\right|=0\) and vectors \(\left(1, a, a^{2}\right)\left(1, b, b^{2}\right)\) and \(\left(1, c, c^{2}\right)\) are non-coplanar, then the product \(a b c\) equals [2003] (A) 2 (B) \(-1\) (C) 1 (D) 0

Step-by-Step Solution

Answer

The product of \(abc\) is 1.

Step 1: Understand the Matrix Condition

The determinant of the given matrix is zero: \[\begin{vmatrix}a & a^2 & 1 + a^3 \b & b^2 & 1 + b^3 \c & c^2 & 1 + c^3 \end{vmatrix} = 0\]This means that the rows (or columns) are linearly dependent.

Step 2: Interpret Coplanarity

The vectors \((1, a, a^2), (1, b, b^2), (1, c, c^2)\) being non-coplanar means that they are linearly independent. Linearly independent vectors can span a 3D space, which contradicts the linear dependence required by the determinant condition unless there's a special relationship among the elements.

Step 3: Set Up the Relation Among Elements

For the vectors to be non-coplanar but still fulfilling the zero determinant condition, the elements of the third column, \(1+a^3, 1+b^3, 1+c^3\), must create dependency among the rows. This happens if these elements are symmetric functions of \(a, b, c\) that sum to zero, suggesting a root-related relation among \(a, b, c\).

Step 4: Hypothesize Possible Values

To satisfy the determinant condition and non-coplanarity, the elements \(a, b, c\) should be roots of a polynomial relation like \(x^3 = 1\), which gives roots \(1, \omega, \omega^2\), where \(\omega\) are the primitive cube roots of unity. Hence \(abc = 1 \cdot \omega \cdot \omega^2 = 1\).

Step 5: Verify Solution

With \(abc = 1\), verify if the original determinant is zero. For cube roots of unity: \(1+\omega^3 = 2\), results in all rows being dependent, fulfilling the deterministic condition and satisfying vector non-coplanarity with the given vectors being independent.

Key Concepts

Matrix Determinant

The determinant is a critical concept in Linear Algebra. It is a scalar value that can be computed from the elements of a square matrix. Determinants help in understanding the properties of a matrix. One important property is determining whether a set of vectors are linearly independent or dependent.
To calculate a determinant of a 3x3 matrix:

• Multiply the diagonals from top-left to bottom-right.
• Subtract the product of diagonals from top-right to bottom-left.

For example, if the determinant is equal to zero, then the vectors (or the rows/columns) of the matrix are linearly dependent, meaning one can be written as a linear combination of others.
This understanding is crucial when examining the given matrix in the problem, as a determinant equal to zero indicates that the rows of this matrix are not independent and share some relationship that makes them dependent.

Coplanarity of Vectors

When talking about vectors, coplanarity refers to whether vectors lie on the same plane. In simpler terms, can they all be contained in a single, flat surface?

• Vectors need to be linearly independent to determine coplanarity.
• If vectors are linearly independent, they are non-coplanar.

In the context of the exercise, the vectors given are \((1, a, a^2), (1, b, b^2), (1, c, c^2)\). They must be non-coplanar to span a three-dimensional space. Non-coplanar vectors can not lie on the same plane and signify the vectors span a 3D space.
The juxtaposition of having a determinant of zero but requiring non-coplanarity indicates some unique conditions among the entries or roots creating dependencies across different parts of the matrix.

Roots of Unity

Roots of unity are special complex numbers representing the solutions of the equation \(x^n = 1\). These are numbers which, when raised to a certain power like 3 (in the case of cube roots), result in 1.
The roots of unity prime chords:

• The principal cube roots of unity are: \(1, \omega, \omega^2\).
• \(\omega = \frac{-1 + i\sqrt{3}}{2}\), where \(i\) is the imaginary unit.

Cube roots of unity have the property that their sum equals zero and the product is 1. This property plays a vital role in forming the polynomial relationships among variables in problems like the one described.
In the exercise, using the roots of unity \(1, \omega, \omega^2\) for \(a, b, c\) results in their product equaling 1, satisfying \(abc = 1\). This makes it possible for these vectors to align with the determinant being zero, considering their particular relationship and roles as roots of unity.