Problem 26 - 71 JEE Main Mathematics Online (2020 - 2012) & Offline (2018 - 2002) Chapterwise + Topicwise Solved Papers

Question

If \(\left|\begin{array}{lll}a & a^{2} & 1+a^{3} \\ b & b^{2} & 1+b^{3} \\ c & c^{2} & 1+c^{3}\end{array}\right|=0 \quad\) 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 abc equals (a) 0 (b) 2 (c) \(-1\) (d) 1

Step-by-Step Solution

Answer

(a) 0

Step 1: Understanding the Determinant Condition

The given matrix is a 3x3 matrix with the elements \( a, a^2, 1+a^3 \), \( b, b^2, 1+b^3 \), and \( c, c^2, 1+c^3 \). The determinant of this matrix is 0, indicating that the rows of the matrix are linearly dependent.

Step 2: Non-Coplanarity Condition

The vectors \((1, a, a^2)\), \((1, b, b^2)\), and \((1, c, c^2)\) are non-coplanar, meaning they form a basis for \( \mathbb{R}^3 \). Hence, they should be linearly independent.

Step 3: Connecting Conditions

For these vectors to be non-coplanar, they must be linearly independent, which contradicts the zero determinant condition unless some values make it trivially true. Since \(abc=0\) leads to a column of zeros in the matrix, causing the determinant to be zero, this is a valid condition that satisfies both non-coplanarity and the determinant condition.

Step 4: Examining Possible Value

Substituting specific values like \( a = 0 \), \( b = 0 \), or \( c = 0 \) implies a multiplication result of zero. Thus, \( abc = 0 \). No other values for \(a, b, c\) would satisfy the condition, given the constraints.

Step 5: Conclusion

Since the product \( abc \) leads to zero without conflicting with the vector independence required for non-coplanarity, the correct answer is \( abc = 0 \).

Key Concepts

Linear Independence

Linear independence is a fundamental concept in linear algebra. It helps determine whether a set of vectors are independent, meaning that no vector in the set can be written as a linear combination of the others. If vectors are linearly independent, the only solution to the equation \( c_1 \mathbf{v}_1 + c_2 \mathbf{v}_2 + ... + c_n \mathbf{v}_n = \mathbf{0} \) is when all scalar coefficients \( c_1, c_2, ..., c_n \) are zero. In simpler terms, each vector adds something unique to the set.

When dealing with matrices, the concept of linear independence can be checked using determinants. If the determinant of a square matrix is zero, the columns (or rows) of the matrix are linearly dependent, indicating redundant information. Herein lies the connection to the given problem: the determinant of the matrix from the exercise is zero, suggesting that if vectors are non-coplanar (which implies linear independence in a 3D space), there must be a value-set causing this to be trivially true, such as \( abc = 0 \).

In practical scenarios, checking for linear independence can help ascertain whether a set of vectors spans a space completely.

Vector Spaces

A vector space is a collection of objects called vectors, which can be added together and multiplied by scalars (real or complex numbers) while satisfying certain axioms. These axioms include associativity, commutativity of addition, distributive properties, and the existence of additive identity and inverses. The set of all possible linear combinations of a given set of vectors forms a vector space.

In the exercise, vectors such as \((1, a, a^2)\), \((1, b, b^2)\), and \((1, c, c^2)\) are examined to see if they form a basis of the vector space in three dimensions. A basis is the minimal set of vectors required to represent every vector in that space uniquely. For the problem, if these vectors are non-coplanar, they span the entire space and form a basis for \( \mathbb{R}^3 \). This implies linear independence, but due to the zero determinant, which inherently shows linear dependence, conditions like \( abc = 0 \) must simplify the determinant calculation, ensuring non-trivial solutions.

Understanding vector spaces help in various applications, such as solutions to differential equations, optimization problems, and more.

Coplanar Vectors

Vectors are deemed coplanar if they lie on the same plane in space. In three-dimensional space, a set of three vectors is coplanar if they can all be expressed as linear combinations of just two of them. This means the vectors do not span an entire three-dimensional space, but instead, their span is restricted to a two-dimensional plane. For vectors to be non-coplanar, they must be linearly independent and form a basis for \( \mathbb{R}^3 \).

In the context of the exercise, the statement about non-coplanarity means that the vectors \((1, a, a^2)\), \((1, b, b^2)\), and \((1, c, c^2)\) ideally should not lie on the same plane. The determinant being zero suggests an initial presumption of coplanarity, which contradicts their provided non-coplanar condition unless something simplifies the determinant, such as \( abc = 0 \).

Understanding whether vectors are coplanar is crucial when working with vector operations, especially in physics and engineering, where spatial configurations are key.