Problem 90 - 71 JEE Main Mathematics Online (2020 - 2012) & Offline (2018 - 2002) Chapterwise + Topicwise Solved Papers

Question

If the system of linear equations \(2 x+2 a y+a z=0\) \(2 x+3 b y+b z=0\) \(2 x+4 c y+c z=0\) where \(a, b, c \in \boldsymbol{R}\) are non-zero and distinct; has a non-zero solution, then: \(\quad\) [Jan. 7, 2020 (I)] (a) \(\frac{1}{a}, \frac{1}{b}, \frac{1}{c}\) are in A.P. (b) \(a, b, c\) are in GP. (c) \(a+b+c=0\) (d) \(a, b, c\) are in A.P.

Step-by-Step Solution

Answer

(c) \(a+b+c=0\).

Step 1: Understand the System of Equations

We have the equations: \(2x + 2ay + az = 0\), \(2x + 3by + bz = 0\), \(2x + 4cy + cz = 0\). Given that \(a, b, c\) are non-zero and distinct and the system has a non-zero solution, we look for conditions on \(a, b, c\).

Step 2: Form the Matrix Representation

Express the system as a matrix equation: \(AX = 0\), where the coefficient matrix \(A\) is \[\begin{pmatrix} 2 & 2a & a \ 2 & 3b & b \ 2 & 4c & c \end{pmatrix}\] and \(X = \begin{pmatrix} x \ y \ z \end{pmatrix}\).

Step 3: Determine Condition for Non-zero Solution

For non-zero solutions, the determinant of the coefficient matrix must be zero. Calculate the determinant of \(A\): \[\text{det}(A) = \begin{vmatrix} 2 & 2a & a \ 2 & 3b & b \ 2 & 4c & c \end{vmatrix}.\]

Step 4: Calculate the Determinant

Calculate the determinant of the matrix: \[\text{det}(A) = 2((3b)(c) - (4c)(b)) - 2a((2)(c) - (4c)(2)) + a((2)(4c) - (2)(3b)).\]Simplify this expression to find the condition on \(a, b, c\).

Step 5: Simplify and Set the Determinant to Zero

Simplify the determinant expression: \[\text{det}(A) = a(b-c) + b(c-a) + c(a-b)\]Set it to zero: \[a(b-c) + b(c-a) + c(a-b) = 0.\] This implies that \(a+b+c = 0\).

Step 6: Verify Conditions

Now verify which condition among the given options is satisfied:- \(a+b+c = 0\) matches condition (c).

Step 7: Conclusion

The correct condition for the non-zero solution of the given system is that \(a+b+c=0\). Therefore, the correct answer is option (c).

Key Concepts

Determinant of a Matrix

When solving a system of linear equations using matrices, the determinant plays a crucial role in determining the type of solution the system may have. The determinant is a unique number derived from a matrix that can indicate whether a system has a unique solution, no solution, or infinitely many solutions.
The system of equations is represented in matrix form, with the determinant of the matrix used to test whether the unknowns have a unique solution or not.
For a system to have a non-zero solution, the determinant must be zero. This may seem counterintuitive initially, but it makes sense because if the determinant is zero, it means that the matrix is singular, indicating that the equations are dependent or linearly related. Thus, a non-zero solution arises when the determinant is zero, signifying some dependency among the equations.
Therefore, solving the determinant, as indicated in the exercise, leads to the conclusion of certain conditions on coefficients, such as the requirement that the sum of certain parameters equals zero, allowing for the possibility of a non-zero solution.

Matrix Representation

Matrix representation is a powerful tool for solving systems of linear equations. In this context, the system is expressed as a matrix equation, making it easier to manipulate and solve using matrix operations.
For a system with equations like \(2x + 2ay + az = 0\), \(2x + 3by + bz = 0\), and \(2x + 4cy + cz = 0\), the matrix representation is structured as \(AX = 0\).
Here, the matrix \(A\) is composed of the coefficients of the equations:\[A = \begin{pmatrix} 2 & 2a & a \ 2 & 3b & b \ 2 & 4c & c \end{pmatrix}\]
The column vector \(X\) includes the variables \((x, y, z)\), and the resulting equation \(AX = 0\) sets up a framework to solve for \(X\).
Using this matrix representation allows simplifying calculations, especially when determining the conditions for non-zero solutions by calculating the determinant of the matrix.

Non-zero Solution

A non-zero solution in the context of a linear system refers to solutions where the variables involved do not all equal zero simultaneously. Finding these solutions typically involves setting the determinant of the coefficient matrix to zero, as shown in the presented problem.
This condition indicates that the equations are linearly dependent, and as such, possess an infinite number of solutions rather than a single, unique solution.
In practical terms, solving for a non-zero solution requires examining relationships between the variables, like ensuring certain sums or differences of coefficients satisfy specific conditions (e.g., \(a + b + c = 0\)).
Ultimately, the quest to determine if a non-zero solution exists and the conditions needed for it requires breaking down the system through various algebraic methods, including matrix transformation and determinant calculations. By delving deeply into these methods, one can uncover the underlying patterns or relationships among the variables that result in non-zero solutions.