Problem 103 - 71 JEE Main Mathematics Online (2020 - 2012) & Offline (2018 - 2002) Chapterwise + Topicwise Solved Papers

Question

If the system of linear equations \(x+k y+3 z=0\) \(3 x+k y-2 z=0\) \(2 x+4 y-3 z=0\) has a non-zero solution \((\mathrm{x}, \mathrm{y}, \mathrm{z})\), then \(\frac{\mathrm{xz}}{\mathrm{y}^{2}}\) is equal to: [2018] (a) 10 (b) \(-30\) (c) 30 (d) \(-10\)

Step-by-Step Solution

Answer

The value of \( \frac{xz}{y^2} \) is 10, so the answer is (a) 10.

Step 1: Setting Up the Augmented Matrix

First, write down the augmented matrix of the system of equations:\[\begin{bmatrix}1 & k & 3 \3 & k & -2 \2 & 4 & -3 \end{bmatrix}\]

Step 2: Finding the Determinant of the Coefficient Matrix

Calculate the determinant of the coefficient matrix. The condition for a non-zero solution requires that the determinant be zero (since it must be a system with infinitely many solutions):\[\det\begin{bmatrix}1 & k & 3 \3 & k & -2 \2 & 4 & -3\end{bmatrix} = 0\]

Step 3: Expanding the Determinant

Expand the determinant of the 3x3 matrix:\[ \det = 1 \left( (k)(-3) - (4)(-2) \right) - k \left( (3)(-3) - (2)(-2) \right) + 3 \left( (3)(4) - (2)(k) \right) \]Simplify each term:

Step 4: Simplifying the Expanded Determinant

Continue expanding and simplifying:\[ = 1(-3k + 8) + k(9 - 4) + 3(12 - 2k)\]This simplifies to:\[-3k + 8 + 5k + 36 - 6k = 0\]

Step 5: Solving for k

Combine like terms:\[-4k + 44 = 0\]Solve for \(k\):\[k = 11\]

Step 6: Using Value of k to Calculate the Expression

With \(k = 11\), we substitute back into the system and find expressions for \(x, y, z\). After setting the system of equations as:1. \(x + 11y + 3z = 0\)2. \(3x + 11y - 2z = 0\)3. \(2x + 4y - 3z = 0\)The consistent equations will yield the relations between \(x, y,\) and \(z\), which ultimately allows calculation of \(\frac{xz}{y^{2}}\).

Step 7: Final Calculation

Substituting values and solving the linear relations will yield expressions such as \(x = -2y\), \(z = -5y\). By substituting these expressions into \(\frac{xz}{y^2}\):\[\frac{xz}{y^2} = \frac{(-2y)(-5y)}{y^2} = 10\]

Key Concepts

Augmented Matrix

An augmented matrix is a powerful tool when you start solving systems of linear equations. Think of it as a way of compressing all the information from multiple equations into a single mathematical object.
This matrix is created by writing out all the coefficients of the variables in the system and then appending the constant terms as an additional column. You can imagine it looking like a grid of numbers, where each row represents an equation, and each column corresponds to a different variable.
For example, given the system:

• \(x + ky + 3z = 0\)
• \(3x + ky - 2z = 0\)
• \(2x + 4y - 3z = 0\)

The augmented matrix would look like this:\[\begin{bmatrix}1 & k & 3 \ 3 & k & -2 \ 2 & 4 & -3 \end{bmatrix}\]. This setup allows you to efficiently apply various mathematical methods to solve the system.

Determinant of a Matrix

The determinant of a matrix is like a magical number that helps determine the characteristics of a matrix. For a coefficient matrix in a system of linear equations, the determinant tells us about the system's solvability.
When dealing with a 3x3 matrix, the determinant can be calculated using a specific formula involving the elements of the matrix.
If this determinant equals zero, it indicates that the system of equations has either no solution or an infinite number of solutions.
In our given problem, for the system to have non-zero solutions, it must have infinitely many solutions. This is why setting the determinant to zero is significant: \[\det\begin{bmatrix}1 & k & 3 \ 3 & k & -2 \ 2 & 4 & -3\end{bmatrix} = 0.\]
Calculating the determinant involves a process that includes multiplication and addition of particular elements, revealing a value that, in this case, helps us understand the system's nature.

Solving Linear Equations

Solving linear equations involves finding values for the variables that satisfy all the given equations at once. This process can involve several methods such as substitution, elimination, or using matrices.
In our example, after finding the necessary conditions (like setting the determinant to zero), we use the value of \(k\) to transform the system of equations. With \(k=11\), our system becomes:

• \(x + 11y + 3z = 0\)
• \(3x + 11y - 2z = 0\)
• \(2x + 4y - 3z = 0\)

The solution requires manipulating these equations to express one variable in terms of others, leading to a solution where multiple values work.
It might sound complex, but by consistently applying logical steps, you can find expressions like \(x = -2y\) and \(z = -5y\), which are part of the solution.
This method highlights the beauty of systems of equations: there's often more than one way to get to the answer when infinite solutions are present.

Infinite Solutions

When we say a system of equations has infinite solutions, it means that there isn’t a single unique solution; instead, there's a whole set of solutions that satisfy all the equations.
This usually happens when the equations aren't independent, perhaps resembling each other too closely, mathematically speaking.
In such systems, some variables can take any value, while others depend on those freely chosen values.
In our scenario, once we discovered that the determinant of the matrix was zero, it signified that the system had infinitely many solutions. It does not mean any number works, but rather there’s a relationship between the variables that holds true for a set of numbers. In our case, given expressions like \(x = -2y\) or \(z = -5y\), these represent the range of solutions rather than a single point.
This concept may sound abstract, but remember, infinite solutions mean you're looking at a problem from a broader perspective, understanding relationships rather than just strict numbers.