Problem 116 - 71 JEE Main Mathematics Online (2020 - 2012) & Offline (2018 - 2002) Chapterwise + Topicwise Solved Papers

Question

Statement 1: If the system of equations \(x+k y+3 z=0\), \(3 x+k y-2 z=0,2 x+3 y-4 z=0\) has a non-trivial solution, then the value of \(k\) is \(\frac{31}{2}\) Statement 2: A system of three homogeneous equations in three variables has a non trivial solution if the determinant of the coefficient matrix is zero. [Online May 26, 2012] (a) Statement 1 is false, Statement 2 is true. (b) Statement 1 is true, Statement 2 is true, Statement 2 is a correct explanation for Statement 1 . (c) Statement 1 is true, Statement 2 is true,, Statement 2 is not a correct explanation for Statement 1 . (d) Statement 1 is true, Statement 2 is false.

Step-by-Step Solution

Answer

(a) Statement 1 is false, Statement 2 is true.

Step 1: Determine the Coefficient Matrix

The given system of equations is \(x + ky + 3z = 0\), \(3x + ky - 2z = 0\), and \(2x + 3y - 4z = 0\). The coefficient matrix is:\[\begin{bmatrix}1 & k & 3 \3 & k & -2 \2 & 3 & -4\end{bmatrix}\]

Step 2: Calculate the Determinant

To check for non-trivial solutions, we compute the determinant of the coefficient matrix. A non-trivial solution exists if the determinant is zero.\[\text{Det} = 1 \begin{vmatrix} k & -2 \ 3 & -4 \end{vmatrix} - k \begin{vmatrix} 3 & -2 \ 2 & -4 \end{vmatrix} + 3 \begin{vmatrix} 3 & k \ 2 & 3 \end{vmatrix}\]Calculate each minor and simplify.

Step 3: Calculate Minors and Determine Condition

Compute each minor matrix as follows:- \(\begin{vmatrix} k & -2 \ 3 & -4 \end{vmatrix} = k(-4) - (-2)(3) = -4k + 6\)- \(\begin{vmatrix} 3 & -2 \ 2 & -4 \end{vmatrix} = (3)(-4) - (-2)(2) = -12 + 4 = -8\)- \(\begin{vmatrix} 3 & k \ 2 & 3 \end{vmatrix} = (3)(3) - (k)(2) = 9 - 2k\)Now substitute these back to the determinant expression:\[\text{Det} = 1(-4k + 6) - k(-8) + 3(9 - 2k)\]

Step 4: Simplify the Determinant Expression

Simplify:\[-4k + 6 + 8k + 27 - 6k = 0\]Combine like terms:\[-2k + 33 = 0\]Thus, solve for \(k\).

Step 5: Solve for k

Rearrange to isolate \(k\):\[-2k = -33\]Divide by -2:\[k = \frac{33}{2}\]

Step 6: Verify Statement Conditions

Given the value \(k = \frac{31}{2}\) for non-trivial solutions and we calculated \(k = \frac{33}{2}\), Statement 1 is false. Statement 2 states the right condition for non-trivial solutions, and it is true.

Key Concepts

Coefficient Matrix

In a system of linear equations, the coefficient matrix plays a crucial role. It's a matrix composed of the coefficients of the variables in the system. For example, consider the system of equations:

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

The coefficients of \(x\), \(y\), and \(z\) are organized into a matrix as follows: \[\begin{bmatrix}1 & k & 3 \3 & k & -2 \2 & 3 & -4\end{bmatrix}\] This coefficient matrix is used to determine the solvability of the system. By analyzing this matrix, one can use its determinant to decide whether there are non-trivial solutions. Its structure directly affects the determinant, which is key in systems of linear equations.

Homogeneous Equations

Homogeneous equations are systems of equations where all the constant terms are zero. This means that the entire system is equated to zero, like in our example:

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

Every linear combination of the variables results in zero. Homogeneous equations inherently include the trivial solution (where all variables equal zero). However, they can also have non-trivial solutions. The presence of these solutions depends heavily on the determinant of the coefficient matrix. Homogeneous equations are foundational in linear algebra, especially when exploring eigenvalues and eigenvectors.

Non-Trivial Solutions

Non-trivial solutions in the context of homogeneous equations refer to solutions where not all of the variables are zero. For these solutions to exist in a system of equations, a specific condition must be met: the determinant of the coefficient matrix is zero. This condition indicates that the system is singular, suggesting that the equations are not linearly independent. If the determinant of the coefficient matrix is non-zero, it implies that the only solution is the trivial one (all variables are zero). In our example, calculating the determinant involves assessing the following minors: - The minor \( \begin{vmatrix} k & -2 \ 3 & -4 \end{vmatrix} \) results in \( -4k + 6 \).- The minor \( \begin{vmatrix} 3 & -2 \ 2 & -4 \end{vmatrix} \) results in \( -8 \).- The minor \( \begin{vmatrix} 3 & k \ 2 & 3 \end{vmatrix} \) results in \( 9 - 2k \). These are used to compute the determinant, guiding whether non-trivial solutions exist. This interplay of determinants and solutions reveals deep insights into the nature of linear systems.