Problem 31 - A Complete Resource Book in Mathematics for JEE Main 2017

Question

The number of values of \(k\) for which the linear equations $$ \begin{array}{r} 4 x+k y+2 z=0 \\ k x+4 y+z=0 \\ 2 x+2 y+z=0 \end{array} $$ possess a non-zero solution is (A) 0 (B) 3 (C) 2 (D) 1

Step-by-Step Solution

Answer

The number of values of \(k\) for which the equations have a non-zero solution is 2.

Step 1: Identify the System of Equations

The given system of linear equations is:\[ 4x + ky + 2z = 0 \]\[ kx + 4y + z = 0 \]\[ 2x + 2y + z = 0 \] and we aim to find the number of values of \(k\) for which there is a non-zero solution. This means the determinant of the coefficient matrix should be zero (since it has a non-trivial solution).

Step 2: Write the Coefficient Matrix

The coefficient matrix \(A\) of the system is:\[A = \begin{bmatrix} 4 & k & 2 \ k & 4 & 1 \ 2 & 2 & 1 \end{bmatrix}\]

Step 3: Find the Determinant of the Coefficient Matrix

The determinant of matrix \(A\) is given by:\[\text{det}(A) = 4(4 \cdot 1 - 1 \cdot 2) - k(k \cdot 1 - 1 \cdot 2) + 2(k \cdot 2 - 4 \cdot 2)\]Simplify:\[= 4(4 - 2) - k(k - 2) + 2(2k - 8)\]\[= 8 - (k^2 - 2k) + (4k - 16)\]\[= 8 - k^2 + 2k + 4k - 16\]\[= - k^2 + 6k - 8\]

Step 4: Determine When Determinant is Zero

Set the determinant equal to zero to find values of \(k\):\[-k^2 + 6k - 8 = 0\]This can be rearranged as:\[k^2 - 6k + 8 = 0\]We need to solve this quadratic equation.

Step 5: Solve the Quadratic Equation

The roots of the equation \(k^2 - 6k + 8 = 0\) can be found using the quadratic formula:\[k = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\]Here, \(a = 1\), \(b = -6\), \(c = 8\) gives:\[k = \frac{6 \pm \sqrt{(-6)^2 - 4 \times 1 \times 8}}{2 \times 1}\]\[k = \frac{6 \pm \sqrt{36 - 32}}{2}\]\[k = \frac{6 \pm 2}{2}\]This results in \(k = 4\) and \(k = 2\).

Step 6: Confirm Non-zero Solutions

Both \(k = 4\) and \(k = 2\) create conditions where the determinant is zero, thus indicating infinitely many solutions exist for each value due to the linear dependence of the equations. As a result, non-zero solutions do exist for these values of \(k\).

Key Concepts

Determinant

The determinant is an essential part of linear algebra involving matrices. For a system of linear equations to have a non-zero solution, its coefficient matrix must have a determinant of zero. This happens because a zero determinant indicates that the matrix is singular, leading to infinitely many solutions (non-trivial solutions) or no solution. You compute the determinant by applying specific operations to the entries of a square matrix. In this exercise, the determinant of the coefficient matrix \(A\) is calculated using the formula:

• \[\text{det}(A) = a(ei-fh) - b(di-fg) + c(dh-eg)\]

This results in a polynomial expression, which we set to zero to find specific values of \(k\) leading to non-zero solutions. Understanding the determinant as a tool that helps us analyze solvability is crucial for solving system equations.

Non-zero solution

In linear algebra, a non-zero solution refers to a solution of the system where at least one of the variables is non-zero. For this to occur, the determinant of the coefficient matrix must equal zero; this is because a determinant of zero implies that the matrix is not invertible and hence, the system may have infinitely many solutions or no solutions at all. Non-zero solutions are also referred to as non-trivial solutions since they differ from the trivial solution, which is where all variables equal zero. Solving for the determinant allows us to identify those values of the parameter \(k\) that result in such solutions. By finding the roots of the determinant equation, we can establish which values of \(k\) lead to non-zero solutions.

Quadratic equation

A quadratic equation is a second-order polynomial equation in a single variable \(x\), with a standard form of \(ax^2 + bx + c = 0\). Solving this type of equation involves finding the roots which satisfy the equation, usually through factoring, completing the square, or the quadratic formula:

• \[x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\]

In our exercise, the polynomial equation \(k^2 - 6k + 8 = 0\) needed solving to find the values of \(k\) where the determinant equals zero. These roots are identified as \(k = 4\) and \(k = 2\), showing the importance of the quadratic formula in deciphering potentially complex root solutions. Understanding how to manipulate and solve quadratic equations is an integral skill within linear algebra.

Coefficient matrix

A coefficient matrix is a matrix constructed from the coefficients of the variables in a system of linear equations. It plays a vital role in matrix algebra and linear equations. This matrix acts as the bridge between equations and their solutions by offering a structured way to organize information. For the given system of linear equations, the coefficient matrix is:

• \[A = \begin{bmatrix} 4 & k & 2 \ k & 4 & 1 \ 2 & 2 & 1 \end{bmatrix}\]

This matrix helps determine properties like the determinant and potential solutions. Analyzing it allows us to explore system behaviors like whether non-zero solutions exist. Creating the coefficient matrix is a fundamental step that showcases the utility of matrices in solving systems of linear equations efficiently and effectively.