Problem 121 - 71 JEE Main Mathematics Online (2020 - 2012) & Offline (2018 - 2002) Chapterwise + Topicwise Solved Papers

Question

Let \(a, b, c\) be any real numbers. Suppose that there are real numbers \(x, y, z\) not all zero such that \(x=c y+b z, y=a z+c x\), and \(z=b x+a y\). Then \(a^{2}+b^{2}+c^{2}+2 a b c\) is equal to (a) 2 (b) \(-1\) (c) 0 (d) 1

Step-by-Step Solution

Answer

The value is (d) 1.

Step 1: Understand the System of Equations

We have the three equations: 1. \( x = c y + b z \)2. \( y = a z + c x \)3. \( z = b x + a y \).These equations indicate relationships between \(x, y, \) and \(z\). We are looking for a non-trivial solution where \(x, y, z\) are not all zero.

Step 2: Matrix Representation

Represent the system of equations using a matrix form: \( A \begin{bmatrix} x \ y \ z \end{bmatrix} = \begin{bmatrix} 0 \ 0 \ 0 \end{bmatrix} \) where the matrix \( A \) is\[ A = \begin{bmatrix} 1 & -c & -b \ -c & 1 & -a \ -b & -a & 1 \end{bmatrix} \].For non-trivial solutions, the determinant of \( A \) must be zero.

Step 3: Compute the Determinant

The determinant of matrix \(A\) is given by:\[ \det(A) = \begin{vmatrix} 1 & -c & -b \ -c & 1 & -a \ -b & -a & 1 \end{vmatrix} \].Calculate this determinant using expansion or properties of determinants.

Step 4: Expand the Determinant

To find \( \det(A) \), use cofactor expansion along the first row:\[ \det(A) = 1(1 \cdot 1 - a \cdot a) - (-c)(-c \cdot 1 + a \cdot b) + (-b)(-c \cdot (-a) + b \cdot 1) \].Simplify: \( = 1 - a^2 - c^2 + a^2b + a^2c - ac^2 + ab^2 \).

Step 5: Simplify the Expression

Simplifying further, we have:\[ \det(A) = 1 - a^2 - b^2 - c^2 + 2abc \].Set the determinant equal to zero for non-trivial solutions:\[ a^2 + b^2 + c^2 + 2abc = 1 \].

Step 6: Analyze the Result

From the determinant condition, \(a^2 + b^2 + c^2 + 2abc = 1\).This matches option (d) in the given choices.

Key Concepts

System of Equations

A system of equations is a set of equations with multiple variables. The primary goal is to find values for these variables that will satisfy all equations. In this particular exercise, we are working with three equations involving the variables \(x\), \(y\), and \(z\):

• \(x = c y + b z\)
• \(y = a z + c x\)
• \(z = b x + a y\)

Each equation shows how the variables are related to each other. This is a crucial step because understanding these relationships is necessary to efficiently convert them into a matrix form. Note that the exercise seeks non-zero solutions, implying that not all values of \(x\), \(y\), and \(z\) can simultaneously be zero. This is known as seeking non-trivial solutions, which often leads us to consider matrix representation and determinant calculations to find valid solutions.

Determinant Calculation

Determinant calculation is vital in solving systems of linear equations, especially for finding non-trivial solutions. In the matrix representation of the given system: \[A = \begin{bmatrix} 1 & -c & -b \-c & 1 & -a \-b & -a & 1 \end{bmatrix}\]To find whether there's a non-trivial solution, you calculate the determinant of matrix \(A\). If the determinant equals zero, the system has non-trivial solutions. The determinant \(\det(A)\) is found using cofactor expansion:\[\det(A) = 1(1 \cdot 1 - a \cdot a) - (-c)(-c \cdot 1 + a \cdot b) + (-b)(-c \cdot (-a) + b \cdot 1) \]After expanding and simplifying, it yields:\[\det(A) = 1 - a^2 - b^2 - c^2 + 2abc\]Setting this determinant to zero is critical because it ensures that the matrix is singular, leading to infinite solutions with at least one being non-zero.

Non-trivial Solutions

Non-trivial solutions are solutions to equations where the variables are not all zero. In linear algebra, finding such solutions often involves ensuring a matrix derived from the system has a zero determinant. In this exercise, the key is the condition:\[ a^2 + b^2 + c^2 + 2abc = 1 \]This condition arises because setting the determinant of matrix \(A\) to zero results in a non-trivial solution for \(x\), \(y\), and \(z\).

• If the determinant were not zero, it would imply only the trivial solution exists (all variables are zero).
• In this problem, since \(a^2 + b^2 + c^2 + 2abc = 1\) holds, we achieve a non-trivial solution, matching option (d) from the choices given in the exercise.

Such problems demonstrate the interplay between equations, matrices, and determinants in solving for variables in a nuanced and substantial way.