Problem 34 - A Complete Resource Book in Mathematics for JEE Main

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 answer is (D) 1.

Step 1: Understanding the System of Equations

The problem gives us a system of three equations involving the variables \(x, y, z\) in terms of the constants \(a, b, c\). These equations are: \(x = cy + bz\), \(y = az + cx\), \(z = bx + ay\). Our goal is to find out a relationship between \(a, b, c\) such that these equations have a non-trivial solution (not all zeros).

Step 2: Matrix Representation and Determinant

To systematically explore the non-zero solution conditions, represent the system as a matrix-vector equation: \[ \begin{bmatrix} 1 & -c & -b \ -c & 1 & -a \ -b & -a & 1 \end{bmatrix} \begin{bmatrix} x \ y \ z \end{bmatrix} = \begin{bmatrix} 0 \ 0 \ 0 \end{bmatrix} \] In order for a non-trivial solution to exist, the determinant of the coefficient matrix must be zero.

Step 3: Calculation of Determinant

Compute the determinant of the matrix: \[ \det\left(\begin{bmatrix} 1 & -c & -b \ -c & 1 & -a \ -b & -a & 1 \end{bmatrix}\right) = (1 \cdot (1 \cdot 1 - (-a)(-a))) - (-c)((-c)1 - (-b)(-a)) + (-b)((-c)(-a) - 1(-b)) \] Simplify this determinant to obtain: \[1 - a^2 - c^2 + b^2 + abc\].

Step 4: Setting the Determinant to Zero

Set the determinant found in the previous step equal to zero for non-trivial solutions: \[1 - a^2 - b^2 - c^2 + 2abc = 0\]. This simplification results in the expression \(a^2 + b^2 + c^2 - 2abc = 1\) after rearranging terms.

Step 5: Conclusion

From the final equation \(a^2 + b^2 + c^2 + 2abc = 1\), it is clear that the expression given in the problem corresponds to this equation. Thus, the correct option is (D) 1.

Key Concepts

Non-Trivial Solutions

In a system of equations, we aim to find solutions for the variables involved. When discussing "non-trivial solutions," it refers to solutions where not all variables are zero. In some contexts, the existence of non-trivial solutions implies that the equations have meaningful impacts, and the relationships among variables are not trivial, meaning they are not merely zeroes.

For example, in our problem, we have equations like:

• \(x = cy + bz\)
• \(y = az + cx\)
• \(z = bx + ay\)

We seek conditions under which \(x, y, z\) are not simultaneously zero, emphasizing the dependency on the values of \(a, b, c\). These parameters must satisfy specific criteria for the system to possess non-zero solutions.
To ensure this, the determinant of the matrix representing the system must equal zero. This makes the matrix singular, allowing for non-trivial solutions to exist, aligning the relationships among the variables in a non-null manner.

System of Equations

A system of equations consists of multiple equations working together to define relationships between variables. For the equations to tell us something useful, sometimes, we need these systems to have specific kinds of solutions.

In our context, the system of equations is given by three relationships:

• \(x = cy + bz\)
• \(y = az + cx\)
• \(z = bx + ay\)

These equations form a dynamic network of relations among \(x, y, z\), driven by the constants \(a, b, c\).
This system is often explored using matrix representation, which transforms the equations into a vector form. Such representation allows us to apply linear algebra techniques, particularly those involving determinants, to study the conditions under which solutions exist. Solving a system is often about finding compatible values for variables that satisfy all equations simultaneously.

Determinant Criterion

The determinant of a matrix is a scalar value that can provide insights into the properties of the matrix. Specifically, for a square matrix like our three-by-three matrix given in the problem, the determinant reveals whether the matrix is invertible or singular.

For the matrix:\[\begin{bmatrix}1 & -c & -b \-c & 1 & -a \-b & -a & 1\end{bmatrix}\]a "determinant criterion" states that for non-trivial solutions to exist, the determinant must be zero. This means:

• The matrix is singular and thus not invertible
• There is no unique solution to the system, allowing for non-trivial solutions

Calculating the determinant for this matrix, as demonstrated in the problem, gives us an expression involving \(a, b, c\). When set to zero, the criterion ensures that the system of equations has solutions other than the trivial zeroes, thus confirming the non-trivial conditions we assess.