Question

If the system of linear equations
x + my + az = 0
2x + ay + mz = 0
ax + 2y - z = 0
with m and a as non-zero constants, admits a non-trivial solution, then which one of the following conditions is correct?

• A. \(m^2 - a^2 = 3\)
• B. \(m^2 - a^2 = -3\)
• C. \(a^2 - 2m^2 = -3\)
• D. \(m^2 - 2a^2 = 3\)

Hint:

Remember that a homogeneous system \(A\mathbf{x} = \mathbf{0}\) always has the trivial solution \(\mathbf{x} = \mathbf{0}\). The existence of a \textit{non-trivial} solution is a special condition that implies the matrix \(A\) is singular, which is equivalent to its determinant being zero.

Answer 1

The Correct Option is B

Step 1: Understanding the Concept:
A system of homogeneous linear equations (where all constant terms are zero) of the form \(A\mathbf{x} = \mathbf{0}\) has a non-trivial solution (a solution other than \(\mathbf{x} = \mathbf{0}\)) if and only if the determinant of the coefficient matrix \(A\) is zero.
Step 2: Key Formula or Approach:
We need to set up the coefficient matrix for the given system of equations and then calculate its determinant. The condition for a non-trivial solution is \(\det(A) = 0\).
Step 3: Detailed Explanation:
The given system of equations is: \[ x + my + az = 0 \] \[ 2x + ay + mz = 0 \] \[ ax + 2y - z = 0 \] The coefficient matrix \(A\) is: \[ A = \begin{pmatrix} 1 & m & a \\ 2 & a & m \\ a & \\ 2 & -1 \end{pmatrix} \] For a non-trivial solution, the determinant of \(A\) must be zero. \[ \det(A) = \begin{vmatrix} 1 & m & a \\ 2 & a & m \\ a &\\ 2 & -1 \end{vmatrix} = 0 \] Let's compute the determinant by expanding along the first row: \[ 1 \begin{vmatrix} a & m \\ 2 & -1 \end{vmatrix} - m \begin{vmatrix} 2 & m\\ a & -1 \end{vmatrix} + a \begin{vmatrix} 2 & a \\ a & 2 \end{vmatrix} = 0 \] \[ 1(a(-1) - m(2)) - m(2(-1) - m(a)) + a(2(2) - a(a)) = 0 \] \[ (-a - 2m) - m(-2 - am) + a(4 - a^2) = 0 \] \[ -a - 2m + 2m + am^2 + 4a - a^3 = 0 \] Combine like terms: \[ 3a + am^2 - a^3 = 0 \] Since it is given that \(a\) is a non-zero constant, we can divide the entire equation by \(a\): \[ 3 + m^2 - a^2 = 0 \] Rearranging the terms to match the form of the options: \[ m^2 - a^2 = -3 \] This condition must be satisfied for the system to have a non-trivial solution. Step 4: Final Answer:
The condition for a non-trivial solution is \(m^2 - a^2 = -3\). This corresponds to option (B).