Question

Let \( \alpha, \beta \in \mathbb{R} \) be such that the system of linear equations} \[ x + 2y + z = 5 \] \[ 2x + y + \alpha z = 5 \] \[ 8x + 4y + \beta z = 18 \] has no solution. Then \( \frac{\beta}{\alpha} \) is equal to:

• A. \(-4\)
• B. \(4\)
• C. \(8\)
• D. \(-8\)

Answer 1

The Correct Option is C

Concept: For a system of three linear equations: \[ AX = B \]

• If \( \det(A) \neq 0 \) → unique solution.
• If \( \det(A) = 0 \) and the augmented matrix has different rank → no solution.
• If \( \det(A) = 0 \) and ranks are equal → infinitely many solutions.

Thus, for the system to have no solution, the determinant of the coefficient matrix must be zero but the equations must remain inconsistent.
Step 1:Form the coefficient matrix. \[ A = \begin{bmatrix} 1 & 2 & 1 \\ 2 & 1 & \alpha\\ 8 & 4 & \beta \end{bmatrix} \] For no solution, \[ |A| = 0 \]
Step 2:Compute the determinant.} \[ |A|= \begin{vmatrix} 1 & 2 & 1\\ 2 & 1 & \alpha \\ 8 & 4 & \beta \end{vmatrix} \] Expanding along the first row: \[ =1 \begin{vmatrix} 1 & \alpha \\ 4 & \beta \end{vmatrix} -2 \begin{vmatrix} 2 & \alpha \\ 8 & \beta \end{vmatrix} +1 \begin{vmatrix} 2 & 1 \\ 8 & 4 \end{vmatrix} \] \[ =(\beta-4\alpha)-2(2\beta-8\alpha)+(8-8) \] \[ =\beta-4\alpha-4\beta+16\alpha \] \[ =12\alpha-3\beta \] For no solution, \[ 12\alpha-3\beta=0 \] \[ 4\alpha=\beta \]
Step 3:Check the consistency condition. Multiply the first equation by \(4\): \[ 4x+8y+4z=20 \] Multiply the second equation by \(2\): \[ 4x+2y+2\alpha z=10 \] The third equation is \[ 8x+4y+\beta z=18 \] Substituting \( \beta=4\alpha \), the left-hand sides become proportional, but the constants are not proportional, hence the system becomes inconsistent. Thus, \[ \beta = 8\alpha \] \[ \frac{\beta}{\alpha}=8 \]