Question

If the system of simultaneous linear equations \(x - 2y + z = 0\), \(2x + 3y + z = 6\) and \(x + 2y + pz = q\) has infinitely many solutions, then

• A. \(p + q = 4\)
• B. \(pq = \frac{48}{49}\)
• C. \(q - p = 3\)
• D. \(\frac{p}{q} = 4\)

Hint:

When establishing infinite solutions, always solve for the principal determinant \(D = 0\) first to find one parameter, then pick the easiest \(D_x, D_y,\) or \(D_z\) containing the second parameter.

Answer 1

The Correct Option is C

Step 1: Understanding the Question:
For a system of three linear equations to have infinitely many solutions, the determinant of the coefficient matrix (\(D\)) must be zero, and the determinants replacing the columns with the constants (\(D_x, D_y, D_z\)) must also evaluate to zero.
Step 2: Key Formula or Approach:
Apply Cramer's Rule condition for infinite solutions: \[ D = 0 \quad \text{and} \quad D_z = 0 \] Step 3: Detailed Explanation:
Let's find the main determinant \(D\): \[ D = \begin{vmatrix} 1 & -2 & 1
2 & 3 & 1
1 & 2 & p \end{vmatrix} \] Expanding along the first row: \[ D = 1(3p - 2) - (-2)(2p - 1) + 1(4 - 3) \] \[ D = 3p - 2 + 4p - 2 + 1 = 7p - 3 \] Setting \(D = 0\) gives: \[ 7p - 3 = 0 \implies p = \frac{3}{7} \] Now, let's find the determinant \(D_z\): \[ D_z = \begin{vmatrix} 1 & -2 & 0
2 & 3 & 6
1 & 2 & q \end{vmatrix} \] Expanding along the first row: \[ D_z = 1(3q - 12) - (-2)(2q - 6) + 0 \] \[ D_z = 3q - 12 + 4q - 12 = 7q - 24 \] Setting \(D_z = 0\) gives: \[ 7q - 24 = 0 \implies q = \frac{24}{7} \] Now we test the given options with \(p = \frac{3}{7}\) and \(q = \frac{24}{7}\): Option (A): \(p + q = \frac{3}{7} + \frac{24}{7} = \frac{27}{7} \neq 4\)
Option (B): \(pq = \left(\frac{3}{7}\right)\left(\frac{24}{7}\right) = \frac{72}{49} \neq \frac{48}{49}\)
Option (C): \(q - p = \frac{24}{7} - \frac{3}{7} = \frac{21}{7} = 3\) (Matches!)
Option (D): \(\frac{p}{q} = \frac{3/7}{24/7} = \frac{3}{24} = \frac{1}{8} \neq 4\)
Step 4: Final Answer:
The correct condition is \(q - p = 3\).