If the system of linear equations \(x + y + z = 6\), \(x + 2y + 5z = 10\), \(2x + 3y + \lambda z = \mu\) has infinitely many solutions, then the value of \(\lambda + \mu\) equals:
- 12
- 16
- 22
- 28
The Correct Option is C
Solution and Explanation
Step 1: Understanding the Concept:
For a system of linear equations to have infinitely many solutions, the determinant of the coefficient matrix (\(\Delta\)) must be zero, and all other determinants (\(\Delta_x, \Delta_y, \Delta_z\)) must also be zero. Alternatively, one equation must be a linear combination of the others.
Step 2: Key Formula or Approach:
Observe that Equation 1 + Equation 2 gives: \[ (x + x) + (y + 2y) + (z + 5z) = (6 + 10) \] \[ \Rightarrow 2x + 3y + 6z = 16 \]
Step 3: Detailed Explanation:
1. The third equation is: \[ 2x + 3y + \lambda z = \mu \] 2. For infinitely many solutions, this equation must match the sum of the first two equations (or be a multiple of it).
3. Comparing: \[ 2x + 3y + 6z = 16 \] with \[ 2x + 3y + \lambda z = \mu \] \[ \lambda = 6, \quad \mu = 16 \] 4. Therefore: \[ \lambda + \mu = 6 + 16 = 22 \]
Step 4: Final Answer:
\[ \boxed{22} \]
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