Question:
If the two systems of equations \( x+y-2z=0, 4x+4y-8z=0, 3x+3y-6z=0 \) and \( x+y+z=3, 2x+2y-z=\lambda, x+y-\mu z=1 \) have the same set of solutions, then \( \lambda+\mu = \)
If the two systems of equations \( x+y-2z=0, 4x+4y-8z=0, 3x+3y-6z=0 \) and \( x+y+z=3, 2x+2y-z=\lambda, x+y-\mu z=1 \) have the same set of solutions, then \( \lambda+\mu = \)
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When two systems share the same solution set, any specific solution of one must satisfy the other. Pick a simple point satisfying the first system and plug it into the second.
Updated On: Oct 7, 2026
- \( 0 \)
- \( -5 \)
- \( 2 \)
- \( 4 \)
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The Correct Option is D
Solution and Explanation
Concept:
Two systems of linear equations have the same solution set if they are equivalent. The first system simplifies to the single plane \( x+y-2z=0 \).
Step 1: Find the intersection of the solution set.
We solve \( x+y-2z=0 \) and \( x+y+z=3 \) simultaneously. Subtracting the equations: \( (x+y+z) - (x+y-2z) = 3 - 0 \Rightarrow 3z = 3 \Rightarrow z = 1 \). Substituting \( z=1 \): \( x+y-2(1)=0 \Rightarrow x+y=2 \).
Step 2: Substitute \( z=1 \) and \( x+y=2 \) into the other equations.
For \( 2x+2y-z = \lambda \): \( 2(x+y) - z = \lambda \Rightarrow 2(2) - 1 = \lambda \Rightarrow \lambda = 3 \). For \( x+y-\mu z = 1 \): \( 2 - \mu(1) = 1 \Rightarrow \mu = 1 \).
Step 3: Calculate \( \lambda+\mu \).
\( \lambda + \mu = 3 + 1 = 4 \). 4
Step 1: Find the intersection of the solution set.
We solve \( x+y-2z=0 \) and \( x+y+z=3 \) simultaneously. Subtracting the equations: \( (x+y+z) - (x+y-2z) = 3 - 0 \Rightarrow 3z = 3 \Rightarrow z = 1 \). Substituting \( z=1 \): \( x+y-2(1)=0 \Rightarrow x+y=2 \).
Step 2: Substitute \( z=1 \) and \( x+y=2 \) into the other equations.
For \( 2x+2y-z = \lambda \): \( 2(x+y) - z = \lambda \Rightarrow 2(2) - 1 = \lambda \Rightarrow \lambda = 3 \). For \( x+y-\mu z = 1 \): \( 2 - \mu(1) = 1 \Rightarrow \mu = 1 \).
Step 3: Calculate \( \lambda+\mu \).
\( \lambda + \mu = 3 + 1 = 4 \). 4
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