Question

For what value of $\lambda$ the system of equations $x+2y+\lambda z=0, x+2y+z=6, x+2y+3z=10$, has no solution.

• A. 2
• B. 3
• C. 4
• D. 5

Hint:

No solution occurs when equations represent parallel planes that do not coincide.

Answer 1

The Correct Option is A

Step 1: Concept
A system of linear equations has no solution if the determinant of the coefficient matrix ($\Delta$) is zero, but the auxiliary determinants ($\Delta_{x}, \Delta_{y}, \Delta_{z}$) are non-zero.

Step 2: Meaning
Examine the coefficients. The first two columns are identical $(1, 1, 1)$ and $(2, 2, 2)$ across all equations, meaning the rank of the coefficient matrix depends on the third column.

Step 3: Analysis
If we subtract the second equation from the third: $(x+2y+3z) - (x+2y+z) = 10 - 6 \implies 2z = 4 \implies z = 2$. Substituting $z=2$ into the second equation: $x+2y+2=6 \implies x+2y=4$.

Step 4: Conclusion
Substitute $z=2$ and $x+2y=4$ into the first equation: $4 + \lambda(2) = 0 \implies 2\lambda = -4 \implies \lambda = -2$. Given the options, if $\lambda=2$ (the marked answer), the first equation becomes $x+2y+2z=0$. With $z=2$, $x+2y=-4$, which contradicts $x+2y=4$ from the other equations, leading to no solution. Final Answer: (A)