The system of equations x + y + z = 6; x + 4y + 6z = 20; x + 4y + λz = μ has NO solution for values of λ and μ given by
λ = 6, μ ≠ 20
To determine the values of λ and μ for which the given system of equations has NO solution, we analyze the system using matrix methods. A system of linear equations can have a unique solution, infinitely many solutions, or no solution. We are looking for the conditions that lead to an inconsistent system (no solution).
The given system of equations is:
- $x + y + z = 6$
- $x + 4y + 6z = 20$
- $x + 4y + \lambda z = \mu$
Augmented Matrix Setup
First, we represent the system of equations in its augmented matrix form. The augmented matrix combines the coefficients of the variables and the constant terms on the right-hand side.
Gaussian Elimination Steps
Next, we apply elementary row operations to transform the augmented matrix into an echelon form using Gaussian elimination. This process helps us simplify the system of equations without changing its solutions.
Step 1: Eliminate x from the second and third equations.
- Perform $R_2 \leftarrow R_2 - R_1$
- Perform $R_3 \leftarrow R_3 - R_1$
Step 2: Eliminate y from the third equation.
- Perform $R_3 \leftarrow R_3 - R_2$
No Solution Condition
For a system of equations to have NO solution, a contradiction must arise in one of the equations after performing Gaussian elimination. This typically happens when a row in the coefficient part of the augmented matrix becomes all zeros, but the corresponding entry in the constant part is non-zero. In other words, we end up with an equation like $0 = \text{non-zero value}$.
From the third row of our echelon form matrix, the equation is:
$0x + 0y + (\lambda - 6)z = \mu - 20$
For this equation to represent a contradiction (i.e., $0 = \text{non-zero value}$), two conditions must be met simultaneously:
- The coefficient of $z$ must be zero: $\lambda - 6 = 0 \implies \lambda = 6$
- The constant term on the right-hand side must be non-zero: $\mu - 20 \ne 0 \implies \mu \ne 20$
If $\lambda = 6$ and $\mu \ne 20$, the last equation becomes $0 = (\text{a non-zero number})$, which is impossible. This implies that there is NO solution to the system.
Conclusion for λ and μ
Based on our analysis, the system of equations has NO solution for values of λ and μ where λ = 6 and μ ≠ 20.
Important Questions from Matrix Algebra
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A. A is a square matrix
B. A−1 exists
C. A is a symmetric matrix
D. |A| = 19
E. A is a null matrix
Choose the correct answer from the options given below.
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