Question

The system of equations, $x+y+z=6, x+2y+3z = 10$ and $x + 2y + \lambda z = \mu$ has no solution, if

• A. λ = 3, μ = 10
• B. λ ≠ 3, μ = 10
• C. λ ≠ 3, μ ≠ 10
• D. λ = 3, μ≠ 10

Answer 1

Answer: (D)

λ = 3, μ≠ 10

Answer 2

The given system of equations is:

1. $x + y + z = 6$
2. $x + 2y + 3z = 10$
3. $x + 2y + \lambda z = \mu$

We can write the augmented matrix of the system as: $[A | B] = \begin{pmatrix} 1 & 1 & 1 & | & 6 \\ 1 & 2 & 3 & | & 10 \\ 1 & 2 & \lambda & | & \mu \end{pmatrix}$

We apply row operations to transform the matrix into row-echelon form. $R_2 \leftarrow R_2 - R_1$ $R_3 \leftarrow R_3 - R_1$ $\begin{pmatrix} 1 & 1 & 1 & | & 6 \\ 0 & 1 & 2 & | & 4 \\ 0 & 1 & \lambda - 1 & | & \mu - 6 \end{pmatrix}$

$R_3 \leftarrow R_3 - R_2$ $\begin{pmatrix} 1 & 1 & 1 & | & 6 \\ 0 & 1 & 2 & | & 4 \\ 0 & 0 & (\lambda - 1) - 2 & | & (\mu - 6) - 4 \end{pmatrix}$ $\begin{pmatrix} 1 & 1 & 1 & | & 6 \\ 0 & 1 & 2 & | & 4 \\ 0 & 0 & \lambda - 3 & | & \mu - 10 \end{pmatrix}$

The corresponding system of equations is: $x + y + z = 6$ $y + 2z = 4$ $(\lambda - 3)z = \mu - 10$

For the system to have no solution, the last equation must be contradictory. This happens when the coefficient of $z$ is zero, but the constant term on the right side is non-zero. So, we must have: $\lambda - 3 = 0 \implies \lambda = 3$ and $\mu - 10 \neq 0 \implies \mu \neq 10$

Thus, the system has no solution if $\lambda = 3$ and $\mu \neq 10$.

Let's check the given options: (α) $\lambda = 3, \mu = 10$: The last equation becomes $0 \cdot z = 0$, which means infinitely many solutions. (b) $\lambda \neq 3, \mu = 10$: The last equation has a non-zero coefficient for $z$, so there is a unique solution for $z$, and consequently for $x$ and $y$. (c) $\lambda \neq 3, \mu \neq 10$: The last equation has a non-zero coefficient for $z$, so there is a unique solution for $z$, and consequently for $x$ and $y$. (d) $\lambda = 3, \mu \neq 10$: The last equation becomes $0 \cdot z = (\text{non-zero number})$, which is a contradiction, meaning no solution exists.

The condition for no solution is $\lambda = 3$ and $\mu \neq 10$.