The system of equations:
$x+y+z=6$
$x + 2y + 3z=10$
$x + 2y + λz=μ$
has no solution for
Answer & explanation
Correct answer: option 2
The given system of equations may be written as
$\begin{bmatrix}1&1&1\\1&2&3\\1&2&λ\end{bmatrix}\begin{bmatrix}x\\y\\z\end{bmatrix}=\begin{bmatrix}6\\10\\μ\end{bmatrix}$
$⇒\begin{bmatrix}1&1&1\\0&1&2\\0&1&λ-1\end{bmatrix}\begin{bmatrix}x\\y\\z\end{bmatrix}=\begin{bmatrix}6\\4\\μ-6\end{bmatrix}$ Applying $R_2 → R_2 - R_1, R_3→R_3-R_1$
$⇒\begin{bmatrix}1&1&1\\0&1&2\\0&0&λ-3\end{bmatrix}\begin{bmatrix}x\\y\\z\end{bmatrix}=\begin{bmatrix}6\\4\\μ-10\end{bmatrix}$ Applying $R_3 → R_3-R_2$
For $λ = 3$ and $μ≠ 10$, we observe that the rank of the coefficient matrix is 2 and that of the augmented matrix is 3. So, the given system of equations has no solution.