Solutions of a linear system of equations with parameters, using matrices

Question

Cheers, I have this system of linear equations $\begin{cases} 6x_1+ 2x_2 + 3 x_3 + 4 x_4= 15 \\ 5 x_1 + 6x_2 + 7 x_3 + 8 x_4 = 26 \\ 9 x_1 + 10x_2 + 11 x_3 + 12 x_4 = 42 \\ λ x_1 + 14x_2 + 15 x_3 + 16 x_4 = λ \end{cases}$ which as you can see has the parameter $λ$ inside. I am asked to find the solution of this set, so I put this into a matrix $A$ and thus far I have $ A = \begin{pmatrix} 6&& 2 && 3 && 4 && 15 \\ 5 && 6 && 7 && 8 && 26 \\ 9 && 10 && 11 && 12 && 42 \\ λ && 14 && 15 && 16 && λ\end{pmatrix}$. My initial thought was to row reduce the matrix in row echelon form, and then conclude on a solution with $λ$. The $rref$ gives me: $\left( \begin{array}{ccccc} 1 & 0 & 0 & 0 & 0 \\ 0 & 1 & 0 & -1 & 0 \\ 0 & 0 & 1 & 2 & 0 \\ 0 & 0 & 0 & 0 & 1 \\ \end{array} \right)$.From the last row we can see that 0 = 1, so I conclude that the system does not have a solution for $λ \in \mathbb{R}$. Is my solution and/or thought process correct? Thanks!

Answer 1

Yes, your train of thought is correct.

The fact that the last line in the matrix is $(0,0,0,0,1)$ means that by adding scaled versions of the equations $\begin{cases} 6x_1+ 2x_2 + 3 x_3 + 4 x_4= 15 \\ 5 x_1 + 6x_2 + 7 x_3 + 8 x_4 = 26 \\ 9 x_1 + 10x_2 + 11 x_3 + 12 x_4 = 42 \\ λ x_1 + 14x_2 + 15 x_3 + 16 x_4 = λ \end{cases}$
we can create an equation of the form $$ 0x_1+0x_2+0x_3+0x_4=1 $$ , which is false for all $x_1,...,x_4\in\mathbb{R}$.

Therefore, the equation system is unsolvable, and since we didn't assume anything about $\lambda$ but that it's a real number (when we were summing and scaling the equations), the equation system is unsolvable for any $\lambda\in\mathbb R$.