Find the general solution of linear system of equations (homework assignement)

Question

Assuming that $\lambda$ is any number, does below set of equations have a solution? If it has, find the general solution for this system. $$\lambda x_1+x_2+x_3=1$$ $$x_1+\lambda x_2+x_3=\lambda$$ $$x_1+x_2+\lambda x_3=\lambda^2$$ I understand how this problem should be solved, however when transform this set to a matrix and then row reduce it to an echolon form I get: $$B=\left(\begin{array}{ccc|c} 1 & 1 & \lambda & \lambda\\ 0 & \lambda-1 & 1-\lambda & \lambda-\lambda^2\\ 0 & 0 & (\lambda-1)(\lambda+2) & \lambda^3+\lambda^2-\lambda-1\end{array}\right)$$ But from this point I start to struggle, don't really know where I'm making mistake.

Answer 1

Your work is good. Now you are at a crossroads, but either $\lambda=1$ or $\lambda\ne1$.

In the latter case, you can divide the second row by $\lambda-1$ $$ \left( \begin{array}{ccc|c} 1 & 1 & \lambda & \lambda\\ 0 & 1 & -1 & -\lambda\\ 0 & 0 & (\lambda-1)(\lambda+2) & (\lambda-1)(\lambda+1)^2 \end{array} \right) $$ and also the third row by $\lambda-1$, to get $$ \left( \begin{array}{ccc|c} 1 & 1 & \lambda & \lambda\\ 0 & 1 & -1 & -\lambda\\ 0 & 0 & \lambda+2 & (\lambda+1)^2 \end{array} \right) $$ Now, if $\lambda=-2$, the system has no solution. If $\lambda\ne-2$ you can find the unique solution by standard methods.

If $\lambda=1$, the matrix becomes $$ \left( \begin{array}{cc} 1 & 1 & 1 & 1\\ 0 & 0 & 0 & 0\\ 0 & 0 & 0 & 0 \end{array} \right) $$ which is easy to deal with.

Answer 2

$$\lambda x_1+x_2+x_3=1$$ $$x_1+\lambda x_2+x_3=\lambda$$ $$x_1+x_2+\lambda x_3=\lambda^2$$ Adding all three equations together gives $$(\lambda+2)(x_1+x_2+x_3)=1+\lambda+\lambda^2$$ If $\lambda=-2$ we have $0=1+(-2)+(-2)^2=3$, which is absurd. Thus we can divide by $\lambda+2$:

$$x_1+x_2+x_3=\frac{1+\lambda+\lambda^2}{\lambda+2}$$ Subtracting this from all three original equations gives $$(\lambda-1)x_1=1-\frac{1+\lambda+\lambda^2}{\lambda+2}$$ $$(\lambda-1)x_2=\lambda-\frac{1+\lambda+\lambda^2}{\lambda+2}$$ $$(\lambda-1)x_3=\lambda^2-\frac{1+\lambda+\lambda^2}{\lambda+2}$$ If $\lambda=1$, all three of the original equations reduce to $x_1+x_2+x_3=1$, whose solution is simply $$x_1=p,x_2=q,x_3=1-p-q\qquad p,q\in\mathbb R\tag1$$ Otherwise, dividing by $\lambda-1$ gives the unique solution of $$x_1=\frac{1-\frac{1+\lambda+\lambda^2}{\lambda+2}}{\lambda-1}=\frac1{\lambda+2}-1\\ x_2=\frac{\lambda-\frac{1+\lambda+\lambda^2}{\lambda+2}}{\lambda-1}=\frac1{\lambda+2}\\ x_3=\frac{\lambda^2-\frac{1+\lambda+\lambda^2}{\lambda+2}}{\lambda-1}=\frac{(\lambda+1)^2}{\lambda+2}\tag2$$ In conclusion, the solutions of the linear system are

• as in $(1)$ if $\lambda=1$
• none if $\lambda=-2$
• as in $(2)$ otherwise.