Consider this linear equations system:
\begin{align*} \begin{pmatrix} a & 1 & 1\\ 1 & a & 1 \\ 1 & 1 & 1
\end{pmatrix} \begin{pmatrix} x\\ y\\ z \end{pmatrix}=
\begin{pmatrix} b\\ c\\ 2 \end{pmatrix} \end{align*}
- Determine the $a,b,c$ values such that the system has a unique solution.
- To those $a,b,c$ values such that the system has a non unique solution, find all the possible solutions.
We have that:
\begin{align*}
ax+y+z&=b\\x+ay+z&=c\\x+y+z&=2
\end{align*}
Then:
\begin{align*}
\begin{pmatrix}
a & 1 & 1 & \vdots & b\\
1 & a & 1 & \vdots & c\\
1 & 1 & 1 & \vdots & 2
\end{pmatrix}\sim \cdots \sim
\begin{pmatrix}
1 & 0 & 0 & \vdots & \frac{b-2}{a-1}\\
0 & 1 & \frac{1}{a} & \vdots & \frac{c(a-1)-(b-2)}{a(a-1)}\\
0 & 0 & 1 & \vdots & \frac{a \left [ 2(a-1)-(b-2) \right ]}{(a-1)^{2}}
\end{pmatrix}
\end{align*}
But I don't know how to continue to solve 1 and 2. What am I doing wrong or how would you solve it? I would really appreciate your help.
Best Answer
How I would solve the first question: A quick check shows that $\det A=(a-1)^2$, so there is a unique solution unless $a=1$.
A hands-on approach without any theory works too: Note that $x+y+z=2$ and so $$b-2=(ax+y+z)-(x+y+z)=(a-1)x,$$ $$c-2=(x+ay+z)-(x+y+z)=(a-1)y,$$ which shows that if $a\neq1$ then $$x=\frac{b-2}{a-1}\qquad\text{ and }\qquad y=\frac{c-2}{a-1},$$ and hence $$z=2-x-y=\frac{2a-b-c+2}{a-1}.$$ So if $a\neq1$ there is a unique solution, for any value of $b$ and $c$.
As for the second question: If $a=1$ then it is necessary and sufficient that $b=c=2$ for a solution to exist. The solutions are all $x$, $y$ and $z$ such that $x+y+z=2$. That is, all triplets of the form $$(x,y,2-x-y).$$