Problem source

Consider the equations
\begin{alignat*}{2}
ax-&y- \ z  && =3 \;,\\
2ax -&y -3z && = 7 \;,\\
3ax-&y-5z   && =b \;,
\end{alignat*}
where $a$ and $b$ are given constants.
\begin{questionparts}
\item In the case $a=0\,$, show that the equations have a solution if and only if
$b=11\,$.
\item In the case $a\ne0$ and $b=11\,$ show that the equations have
a solution with $z=\lambda$ for any given number $\lambda\,$.
\item  In the case $a=2$ and $b=11\,$ find the solution
for which $x^2+y^2+z^2$ is least.
\item Find a value for $a$ for which there is a solution such that
$x>10^6$ and $y^2+z^2<1\,$.
\end{questionparts}

Solution source

\begin{questionparts}
\item If $a = 0$, then then the LHS second equation is the average of the first and last equations, ie $7 = \frac{b+3}{2}$ so $b = 11$. This clearly has solutions, say $x = 0, y = -1, z = -2$.

\item If $a \neq 0$ and $b = 11$, it is still the case that the third equation a linear combination of the first two. Therefore we can consider the linear system:

\begin{cases}
ax - y &= 3 + \lambda \\
2ax - y &= 7 + 3\lambda
\end{cases} and since $-a+2a = a \neq 0$ the solution has a unique solution for $x$ and $y$.

\item 
\begin{align*}
\begin{cases}
2x - y &= 3 + \lambda \\
4x - y &= 7 + 3\lambda
\end{cases} \Rightarrow x = 2  +\lambda, y = 1 + \lambda \\
x^2 + y^2 + z^2 &= (2 + \lambda)^2 + (1+\lambda)^2 + \lambda^2 \\
&= (4 + 1) + (4+2)\lambda + 3\lambda^2 \\
&= 5 + 3((\lambda+1)^2 - 1) \\
&= 3(\lambda + 1)^2 + 2
\end{align*}

Therefore the solution is minimized when $\lambda = -1, x = 1, y = 0, z = -1$

\item \begin{align*}
\begin{cases}
ax - y &= 3 + \lambda \\
2ax - y &= 7 + 3\lambda
\end{cases} \Rightarrow x = \frac{4  +2\lambda}{a}, y = 1 + \lambda
\end{align*}

We want say $\lambda = -\frac12$ then we have $y^2 + z^2 = \frac12$ and $x = \frac{3}{a}$, so choose $a < \frac{3}{10^6}$

\end{questionparts}