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.
In the case \(a=0\,\), show that the equations have a solution if and only if
\(b=11\,\).
In the case \(a\ne0\) and \(b=11\,\) show that the equations have
a solution with \(z=\lambda\) for any given number \(\lambda\,\).
In the case \(a=2\) and \(b=11\,\) find the solution
for which \(x^2+y^2+z^2\) is least.
Find a value for \(a\) for which there is a solution such that
\(x>10^6\) and \(y^2+z^2<1\,\).
Solution:
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 \(x = 0, y = -1, z = -2\).
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 \(y\).
\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 \(\lambda = -1, x = 1, y = 0, z = -1\)
\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 \(a < \frac{3}{10^6}\)