\section*{Problem 6 - 061226}

a) Give all triples of real numbers \((x, y, z)\) that satisfy the system of equations (1)

\[
\begin{array}{r}
2 x+3 y+z=1 \\
4 x-y+2 z=2 \\
8 x+5 y+3 z=4
\end{array}
\]

b) Form all systems of equations that differ from system (1) in exactly one coefficient and have infinitely many solutions!

Here, "coefficients" refer to both the "coefficients" of the variables on the "left sides" and the "constant terms" on the "right sides."

Also, in these cases, give all triples of real numbers that satisfy the respective systems of equations!

c) Form a system of equations that differs from (1) in exactly two coefficients and is not satisfied by any triple of real numbers!
}

a) Adding twice the first equation to the second yields \(8 x+5 y+4 z=4\), from which with the third equation \(z=0\) follows. Substituting this and subtracting the second equation from twice the first, one gets \(y=0\) and finally \(x=\frac{1}{2}\). Thus, \(\left(\frac{1}{2} ; 0 ; 0\right)\) is the only solution triplet, which is also confirmed by substitution.

b) The system of equations has infinitely many solutions only if there is a combination of the equations that reduces to \(0=0\). (The equations are linearly dependent.) For this, we distinguish two cases:

1. Case: There is a real number \(k\) such that the \(k\)-fold of the first equation equals the second. This yields the following relationship: \(2 k=4\), if no \(x\)-coefficient has been changed, or \(k=2\), if no \(z\)-coefficient has been changed.

Since at least one of these two cases must occur, \(k=2\). Therefore, the double of the coefficient of \(y\) in the first equation must match the coefficient of \(y\) in the second equation, so either the coefficient of \(y\) in the first equation must be changed to \(-\frac{1}{2}\), or the coefficient of \(y\) in the second equation must be changed to +6.

In the first variant, the system of equations now has the form

\[
\begin{array}{r}
2 x-\frac{1}{2} y+z=1 \\
4 x-y+2 z=2 \\
8 x+5 y+3 z=4
\end{array}
\]

and has (subtracting twice the second from the third equation) the solutions

\[
\left\{\left.\left(\frac{1}{2}-\frac{13}{4} \cdot t, t, 7 \cdot t\right) \right\rvert\, t \in \mathbb{R}\right\}
\]

In the second variant, the system of equations has the form

\[
\begin{array}{r}
2 x+3 y+z=1 \\
4 x+6 y+2 z=2 \\
8 x+5 y+3 z=4
\end{array}
\]

and has (subtracting the third equation from twice the second) the solutions

\[
\left\{\left.\left(\frac{1}{2}+2 \cdot t, t,-7 \cdot t\right) \right\rvert\, t \in \mathbb{R}\right\}
\]

2. Case: There are real numbers \(k\) and \(\ell\) such that the sum of the \(k\)-fold of the first and the \(\ell\)-fold of the second equation equals the third. This yields the following relationships:
\(2 k+4 \ell=8\), if no \(x\)-coefficient has been changed,

\(3 k-\ell=5\), if no \(y\)-coefficient has been changed,

\(k+2 \ell=4\), if no right-hand side has been changed, and finally

\(k+2 \ell=3\), if no \(z\)-coefficient has been changed.

Since not both the \(x\)-coefficients and the right-hand sides have been modified, in each case \(k+2 \ell=4\) holds. This contradicts the condition that would hold if no \(z\)-coefficient had been changed. Therefore, only by changing one of these coefficients of the variable \(z\) can a system of equations be constructed that has infinitely many solutions.

In particular, besides the \(x\)-coefficients, the \(y\)-coefficients remain unchanged, and it follows that \(k=2\) and \(\ell=1\). The sum of twice the first equation and the second equation must therefore equal the third. There are three possibilities to adjust the coefficient of \(z\) in each of the three equations:

In the first variant, the coefficient of \(z\) in the first equation is set to \(\frac{3-2}{2}=\frac{1}{2}\), yielding the new system of equations

\[
\begin{array}{r}
2 x+3 y+\frac{1}{2} z=1 \\
4 x-y+2 z=2 \\
8 x+5 y+3 z=4
\end{array}
\]

which has the solutions \(\left\{\left.\left(\frac{1}{2}-\frac{13}{4} \cdot t, t, 7 \cdot t\right) \right\rvert\, t \in \mathbb{R}\right\}\).

In the second variant, the coefficient of \(z\) in the second equation is changed to \(3-2 \cdot 1=1\), yielding the new system of equations

\[
\begin{array}{r}
2 x+3 y+z=1 \\
4 x-y+z=2 \\
8 x+5 y+3 z=4
\end{array}
\]

which has the solutions \(\left\{\left.\left(\frac{1}{2}+2 \cdot t, t,-7 \cdot t\right) \right\rvert\, t \in \mathbb{R}\right\}\).

And finally, in the third variant, the coefficient of \(z\) in the third equation is set to \(2 \cdot 1+2=4\), yielding the new system of equations

\[
\begin{array}{r}
2 x+3 y+z=1 \\
4 x-y+2 z=2 \\
8 x+5 y+4 z=4
\end{array}
\]

which has the solutions \(\{(t, 0,1-2 \cdot t) \mid t \in \mathbb{R}\}\).

In total, there are thus five different ways to adjust one of the coefficients of the originally given system of equations so that the respective new system of equations then has infinitely many solutions.

c) Change the coefficient of \(y\) in the second equation to +6 and the right-hand side of the second equation to 0. Then one obtains the new system of equations

\[
\begin{aligned}
2 x+3 y+z & =1 \\
4 x+6 y+2 z & =0 \\
8 x+5 y+3 z & =4
\end{aligned}
\]

where from the first equation \(4 x+6 y+2 z=2 \neq 0\) follows, which contradicts the second equation. Thus, this system of equations has no solution.