Problem. For each \(a\in\mathbb{R}\) we will define a system of linear equations over \(\mathbb{R}\) by \[\begin{cases} 7x+2ay+z & =a\\ 8x+3ay+z & =a\\ 10x+5ay+2z & =2a \end{cases}\]
Determine, for each value of \(a\), whether the system has no solution, a unique solution, or infinitely many solutions.
In the case where at least one solution exists, write the solution set in parametric form.
We shall represent the system by its augmented matrix and reduce it to RREF using elementary row operations, keeping \(a\) as a parameter.
\[\begin{bmatrix}\begin{array}{ccc|c} 7 & 2a & 1 & a\\ 8 & 3a & 1 & a\\ 10 & 5a & 2 & 2a \end{array}\end{bmatrix}\overset{R_{1}\leftrightarrow R_{3}}{\Longrightarrow}\begin{bmatrix}\begin{array}{ccc|c} 10 & 5a & 2 & 2a\\ 8 & 3a & 1 & a\\ 7 & 2a & 1 & a \end{array}\end{bmatrix}\overset{R_{1}\rightarrow\frac{1}{10}R_{1}}{\Longrightarrow}\begin{bmatrix}\begin{array}{ccc|c} 1 & \frac{1}{2}a & \frac{1}{5} & \frac{1}{5}a\\ 8 & 3a & 1 & a\\ 7 & 2a & 1 & a \end{array}\end{bmatrix}\]
\[\overset{R_{2}\rightarrow R_{2}-8R_{1}}{\underset{R_{3}\rightarrow R_{3}-7R_{1}}{\Longrightarrow}}\begin{bmatrix}\begin{array}{ccc|c} 1 & \frac{1}{2}a & \frac{1}{5} & \frac{1}{5}a\\ 0 & (3-\frac{8}{2})a & 1-\frac{8}{5} & (1-\frac{8}{5})a\\ 0 & (2-\frac{7}{2})a & 1-\frac{7}{5} & (1-\frac{7}{5})a \end{array}\end{bmatrix}=\begin{bmatrix}\begin{array}{ccc|c} 1 & \frac{1}{2}a & \frac{1}{5} & \frac{1}{5}a\\ 0 & -a & -\frac{3}{5} & -\frac{3}{5}a\\ 0 & -\frac{3}{2}a & -\frac{2}{5} & -\frac{2}{5}a \end{array}\end{bmatrix}\overset{R_{2}\rightarrow-\frac{1}{a}R_{2}}{\underset{R_{3}\rightarrow-\frac{1}{a}R_{3}\ (*)}{\Longrightarrow}}\begin{bmatrix}\begin{array}{ccc|c} 1 & \frac{1}{2}a & \frac{1}{5} & \frac{1}{5}a\\ 0 & 1 & \frac{3}{5a} & \frac{3}{5}\\ 0 & \frac{3}{2} & \frac{2}{5a} & \frac{2}{5} \end{array}\end{bmatrix}\]
\((*)\) this step assumes that \(a\neq0\), since we divide by \(a\). The case \(a=0\) is handled separately below.
\[\overset{R_{1}\rightarrow R_{1}-\frac{1}{2}aR_{2}}{\underset{R_{3}\rightarrow R_{3}-\frac{3}{2}R_{2}}{\Longrightarrow}}\begin{bmatrix}\begin{array}{ccc|c} 1 & 0 & \left(\frac{1}{5}-\frac{a}{2}\cdot\frac{3}{5a}\right) & \left(\frac{1}{5}a-\frac{a}{2}\cdot\frac{3}{5}\right)\\ 0 & 1 & \frac{3}{5a} & \frac{3}{5}\\ 0 & 0 & \left(\frac{2}{5a}-\frac{3}{2}\cdot\frac{3}{5a}\right) & \left(\frac{2}{5}-\frac{3}{2}\cdot\frac{3}{5}\right) \end{array}\end{bmatrix}=\begin{bmatrix}\begin{array}{ccc|c} 1 & 0 & \left(\frac{1}{5}-\frac{3a}{10a}\right) & \left(\frac{1}{5}a-\frac{3a}{10}\right)\\ 0 & 1 & \frac{3}{5a} & \frac{3}{5}\\ 0 & 0 & \left(\frac{2}{5a}-\frac{9}{10a}\right) & \left(\frac{2}{5}-\frac{9}{10}\right) \end{array}\end{bmatrix}=\begin{bmatrix}\begin{array}{ccc|c} 1 & 0 & \left(-\frac{1}{10}\right) & \left(-\frac{1}{10}a\right)\\ 0 & 1 & \frac{3}{5a} & \frac{3}{5}\\ 0 & 0 & \left(-\frac{1}{2a}\right) & \left(-\frac{5}{10}\right) \end{array}\end{bmatrix}\]
\[\overset{R_{3}\rightarrow-2a\cdot R_{3}}{\underset{}{\Longrightarrow}}\begin{bmatrix}\begin{array}{ccc|c} 1 & 0 & \left(-\frac{1}{10}\right) & \left(-\frac{1}{10}a\right)\\ 0 & 1 & \frac{3}{5a} & \frac{3}{5}\\ 0 & 0 & 1 & a \end{array}\end{bmatrix}\overset{R_{1}\rightarrow R_{1}+\frac{1}{10}R_{3}}{\underset{R_{2}\rightarrow R_{2}-\frac{3}{5a}R_{3}}{\Longrightarrow}}\begin{bmatrix}\begin{array}{ccc|c} 1 & 0 & 0 & 0\\ 0 & 1 & 0 & 0\\ 0 & 0 & 1 & a \end{array}\end{bmatrix}\]
Which gives the solution \[\left\{ \begin{pmatrix}0\\ 0\\ a \end{pmatrix}\right\}\]
Therefore, for each \(a\in\mathbb{R}\setminus\{0\}\) , there is a unique solution.
Case \(a=0\). we return to the matrix obtained just before \((*)\) and substitute \(a=0\):
\[\begin{bmatrix}\begin{array}{ccc|c} 1 & \frac{1}{2}a & \frac{1}{5} & \frac{1}{5}a\\ 0 & -a & -\frac{3}{5} & -\frac{3}{5}a\\ 0 & -\frac{3}{2}a & -\frac{2}{5} & -\frac{2}{5}a \end{array}\end{bmatrix}\underset{a=0}{\Longrightarrow}\begin{bmatrix}\begin{array}{ccc|c} 1 & 0 & \frac{1}{5} & 0\\ 0 & 0 & -\frac{3}{5} & 0\\ 0 & 0 & -\frac{2}{5} & 0 \end{array}\end{bmatrix}\overset{R_{2}\rightarrow-5R_{2}}{\underset{R_{3}\rightarrow-5R_{3}}{\Longrightarrow}}\begin{bmatrix}\begin{array}{ccc|c} 1 & 0 & \frac{1}{5} & 0\\ 0 & 0 & 3 & 0\\ 0 & 0 & 2 & 0 \end{array}\end{bmatrix}\overset{R_{1}\rightarrow R_{1}-\frac{1}{15}R_{2}}{\underset{R_{3}\rightarrow R_{3}-\frac{2}{3}R_{2}\text{ then }R_{2}\rightarrow\frac{1}{3}R_{2}}{\Longrightarrow}}\begin{bmatrix}\begin{array}{ccc|c} 1 & 0 & 0 & 0\\ 0 & 0 & 1 & 0\\ 0 & 0 & 0 & 0 \end{array}\end{bmatrix}\]
and the set of solutions is therefore \[\left\{ \begin{pmatrix}0\\ y\\ 0 \end{pmatrix}:y\in\mathbb{R}\right\}\]
so the system has infinitely many solutions.
To conclude, \[\begin{cases} 0\text{ solutions} & \text{No such }a\\ 1\text{\text{ solution}} & a\in\mathbb{R}\setminus\{0\}\\ \infty\text{ solutions} & a\in\{0\} \end{cases}\]
This is an exam problem I translated and solved. The original exam was written by the Linear Algebra 1 course lecturers at HUJI.