<s>[INST] Matrix with unknown coefficient $a$ - determine number of solutions for each $a$ I have a matrix that looks like this:
\begin{array}{cccc|c}
a & 1 & 0 & 0 & 1 \\
1 & a & 0 & 0 & 2 \\
0 & 0 & a & 2 & 1 \\
0 & 0 & 2 & a & 1
\end{array}
I then proceeded to row reduce the matrix so it looks like this:
\begin{array}{cccc|c}
1 & 0 & 0 & 0 & a-2/a^2-1 \\
0 & 1 & 0 & 0 & 2a-1/a^2-1 \\
0 & 0 & 1 & 0 & 1/a+2 \\
0 & 0 & 0 & 1 & 1/a+2
\end{array}
I need to determine the number of solutions for each $a$ (one, infinite or none).
I can clearly see that there are no solutions when a is either $1$, $-1$ or $2$ because than you would have a division by zero in atleast one of the expressions on the right hand side. But how do I know which $a$ will have an infinite number of solutions. I have never worked with matrices that have unknown coefficients so I have no idea if I am doing anything correctly here.
The original equation system looked like this:
$$ax+y+0+0=1$$
$$x+ay+0+0=2$$
$$0+0+az+2w=1$$
$$0+0+2z+aw=1$$
Maybe I am missing something really obvious. This looks like a very simple task...
 [/INST] Swap rows 1 and 2, and rows 2 and 4. You obtain an augmented matrix for which it's easier to apply pivot's method:
\begin{align}
\begin{bmatrix}
1&a&0&0&2\\a&1&0&0&1\\0&0&2&a&1\\0&0&a&2&1
\end{bmatrix}\rightsquigarrow
\begin{bmatrix}
1&a&0&0&2\\0&1-a^2&0&0&1-2a\\0&0&2&a&1\\0&0&0&4-a^2&2-a
\end{bmatrix}
\end{align}
Thus, if $a\ne \pm1,\pm2$, the matrix of the system and the augmented matrix have the same, maximal rank ($4$). This means there is exactly one solution.
If $a=\pm1$, the augmented matrix is
$$\begin{bmatrix}
1&\pm1&0&0&2\\0& 0&0&0& \scriptstyle\begin{cases}-1\\3\end{cases}\\0&0&2&\pm1&1\\0&0&0&3&  \scriptstyle\begin{cases}1\\3\end{cases}
\end{bmatrix}$$
The second row shows there's no solution.
If $a=\pm2$, the augmented matrix is
$$\begin{bmatrix}
1&\pm2&0&0&2\\0&-3&0&0& \scriptstyle\begin{cases}-3\\5\end{cases}\\0&0&2&\pm2&1\\
0&0&0&0& \scriptstyle\begin{cases}0\\4\end{cases}
\end{bmatrix}$$
The last row shows there is no solution if $a=2$ and is an affine subspace of dimension $1$ if $a=-2$, because both matrices have the same rank ($3$).
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