Finding multiple solution of a matrix I have this matrix from a book's exercise.
$$
\left[
\begin{array}{@{}cccc@{}}
a&0&b & 2 \\
a& a& 4 & 4 \\
0&a& 2 & b\\
\end{array}
\right]
$$
be the augmented matrix for linear system. Find for what values of ${a}$ and ${b}$ the system has:


*

*a unique solution

*a one parameter solution

*a two parameter solution

*no solution

My questions:


*

*How do i solve this? what should i use? gauss-jordan elimination? row operation?
crammer's rule?

*Can someone please explain what one parameter means? i have hard time comprehending it


Please help.
Thanks.
 A: I like to think of $a$ and $b$ as being real numbers I don't happen to know.
Your task is to write something like "There is a unique solution if and only if [some conditions on $a$ and $b$]" and so on for the other cases.
You can determine the number of solutions from the row echelon form (found via Gaussian elimination, i.e., row operations).


*

*It might not be consistent (no solutions).  This occurs if and only if there is a row such as $[0\ 0\ 0\ 1]$ in the row echelon form (which is equivalent to the equation $0=1$).

*There might be a unique solution.  In this case, there will be three leading entries (one in each row) in row echelon form.

*There might be a one-parameter solution.  In this case, there will be two leading entries (and thus a row of zeroes) in row echelon form.
And so on.
A "one-parameter solution" is when the solution space is a line.  As a more simple example
$$
\left(
\begin{array}{cc|c}
1 & -1 & 0 \\
2 & -2 & 0 \\
\end{array}
\right)
$$
has infinitely many solutions: $(x,y)=(t,t)$ for all $t \in \mathbb{R}$.  Here we have one parameter: $t$.  The solution space is the line $\{(t,t):t \in \mathbb{R}\}$.
The system of equations
$$
\left(
\begin{array}{ccc|c}
1 & -1 & 0 & 0 \\
2 & -2 & 0 & 0 \\
\end{array}
\right)
$$
would have a two-parameter solution space: $\{(t,t,u):t,u \in \mathbb{R}\}$.
A: By preformin row operations we reduce the matrix to 
$$\begin{pmatrix}
a&0&2&4-b\\
0&a&2&b\\
0&0&b-2&b-2\\
\end{pmatrix}$$
Now the system has a unique solution iff the matrix has rank $3$ and this happens when 
$a\neq 0$ and $b\neq 2$.
For a one parameter solution the matrix must have rank $2$ and its easy to see that this happens only when $b=2$ and $a\neq 0$.
For a two parameter solution we need rank of $1$ and if we set $a=0$ we see that 
$$\begin{pmatrix}
0&0&2&4-b\\
0&0&2&b\\
0&0&b-2&b-2\\
\end{pmatrix}$$
and in order for this to be consistent we need $4-b=b$ so $b=2$
and we have 
$$\begin{pmatrix}
0&0&2&2\\
0&0&2&2\\
0&0&0&0\\
\end{pmatrix}$$ which has rank of $1$. In order for no solution as remarked we have $a=0$ and $b\neq 2$.
