Unique, Many or no solutions Determine the following system of equations has 'a unique solution', 'many solutions' or 'no solution':
$$\begin{cases}
& x   +  2y +   z &=  1\\    
&2x  +  2y  - 2z   &= 4\\    
&-x  +  2y  - 3z   &=  5
\end{cases}
$$
Answer = A unique solution
How is it a unique solution? Could anyone explain clearly? 
Thanks
 A: The easiest way is to show that the corresponding coefficient matrix is non-singular. 
Define A = $\begin{bmatrix}1&2&1\\2&2&-2\\-1&2&-3\end{bmatrix}$
It is a property that if this matrix is non-singular, that the system of linear equations corresponding to this matrix has exactly one solution for any combination of outcomes.
The logic behind that it would have one solution is that you have three variables, but also three different ways these variables are described. When they are linear you can combine them to eventually find one solution for every variable.
But if you were too add some fourth equation that is different from the other three, for example 2x + y - 2z = 1, then you couldn't find any solutions since you cannot find a combination of three variables too satisfy these four lines simultaneously.
On the other hand, if you ignored one of the equations, lets say the third, then you could find infinity many solutions. Since for every x you choose, you can always find some y and z such that only the top two equations hold.
Bottem line: If you have equal variables and equations you only have one unique solution. If you have fewer equations than variables you have infinite solutions. If you have more equation than variables you have no solutions.
Note: It is important that we are talking about different lines, for example:
$x+y+z = 3$
$2x+2y+2z=6 $
are the same constraint, and do not yield additional information. Therefore they are counted as one line.
A: 
$$\begin{cases}
& x   +  2y +   z &=  1\\    
&2x  +  2y  - 2z   &= 4\\    
&-x  +  2y  - 3z   &=  5
\end{cases}$$

If you let:
$$A=\begin{bmatrix}1&2&1\\2&2&-2\\-1&2&-3\end{bmatrix}\;,\quad X=\begin{bmatrix}x\\y\\z\end{bmatrix}\;,\quad B=\begin{bmatrix}1\\4\\5\end{bmatrix}$$
then the system of equations can be written in matrix form as $AX=B$.
This system has no solutions if the rank of $A$ is less than the rank of the augmented matrix $(A\vert B)$. If these ranks are equal, the system has at least one solution.
In your case, with the same number of equations as variables and thus a square coefficient matrix $A$, this solution is unique if the rank of $A$ is $3$. A matrix of full rank has a non-zero determinant, so that would be one quick way to verify that this system indeed has a unique solution:
$$\det A=\begin{vmatrix}1&2&1\\2&2&-2\\-1&2&-3\end{vmatrix} 
= \begin{vmatrix}2&3&1\\0&0&-2\\-4&-1&-3\end{vmatrix}
=-(-2)\begin{vmatrix}2&3\\-4&-1\end{vmatrix}= 2\left( -2+12 \right)=20$$
There are other ways but perhaps you should clarify what you have already learned (relevant theory, properties) about this topic.
