In a row echelon form, different pivot entries are in different rows and different columns. In other words, any row or column can have at most one pivot entry. As a result, we have the following equality (see this exercise).
With the help of the equality, we have
Ax = b has solutions for any right side b
⇔ All rows of A are pivot (by this criterion)
⇔ number of rows = number of pivots
⇔ number of rows = number of pivot columns (by the equality above)
Since number of pivot columns ≤ number of columns, we conclude the numerical consequence of always existence.
For a special case for the statement above, see this exercise. The numerical consequence can also be rephrased.
A system has solutions for any right side ⇒ number of equations ≤ number of variables
Example Without any computation, we know the following system
|3.547x1||+ 5.782x2||+ 0.003x3||=||b1|
|-2.358x1||- 2.683x2||- 3.907x3||=||b2|
|-11.291x1||+ 4.902x2||+ 2.013x3||=||b3|
|7.912x1||+ 3.091x2||+ 7.892x3||=||b4|
does not always have solutions, simply because 4 (number of rows) > 3 (number of columns).
Similarly, in an earlier exercise, we used row operations to conclude that the second and the third systems in the exercise do not always have solutions. In fact, the second system has more equations (number = 4) than variables (number = 3). Therefore without any computation, we may immediately conclude that the second system does not always have solutions.
Note that for the first and the third systems in the earlier exercise, we do have number of equations ≤ number of variables. In this case, we have to do row operations to determine whether solutions always exist.
Now we carry out a similar argument for the uniqueness case.
Ax = b has a unique solution
⇔ All columns of A are pivot (by this criterion)
⇔ number of columns = number of pivots
⇔ number of columns = number of pivot rows (by the equality above)
Since number of pivot rows ≤ number of rows, we conclude the numerical consequence of uniqueness.
A special case can be found in this exercise. The numerical consequence can also be rephrased.
A system has a unique solution ⇒ number of equations ≥ number of variables
Example Because 3 < 4, the following complicated system
|0.75x1||+ 3.55x2||+ 2.73x3||+ 0.39x4||=||1.22|
|- 3.24x1||- 0.01x2||- 6.01x3||+ 0.03x4||=||3.71|
|5.32x1||+ 4.22x2||+ 0.36x3||- 7.56x4||=||-0.19|
cannot have a unique solution. In other words, it either has no solution, or has infinitely many solutions. Of course to find out which case happens, we need to do some complicated row operations.
Without any computation, we know that the solution to the following consistent system (see this example)
|x1||+ 3x2||+ 2x3||+ x5||=||0|
|- x1||- x2||- x3||+ x4||=||1|
|4x2||+ 2x3||+ 4x4||+ 3x5||=||2|
|x1||+ 3x2||+ 2x3||- 2x4||=||0|
is not unique, simply because 4 (number of rows) < 5 (number of columns).