Suppose a $4×7$ coefficient matrix for a system of equations has $4$ pivots. (1) Is the system consistent? (2) If the system is consistent, how many solutions are there?
(1) Yes. The matrix is the coefficient matrix and each 4 row of the coefficient matrix has a pivot, it means for the echelon form of the augmented matrix, no row is of the form $[0 0 0 0 0 0 b]$
(2) I have no idea. How should I solve this?
[EDIt Oh I checked my book again]
"If a linear system is consistent, then the solution set contains either (i) a unique solution, when there are no free variables, or (ii) infinitely many solutions, when there is at least one free variable."
Variables corresponding to pivot colums in a matrix are called leading or basic variables, the other variables are called free variables.
So it would be one solution for (2) since there are no free variables because free variables correspond to non-pivot colums. Right?
(2) There are 4 pivot columns in each 4 row of the $4×7$ coefficient matrix, so the other 3 pivot colums of the coefficient matrix are non-pivot colums. This means three variables are free variables. So the solution set is infinitely many.