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?

My answer
(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?

[corrected again]

(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.

  • $\begingroup$ If the system is consistent there are infinitely many solutions. Having a unique solution is equivalent to having the identity matrix as your reduced row echelon form matrix, which is not possible with a 4x7 matrix. $\endgroup$ – ÍgjøgnumMeg Mar 29 '16 at 12:05
  • $\begingroup$ Free variables correspond to non-pivot columns. In your question you have pointed out that there are 4 pivots. What does that make the other 3 columns? $\endgroup$ – ÍgjøgnumMeg Mar 29 '16 at 12:29
  • 2
    $\begingroup$ @Ed_4434 Oh I confused the colums. The other 3 colums in the coefficient matrix are non-pivot colums, so three variables are free variables. So the solution sets are infinitely many. $\endgroup$ – buzzee Mar 29 '16 at 12:50
  • $\begingroup$ Assuming the linear system is consistent (has at least one solution), then yes, the presence of at least one "free variable" produces infinitely many solutions. Perhaps you could edit the title of your Question as well? $\endgroup$ – hardmath Mar 29 '16 at 13:42
  • $\begingroup$ @hardmath I edited it. $\endgroup$ – buzzee Mar 29 '16 at 14:02

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