Option A: Lets have a echelon matrix A which 2x3 (n > m)
$$\begin{bmatrix} a & b & c\\ 0 & e & f \end{bmatrix}$$
In this example, c1 and c2 are pivot variables. We can see that c3 is a free variable,
that is, when n>m, not all the columns can become pivot variables.
Therefore, A is False.
Option B: Consistent means that the equation has atleast one solution for all b.
In other words, it means that the solutions exists for all b.
Now, the condition for No solution is that for an Augmented Matrix [ A | b ]
If any row of the matrix looks like this:
$$\begin{bmatrix} 0 & 0 & x \end{bmatrix}$$
where x != 0
But in the options, it has been given that every row has a pivot. In other words, every row has a non-zero variable.
Therefore, the condition for No solutions fails and the equation is consistent.
B is True.
Option C: The matrix it is talking about is A which is 25 x 13, that is m = 25 and n = 13.
Since, it has only 13 variables, that means only 13 pivots are possible.
25-13 = 12 rows have to be 0 in echelon form.
but it's not given that b in (Ax = b), will be zero or not.
So, whether the solution exists or not, will depend on b for that row.
If b != 0, then it will satisfy the No solution condition otherwise it will not.
Option C is False.
Option D: For the same matrix in Option C. Now, it has been given that atleast one solutions exists for all b.
So, b = 0 for the rows which are 0 in echelon form.
For the solution to be infinite, there has to be a free variable in row echelon form, but since the matrix has 25 rows and only 13 columns. Therefore, all the 13 columns (or variables) will become pivot and no free variables will be there.
Therefore, the solution will be unique if it exists.
D is False