Given Information
We are given with some statement that we have to prove if they are True or False.

Step-1: (a)
Every matrix is row equivalent to a unique matrix in only in reduced echelon form but not echelon form. Therefore,

The statement is False



Step-2: (b)
In case there are free variables, then the system can also have infinite solutions.

The statement is False



Step-3: (c)
If the system has more than one solution, then it has at least one free variable. Hence the system will have infinite solutions.

The statement is True



Step-4: (d)
Even if the system has no free variable, there can be cases when system has no solution (inconsistent)

The statement is False



Step-5: (e)
The row operations used for reducing the augmented matrix does not change the solution set. Hence

The statement is True



Step-6: (f)
The solution sets of the Non-homogeneous system ($Ax=b$) can be obtained by translating the solution set of corresponding homogeneous system ($Ax=0$). The number of solutions remain same. Therefore,

The statement is True



Step-7: (g)
For the columns of $A$ to span $R^m$ the equation $A x=b$ must be consistent for all $b$ not for just one $b$. Therefore,

The statement is False



Step-8: (h)
Even if a matrix can be transformed to reduced echelon, it may be possible that the system may have no solutions.

The statement is False



Step-9: (i)
We know that reduced echelon form of a matrix is unique. Hence if two matrices are row equivalent, then they have same reduced echelon form. Therefore,

The statement is True



Step-10: (j)
If the system has free variables, then it can not have trivial solution.

The statement is True



Step-11: (k)
By Theorem-4, if the equation $Ax=b$ is consistent for every b in $R^m$, then A must have a pivot position in each row.

The statement is True



Step-12: (l)
If n is greater than m, and there are pivot elements in each row, then not all the columns have pivot elements. Hence, there are free variables and system has infinite solutions

The statement is False



Step-13: (m)
If there are n pivot elements in each row, then not the columns have pivot elements. Hence, there are no free variables and system is identity matrix in reduced echelon form.

The statement is True



Step-14: (n)
Both matrices A and B can be converted to identity matrix $I_3$ in reduced echelon form, as both have 3 pivot elements. Therefore

The statement is True



Step-15: (o)
If a system of linear equations (Ax=b) has two different solutions, it must have infinite solutions. So, the equation ($Ax=c$) should also have infinite solutions. Therefore

The statement is True



Step-16: (p)
If columns of A span $R^m$, then reduced echelon form of matrix A has pivot element in each row. Also since B is row equivalent to A, then B can be converted to A by row operations. Hence columns of B also span $R^m$

The statement is True



Step-17: (q)
Even if individual vector is not a multiple of other vector, it is possible that some vector can be written as a linear combination of remaining vectors. Hence

The statement is False



Step-18: (r)
If u, v and w are linearly independent, then they must be in M Hence,

The statement is True



Step-19: (s)
Four vectors can not span $R^5$. Even if the vectors have five rows but they will have only four columns, which restricts them to span $R^4$ Hence,

The statement is False