Answer:
a) Let's consider each type of elementary row operation one by one:
1. **Row Replacement Operation**: If B is obtained from A by replacing one row of A with a scalar multiple of itself, say, R_i → kR_i, where k is a nonzero scalar, then it's clear that the new row in B is a linear combination of the rows of A, where all coefficients are zero except for the coefficient of the replaced row, which is k.
2. **Row Interchange Operation**: If B is obtained from A by interchanging two rows of A, say, R_i ↔ R_j, then the rows of B are still linear combinations of the rows of A because interchanging rows does not change the linear dependence relationships between the rows.
3. **Row Scaling Operation**: If B is obtained from A by scaling a row of A, say, R_i → kR_i, where k is a nonzero scalar, then the rows of B are also linear combinations of the rows of A. Each row in B is obtained by multiplying the corresponding row in A by k.
In all these cases, we see that every row of B is indeed a linear combination of rows of A.
b) To show that every row of A is a linear combination of rows of B, we can use the fact that the inverse of an elementary row operation is also an elementary row operation of the same type:
1. **Row Replacement Operation**: If B is obtained from A by a row replacement operation, then performing the inverse operation on B (which is dividing the row by the scalar k) will give us A. Thus, every row of A can be expressed as a linear combination of rows of B.
2. **Row Interchange Operation**: The inverse of a row interchange operation is the same operation itself. So, if B is obtained from A by interchanging rows, then A can be obtained from B by the same operation, and every row of A is a linear combination of rows of B.
3. **Row Scaling Operation**: The inverse of a row scaling operation (multiplying a row by k) is to multiply the row by 1/k. So, if B is obtained from A by scaling rows, then A can be obtained from B by the same type of scaling operation, and every row of A is a linear combination of rows of B.
In all these cases, we've shown that every row of A is indeed a linear combination of rows of B.