Question

I know the answer is True, but I need an explanation.

Suppose that A is an m x n matrix such that the solution AX=b, when it exists, is unique. Then A^tX = B has a solution for all B belongs R^n.

Answer 1

Solution: If AX = b has a unique solution then it implies that all the columns of matrix A are linearly independent. So, the all rows of the matrix A^t are linearly independent. Hence for any vector B in R^n, the echleon form of the augmented matrix [ A^t | B ] , coresponds to the system A^t x = B , has a pivot entry in each row and last column is not pivot column.Hence, this system is consistent an has solution for any vector B in R^n.

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