math111_logo Composition and Matrix Multiplication


Exercise Let

A = [ 1 -1 1 0 2 ],     B = [ 1 3 1 ]
2 -2 0 2 2 0 2 0
-1 1 2 -3 1 0 3 2
-2 2 1 -3 -1 0 0 1

Can you find a matrix X such that AX = B? In general, what is the criterion for the existence of X such that AX = B? What about the uniqueness of X?

Answer As we have seen in the previous exercise, the equation AX = B is equivalent to solving three separate systems Ax = b1, Ax = b2, Ax = b3, where b1, b2, b3 are the three columns of b. In an earlier exercise, we have argued that the system Ax = b1 has no solution. Therefore the equation AX = B has no solution.

In general, AX = B has solutions if and only if no column of B is a pivot column of the "augmented matrix" [A B]. The solution is unique if and only if all columns of A are pivot.