Question

Suppose A is a 3 times 33×3 matrix and y is a vector in set of real numbers R cubedℝ3 such that the equation Upper A Bold x equals Bold yAx=y does not have a solution. Does there exist a vector z in set of real numbers R cubedℝ3 such that the equation Upper A Bold x equals Bold zAx=z has a unique​ solution?

Answer 1

No

Explanation:

If A were reversible, then such y coudnt not exist, because A⁻¹y would be a solution of the equation Ax=y. This means that A in not reversible, thus, there exist a non zero vector w in R³ such that A*w = 0. If x' were a solution of Ax =z, then x'' = x'+w is a different solution of the equation: Ax'' = A(x'+w) = Ax'+Aw = z+0 = z. This means that a vector z such that there exist a unique solution for the equation Ax=z cant exist.