Recall the definition of linear independence. The columns of a matrix [M] are said to be linearly dependent if there exists a vector [vecv!=0] with [Mvecv = 0] . We will say that the columns of [M] are linearly independent if [Mvecv = 0] implies [vecv = 0] . Let [A] be a square matrix. Show that if the columns of [A] are linearly dependent, [A^(-1)] cannot exist. Hint: [vecv] cannot be both zero and not zero at the same time. Solution If the columns of [A] are dependent, then there is some [vecv!=0] such that [Avecv=0.] If [A^(- 1)] existed, then by definition of [A^(-1)] we would have [A^(-1)Avecv=Ivecv=vecv,] where [I] is the appropriately sized identity matrix. But [Avecv=0,] so [A^(-1)Avecv=A^(-1)(0)=0,] because [A^(-1)] is also a linear transformation, so it takes the zero vector to the zero vector. So we would have [vecv=A^(-1)Avecv=0,] but this contradicts what we said above. Therefore, [A^(-1)] can\'t exist..