# Linear algebra, unique solutions

1. Mar 3, 2013

### Nikitin

1. The problem statement, all variables and given/known data
"Suppose A is an n x n matrix with the property that the equation Ax = b has at least one solution for each b in |Rn. Explain why each equation Ax = b has in fact exactly one solution."

3. Attempt at solution

A*x=b

=> [Ax1 Ax2 Ax3 ... Axm] = [b1 b2 b3 ...bm], where the x signs represent the column vectors making up x.

Why should this imply A has a pivot in each row? Let's say b1= [3,4,0] and A = [1,0,0:0,1,0:0,0,0]. Then x1 = [3,4,k], where k is a free variable. This means you can find a solution for b without it being unique..

Edit: I thought the problem text said "at least one solution for any b", while it was saying "at least one solution for each b". So nevermind the last part. But still, can you guys assist me? I think I understand it now, but I would still appreciate an intuitive explanation from one of you math-experts.

Last edited: Mar 3, 2013
2. Mar 3, 2013

### Dick

I would think of it abstractly instead of in terms of the matrix. If {e1,...,en} is the usual basis for R^n then the range of A is the span of {Ae1,...,Aen}. Then if Ax=b has at least one solution for any b, that means that span must be R^n. Which in turn means {Ae1,...,Aen} is also a basis for R^n. Now any vector can be expressed UNIQUELY as a linear combination of basis vectors. See how that implies that the expression Ax=b has exactly one solution?