Let A = [ 1 − 3 − 4 − 3 2 6 5 −...
Show that the equation Ax=b does not have a solution for all possible b, anddescribe the set of all b for which Ax=b does have a solution.
Please explain in steps and especially the finall step towhere you explain for which values b does have a solution and whenthere is none.
Answer & Explanation
First, I think that Linear Algebra should be in the "Advanced Math"part of the Q&A board. Usually "Algebra" means 100-level college courses and sometimes high school algebra. That isjust my opinion (I'm new here), but I wasn't expecting to find thistype of question in
"Algebra to Pre-Calc."
There is a solution when the Augmented Matrix [ A | b ] doesn'thave a contradiction - i.e. doesn't have a row that says somethinglike:
[ 0 0 0 | 1 ] (Saying that , or 0=1 )
So row reduce your matrix : (it is just me, or is it hard to make a3x4 matrix?)
The first two rows are fine, but the third row will only allow asolution to exist when there is no contradiction.
In other words, there are solutions ONLY when:
(from the augmented matrix afterthe mostly zero row)
So if b = [ 2 3 4 ], there would be no solution.
There ARE solutions when
and and are allowed to be anynumbers.