Let A = [ 1 − 3 − 4 − 3 2 6 5 −...

Elsa Brewer

Elsa Brewer

Answered

2022-07-26

Let A = [ 1 3 4 3 2 6 5 1 8 ] and  b = [ b 1 b 2 b 3 ]
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

Danica Ray

Danica Ray

Expert

2022-07-27Added 15 answers

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 0 x 1 + 0 x 2 + 0 x 3 = 1, or 0=1 )
So row reduce your matrix : (it is just me, or is it hard to make a3x4 matrix?)
[ 1 3 4 b 1 3 2 6 b 2 5 1 8 b 3 ] [ 1 3 4 b 1 0 7 6 b 2 + 3 b 1 0 14 12 b 3 5 b 1 ] [ 1 3 4 b 1 0 7 6 b 2 + 3 b 1 0 0 ( b 3 5 b 1 ) + 2 ( b 2 + 3 b 1 ) ]
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:
b 3 5 b 1 + 2 b 2 + 6 b 1 = 0
(from the augmented matrix afterthe mostly zero row)
b 1 + 2 b 2 + b 3 = 0
So if b = [ 2 3 4 ], there would be no solution.
There ARE solutions when
b 1 = 2 b 2 b 3
and b 2 and b 3 are allowed to be anynumbers.

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