Elsa Brewer

2022-07-26

Let
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.

Danica Ray

Expert

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?)
$\left[\begin{array}{cccc}1& -3& -4& {b}_{1}\\ -3& 2& 6& {b}_{2}\\ 5& 1& -8& {b}_{3}\end{array}\right]\to \left[\begin{array}{cccc}1& -3& -4& {b}_{1}\\ 0& -7& -6{b}_{2}& +3{b}_{1}\\ 0& 14& 12{b}_{3}& -5{b}_{1}\end{array}\right]\to \left[\begin{array}{cccc}1& -3& -4& {b}_{1}\\ 0& -7& -6{b}_{2}& +3{b}_{1}\\ 0& 0& \left({b}_{3}-5{b}_{1}\right)& +2\left({b}_{2}+3{b}_{1}\right)\end{array}\right]$
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
$b1=-2{b}_{2}-{b}_{3}$
and ${b}_{2}$ and ${b}_{3}$ are allowed to be anynumbers.

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