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.

Question

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.

Danica Ray · Accepted Answer

First, I think that Linear Algebra should be in the &quot;Advanced Math&quot;part of the Q&amp;amp;A board. Usually &quot;Algebra&quot; means 100-level college courses and sometimes high school algebra. That isjust my opinion (I&#039;m new here), but I wasn&#039;t expecting to find thistype of question in &quot;Algebra to Pre-Calc.&quot;There is a solution when the Augmented Matrix [ A | b ] doesn&#039;thave a contradiction - i.e. doesn&#039;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    =  0So 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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