Prove that the system { a , b , d , e , f , g , h , i &gt; 0 − a + e − i &gt; 0 − a e + a i + b d − e i + f h &gt; 0 a e i − h f a − b d i − g b f &gt; 0 is inconsistent.I tried using some previously acquired techniques (such as factoring, or multiplying an equality and adding it to another equality) that worked for many families of such systems, but I have only proven it for the case when i ≤ a. Here is what I have so far.From the first inequality, we get e &gt; a + i ⇒ e &gt; a . ( ∗ )Multiplying the second inequality by i and adding it to the third gives a i 2 − e i 2 + f h i − h f a − g b f = i 2 ( a − e ) + f h ( i − a ) − g b f &gt; 0.But a − e is negative by ( ∗ ), and i − a ≤ 0 if i ≤ a. This leads to a contradiction.How do I prove this if a &gt; i?

Question

Prove that the system      {                            a          ,          b          ,          d          ,          e          ,          f          ,          g          ,          h          ,          i          &amp;gt;          0                                      −          a          +          e          −          i          &amp;gt;          0                                      −          a          e          +          a          i          +          b          d          −          e          i          +          f          h          &amp;gt;          0                                      a          e          i          −          h          f          a          −          b          d          i          −          g          b          f          &amp;gt;          0                        is inconsistent.I tried using some previously acquired techniques (such as factoring, or multiplying an equality and adding it to another equality) that worked for many families of such systems, but I have only proven it for the case when   i  ≤  a. Here is what I have so far.From the first inequality, we get  e  &amp;gt;  a  +  i  ⇒  e  &amp;gt;  a  .  (  ∗  )Multiplying the second inequality by   i and adding it to the third gives  a      i    2    −  e      i    2    +  f  h  i  −  h  f  a  −  g  b  f  =      i    2    (  a  −  e  )  +  f  h  (  i  −  a  )  −  g  b  f  &amp;gt;  0.But   a  −  e is negative by   (  ∗  ), and   i  −  a  ≤  0 if   i  ≤  a. This leads to a contradiction.How do I prove this if   a  &amp;gt;  i?

britspears523jp · Accepted Answer

You&#039;ve already proven it when   i  ≤  a and the   i  ≥  a situation is symmetric and also gives a contradiction:  0  &amp;lt;  a  (  −  a  e  +  a  i  +  b  d  −  e  i  +  f  h  )  +  a  e  i  −  h  f  a  −  b  d  i  −  g  b  f  =      a    2    (  i  −  e  )  +  b  d  (  a  −  i  )  −  g  b  f

Prove that the system { <mtable columnalign="left left" rowspacing=".2em" columnspacing

Answered question

Answer & Explanation

New Questions in Algebra I