Are the following two systems of linear equations equivalent? − x 1 + x 2 + 4 x 3 = 0 x 1 + 3 x 2 + 8 x 3 = 0 1 2 x 1 + x 2 + 5 2 x 3 = 0and x 1 − x 3 = 0 x 2 + 3 x 3 = 0For the linear combination of the first equation of the second system, I got: x 1 − x 3 = x 1 ( − a + b + 1 2 c ) + x 2 ( a + 3 b + c ) + x 3 ( 4 a + 8 b + 5 2 c )which results in this system of equations in order to solve for a , b , c: ( 1 ) − a + b + 1 2 c = 1 ( 2 ) a + 3 b + c = 0 ( 3 ) 4 a + 8 b + 5 2 c = − 1Simplifying (1) and (2) gives you 4 b + 3 2 c = 1, and simplifying (2) and (3) gives you the same equation, which is why I figured there were infinite solutions. I used a different technique and different combination of equations, with the same result. (And the same thing happened with the second equation of the second system.)Where did I go wrong?

Question

Are the following two systems of linear equations equivalent?  −      x    1    +      x    2    +  4      x    3    =  0      x    1    +  3      x    2    +  8      x    3    =  0      1    2        x    1    +      x    2    +      5    2        x    3    =  0and      x    1    −      x    3    =  0      x    2    +  3      x    3    =  0For the linear combination of the first equation of the second system, I got:      x    1    −      x    3    =      x    1    (  −  a  +  b  +      1    2    c  )  +      x    2    (  a  +  3  b  +  c  )  +      x    3    (  4  a  +  8  b  +      5    2    c  )which results in this system of equations in order to solve for   a  ,  b  ,  c:  (  1  )    −  a  +  b  +      1    2    c  =  1  (  2  )    a  +  3  b  +  c  =  0  (  3  )    4  a  +  8  b  +      5    2    c  =  −  1Simplifying (1) and (2) gives you   4  b  +      3    2    c  =  1, and simplifying (2) and (3) gives you the same equation, which is why I figured there were infinite solutions. I used a different technique and different combination of equations, with the same result. (And the same thing happened with the second equation of the second system.)Where did I go wrong?

Jerome Page · Accepted Answer

You went wrong in that you solved the system and considered that the answer to the question. Both of these systems have infinite solutions but that is not what the question is asking. It is asking are the systems equivalent. One way to do that is solve them both and see that they give the same solution set. Another way to to transform one set of equations into the other using algebra.                    −                  x          1                +                  x          2                +        4                  x          3                                    =        0                            +                  x          1                +        3                  x          2                +        8                  x          3                                    =        0                            4                  x          2                +        12                  x          3                                    =        0            Divide the last equation by   4 and you get       x    2    +  3      x    3    =  0, which can also be viewed as       x    3    =  −  3      x    3  . Substituting this into one of the equations (say the second), we get       x    1    +  3      x    2    +  8      x    3    =  0    ⟺        x    1    +  3  (  −  3      x    3    )  +  8      x    3    =  0    ⟺        x    1    −      x    3    =  0Sine we can transform the one system into the other, they are equivalent. This is often done using the augmented matrix version of an equation and Gaussian elimination (also called Gauss-Jordan elimination).In response to the comment, label the first system&#039;s equations   a  ,  b  ,  c, in order, and the second system&#039;s equations   d  ,  e, in order. The work above shows that   e  =      1    4    a  +      1    4    b. Since we used   a and   b to get   e, we have to pick a different pair to try and get   d. It turns out that   d  =      2    3    c  −      2    3    a (though this is not the only possible answer).The other direction is much simpler. Since   d has       x    1   but not       x    2  , any       x    1   terms in   a  ,  b  ,  c have to have come from   d. The same is true for e and       x    2  . Since a has   −      x    1  , we need a   −  d. Since   a has       x    2  , we need   e. It turns out that this is enough to get the answer as the       x    3   takes care of itself. So   a  =  e  −  d. Similarly,   b  =  d  +  3  e and   c  =      1    2    d  +  e.

Are the following two systems of linear equations equivalent? − x 1

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