Made an error solving a linear system












1












$begingroup$


Can anybody explain to me what I did wrong here?
I am working on practice problems for my linear algebra course. The problem in question is as follows:



Suppose the system below is consistent for all possible values of f and g. What can you say about the coefficients c and d? Justify your answer.



$x_1+3x_2=f$



$cx_1+dx_2=g$



My first idea was to simply solve for d and c in terms of f and g:



$x_1+3x_2=f implies frac{x_1}{f}+frac{3x_2}{f}=1$



$cx_1+dx_2=g implies frac{cx_1}{g}+frac{dx_2}{g}=1$



Then these can be set equal to each other:



$frac{x_1}{f}+frac{3x_2}{f}=frac{cx_1}{g}+frac{dx_2}{g}$



$implies frac{1}{f}=frac{c}{g}, frac{3}{f}=frac{d}{g}$



$implies frac{g}{f}=c, frac{3g}{f}=d$



$therefore d=3c$



I believed this answer would hold true, at least for nonzero values of f and g. However, row reduction on the system shows the exact opposite:



$ begin{array}{lcr}
mbox{1} & 3 & f \
mbox{c} & d & g \
end{array} $



$implies$



$ begin{array}{lcr}
mbox{1} & 3 & f \
mbox{0} & d-3c & g-fc \
end{array} $



This matrix suggests that $gneq3c$ except when f and g are zero. Can anyone tell me what is wrong with my first approach? Looking at it, I can tell that my answer is wrong (unless $g=cf$), but I don't see what is wrong with the algebra.






linear-algebra matrices systems-of-equations







share|cite|improve this question









$endgroup$












  • $begingroup$
    You’ve reached the opposite of the correct conclusion with your first method: when $d=3c$, the system cannot be consistent for all $f$ and $g$. You’ve also got some potential divisions by zero that you haven’t dealt with.
    $endgroup$
    – amd
    Jan 10 at 9:11
















1












$begingroup$


Can anybody explain to me what I did wrong here?
I am working on practice problems for my linear algebra course. The problem in question is as follows:



Suppose the system below is consistent for all possible values of f and g. What can you say about the coefficients c and d? Justify your answer.



$x_1+3x_2=f$



$cx_1+dx_2=g$



My first idea was to simply solve for d and c in terms of f and g:



$x_1+3x_2=f implies frac{x_1}{f}+frac{3x_2}{f}=1$



$cx_1+dx_2=g implies frac{cx_1}{g}+frac{dx_2}{g}=1$



Then these can be set equal to each other:



$frac{x_1}{f}+frac{3x_2}{f}=frac{cx_1}{g}+frac{dx_2}{g}$



$implies frac{1}{f}=frac{c}{g}, frac{3}{f}=frac{d}{g}$



$implies frac{g}{f}=c, frac{3g}{f}=d$



$therefore d=3c$



I believed this answer would hold true, at least for nonzero values of f and g. However, row reduction on the system shows the exact opposite:



$ begin{array}{lcr}
mbox{1} & 3 & f \
mbox{c} & d & g \
end{array} $



$implies$



$ begin{array}{lcr}
mbox{1} & 3 & f \
mbox{0} & d-3c & g-fc \
end{array} $



This matrix suggests that $gneq3c$ except when f and g are zero. Can anyone tell me what is wrong with my first approach? Looking at it, I can tell that my answer is wrong (unless $g=cf$), but I don't see what is wrong with the algebra.






linear-algebra matrices systems-of-equations







share|cite|improve this question









$endgroup$












  • $begingroup$
    You’ve reached the opposite of the correct conclusion with your first method: when $d=3c$, the system cannot be consistent for all $f$ and $g$. You’ve also got some potential divisions by zero that you haven’t dealt with.
    $endgroup$
    – amd
    Jan 10 at 9:11














1












1








1


1



$begingroup$


Can anybody explain to me what I did wrong here?
I am working on practice problems for my linear algebra course. The problem in question is as follows:



Suppose the system below is consistent for all possible values of f and g. What can you say about the coefficients c and d? Justify your answer.



$x_1+3x_2=f$



$cx_1+dx_2=g$



My first idea was to simply solve for d and c in terms of f and g:



$x_1+3x_2=f implies frac{x_1}{f}+frac{3x_2}{f}=1$



$cx_1+dx_2=g implies frac{cx_1}{g}+frac{dx_2}{g}=1$



Then these can be set equal to each other:



$frac{x_1}{f}+frac{3x_2}{f}=frac{cx_1}{g}+frac{dx_2}{g}$



$implies frac{1}{f}=frac{c}{g}, frac{3}{f}=frac{d}{g}$



$implies frac{g}{f}=c, frac{3g}{f}=d$



$therefore d=3c$



I believed this answer would hold true, at least for nonzero values of f and g. However, row reduction on the system shows the exact opposite:



$ begin{array}{lcr}
mbox{1} & 3 & f \
mbox{c} & d & g \
end{array} $



$implies$



$ begin{array}{lcr}
mbox{1} & 3 & f \
mbox{0} & d-3c & g-fc \
end{array} $



This matrix suggests that $gneq3c$ except when f and g are zero. Can anyone tell me what is wrong with my first approach? Looking at it, I can tell that my answer is wrong (unless $g=cf$), but I don't see what is wrong with the algebra.






linear-algebra matrices systems-of-equations







share|cite|improve this question









$endgroup$




Can anybody explain to me what I did wrong here?
I am working on practice problems for my linear algebra course. The problem in question is as follows:



Suppose the system below is consistent for all possible values of f and g. What can you say about the coefficients c and d? Justify your answer.



$x_1+3x_2=f$



$cx_1+dx_2=g$



My first idea was to simply solve for d and c in terms of f and g:



$x_1+3x_2=f implies frac{x_1}{f}+frac{3x_2}{f}=1$



$cx_1+dx_2=g implies frac{cx_1}{g}+frac{dx_2}{g}=1$



Then these can be set equal to each other:



$frac{x_1}{f}+frac{3x_2}{f}=frac{cx_1}{g}+frac{dx_2}{g}$



$implies frac{1}{f}=frac{c}{g}, frac{3}{f}=frac{d}{g}$



$implies frac{g}{f}=c, frac{3g}{f}=d$



$therefore d=3c$



I believed this answer would hold true, at least for nonzero values of f and g. However, row reduction on the system shows the exact opposite:



$ begin{array}{lcr}
mbox{1} & 3 & f \
mbox{c} & d & g \
end{array} $



$implies$



$ begin{array}{lcr}
mbox{1} & 3 & f \
mbox{0} & d-3c & g-fc \
end{array} $



This matrix suggests that $gneq3c$ except when f and g are zero. Can anyone tell me what is wrong with my first approach? Looking at it, I can tell that my answer is wrong (unless $g=cf$), but I don't see what is wrong with the algebra.






linear-algebra matrices systems-of-equations




linear-algebra matrices systems-of-equations






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asked Jan 10 at 5:58









Jamman00Jamman00

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  • $begingroup$
    You’ve reached the opposite of the correct conclusion with your first method: when $d=3c$, the system cannot be consistent for all $f$ and $g$. You’ve also got some potential divisions by zero that you haven’t dealt with.
    $endgroup$
    – amd
    Jan 10 at 9:11


















  • $begingroup$
    You’ve reached the opposite of the correct conclusion with your first method: when $d=3c$, the system cannot be consistent for all $f$ and $g$. You’ve also got some potential divisions by zero that you haven’t dealt with.
    $endgroup$
    – amd
    Jan 10 at 9:11
















$begingroup$
You’ve reached the opposite of the correct conclusion with your first method: when $d=3c$, the system cannot be consistent for all $f$ and $g$. You’ve also got some potential divisions by zero that you haven’t dealt with.
$endgroup$
– amd
Jan 10 at 9:11




$begingroup$
You’ve reached the opposite of the correct conclusion with your first method: when $d=3c$, the system cannot be consistent for all $f$ and $g$. You’ve also got some potential divisions by zero that you haven’t dealt with.
$endgroup$
– amd
Jan 10 at 9:11










1 Answer
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1












$begingroup$

You can't jump from$$frac{x_1}f+frac{3x_2}f=frac{cx_1}g+frac{dx_2}gtag1$$to$$frac1f=frac cgtext{ and }frac3f=frac dg.$$You could if the equality $(1)$ was true for every $x_1$ and every $x_2$, but you are assuming it only for some $x_1$ and some $x_2$.






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    $begingroup$

    You can't jump from$$frac{x_1}f+frac{3x_2}f=frac{cx_1}g+frac{dx_2}gtag1$$to$$frac1f=frac cgtext{ and }frac3f=frac dg.$$You could if the equality $(1)$ was true for every $x_1$ and every $x_2$, but you are assuming it only for some $x_1$ and some $x_2$.






    share|cite|improve this answer









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      1












      $begingroup$

      You can't jump from$$frac{x_1}f+frac{3x_2}f=frac{cx_1}g+frac{dx_2}gtag1$$to$$frac1f=frac cgtext{ and }frac3f=frac dg.$$You could if the equality $(1)$ was true for every $x_1$ and every $x_2$, but you are assuming it only for some $x_1$ and some $x_2$.






      share|cite|improve this answer









      $endgroup$
















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        1





        $begingroup$

        You can't jump from$$frac{x_1}f+frac{3x_2}f=frac{cx_1}g+frac{dx_2}gtag1$$to$$frac1f=frac cgtext{ and }frac3f=frac dg.$$You could if the equality $(1)$ was true for every $x_1$ and every $x_2$, but you are assuming it only for some $x_1$ and some $x_2$.






        share|cite|improve this answer









        $endgroup$



        You can't jump from$$frac{x_1}f+frac{3x_2}f=frac{cx_1}g+frac{dx_2}gtag1$$to$$frac1f=frac cgtext{ and }frac3f=frac dg.$$You could if the equality $(1)$ was true for every $x_1$ and every $x_2$, but you are assuming it only for some $x_1$ and some $x_2$.







        share|cite|improve this answer












        share|cite|improve this answer



        share|cite|improve this answer










        answered Jan 10 at 6:20









        José Carlos SantosJosé Carlos Santos

        155k22124227




        155k22124227






























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