Okay, I ended up making sense of his solution. Hell, I like that solution, it makes sense but goddamn, that was a mess of doing elimination. For anyone not in the know this method was referred to Gauss Jordan or Reduced Row Echelon Form (RREF). The echelon bit is French for stairs. This essentially means from the first row, which should have an entry zero, every variable one step right and one step down (diagonal at 45 degrees) should be one. This makes the coefficients one and when written as an eq again just allows the equation of each row to be a variable equals some number, in this case it’s a variable. The variable d is called a parameter. Essentially if the parameter (side measure) takes a step, each variable maintains some relationship to it. This allows for you to model the behavior of the other 3 with only using d. Here you need an integer, so each number being divided by 4 needed a multiple of d that would eliminate the 4, i.e. 4 and so they are all multiplied by the constant accordingly. I’ll make a page on this soon to make it clearer. Back to the problem with elimination, this was how I learned it for the SAT. This is not how I learned it for linear algebra. The reason it looked funky was because the guy took the liberty to start essentially swap the row order which with his notation was more confusing, essentially putting you in a position to lose track. There was also no notation for the progressive execution of row operations (for those not in the know it’s the manipulations of the numbers in the matrix he did). He also kept referring to each row by its old name, adding further confusion. All in all conceptually a good video, but poor execution there. I don’t like being this guy, but I had to say it.
Yes, as the rank of the matrix is 3, however the number of variables is 4, so the solutions would be infinite and defined in terms of 1 parameter. This is useless in balancing chemical equations however; it is analogous with scaling all of the coefficients of a balanced equation - this would be technically correct as the mole ratio would remain constant however, likely invalid in a test as the point is to simplify these coefficients to the smallest set of numbers that represents that ratio. When defining solutions in terms of one parameter in RREF where that parameter is assigned to the variable that falls to the right of all other leading entries (d), all other variables will be defined in proportionate terms of d.
Thanks, this really helped to reintroduce me to a method I have long since forgotten. It is really thoroughly and well explained, just watch for those typos ;P.
Well up untill the writing all the lines into each other it was pretty straight forward. I guess, its someting i will need to learn, while its really hard to get some equations done by trial and error. Thank you! Greetings, Jeff
@3:23, by interchanging row 2 and 3, it makes my goal of converting it into row echelon form occur faster, although it is not necessary. If you could keep it the way it is, and still manage to make it into row-echelon form, than you should be fine as well.
Why is your second equation not 8a-2d=0 ???
@2:25... Yes, you're right. I think it was corrected afterwards, just a typo 🤭
Where did you get 3x8
It's interesting to see it's application in balancing chemical equations. However, it feels excessive for this kind of exercise.
Okay, I ended up making sense of his solution. Hell, I like that solution, it makes sense but goddamn, that was a mess of doing elimination. For anyone not in the know this method was referred to Gauss Jordan or Reduced Row Echelon Form (RREF). The echelon bit is French for stairs. This essentially means from the first row, which should have an entry zero, every variable one step right and one step down (diagonal at 45 degrees) should be one. This makes the coefficients one and when written as an eq again just allows the equation of each row to be a variable equals some number, in this case it’s a variable. The variable d is called a parameter. Essentially if the parameter (side measure) takes a step, each variable maintains some relationship to it. This allows for you to model the behavior of the other 3 with only using d. Here you need an integer, so each number being divided by 4 needed a multiple of d that would eliminate the 4, i.e. 4 and so they are all multiplied by the constant accordingly. I’ll make a page on this soon to make it clearer. Back to the problem with elimination, this was how I learned it for the SAT. This is not how I learned it for linear algebra. The reason it looked funky was because the guy took the liberty to start essentially swap the row order which with his notation was more confusing, essentially putting you in a position to lose track. There was also no notation for the progressive execution of row operations (for those not in the know it’s the manipulations of the numbers in the matrix he did). He also kept referring to each row by its old name, adding further confusion. All in all conceptually a good video, but poor execution there. I don’t like being this guy, but I had to say it.
this was lowkey confusing..
Would this system be considered to have "INFINITELY MANY SOLUTIONS" since the value of "d" can be any number?
Yes, as the rank of the matrix is 3, however the number of variables is 4, so the solutions would be infinite and defined in terms of 1 parameter. This is useless in balancing chemical equations however; it is analogous with scaling all of the coefficients of a balanced equation - this would be technically correct as the mole ratio would remain constant however, likely invalid in a test as the point is to simplify these coefficients to the smallest set of numbers that represents that ratio. When defining solutions in terms of one parameter in RREF where that parameter is assigned to the variable that falls to the right of all other leading entries (d), all other variables will be defined in proportionate terms of d.
Yes, but in chemical reactions you want the smallest positive integers. So choose the free variable so that it satisfies the need.
Couldn’t you just find the common factor of all the numbers, no matter the input of d, assuming choosing mostly simple numbers?
Well explained
Thanks, this really helped to reintroduce me to a method I have long since forgotten. It is really thoroughly and well explained, just watch for those typos ;P.
Glad to help
Well up untill the writing all the lines into each other it was pretty straight forward.
I guess, its someting i will need to learn, while its really hard to get some equations done by trial and error.
Thank you!
Greetings,
Jeff
Great Explaining!!
THANKYOU!
Hi! Why did you interchange the O and H in 3:23..is there a certain reason for it?
@3:23, by interchanging row 2 and 3, it makes my goal of converting it into row echelon form occur faster, although it is not necessary. If you could keep it the way it is, and still manage to make it into row-echelon form, than you should be fine as well.
What if there are more than 3 elements?
The matrix gets more complicated
king
Why (0 x 8) + (-2 x 3) becomes 6?
Hi Jerold, please mention the minute and second to review.
@@StudyForceOnline 4:19 but its my bad i thought its R1 + R2. Sorry!
Idk theres many typos
quick maffs