00:01
Hi, so for this exercise we have a chemical reaction and we should work with the balancing problem.
00:07
So technically we can apply linear algebra to the balancing problem because we are trying to find the proportions that multiplies to these molecules x2, x3 and x4 in this case four unknowns that corresponds to these integral values that multiplies to the molecules to obtain the same number of atoms on the left and a number of atoms.
00:30
On the right okay so based on these property of the chemical chemical equation here that we should preserve the atoms we should work to construct our equations or set of linear equations so let's focus first on the carbon so what happened here is that the carbon we have only one atom of carbon multiplying x1 so we have x1 we have any other carbon here on the left part so this should be equal to the six atoms that are multiplying to x3.
01:07
That means six times x3.
01:12
Then we have the atom of hydrogen.
01:17
So the hydrogen on the left part, we have two atoms multiplying x2.
01:23
So two times x2.
01:26
And this should be equals to the 12 atoms that are multiplying to x3.
01:35
And finally we have the oxygen.
01:39
So on the left we have two atoms of oxygen that are multiplying x1, that means 2 times x1 plus the oxygen multiply in x2 so we have here x2 and this should be equals to 6 times x3 plus 2 times x4.
02:08
Okay so we can rewrite equations as x1 minus 6x3 equals to 0 to x2 minus 12 x3 equals to 0 and 2x1 was x2 minus 6x3 and minus 2x4 equals to 0 so you can see that we have an homogeneous system of linear equations so to solve that we have also we have four unknowns and three equations that means that this system will have an infinitely many solutions mathematically speaking of course so having this set of linear equations we can consider the extended matrix associated to this system to solve the problem but first you can see that from this equation from this yeah from this equation you can divide by two so we have this equation here okay so let's put this on the matrix one zero minus six zero zero one minus six zero and two one minus six minus two and the extended matrix so we have this matrix here.
03:49
So let's put a zero on this position...