00:01
We are going to give the parametric vector form of the general solution of the following system of equations.
00:07
We have 3 equations, 3 linear equations and we have 6 variables.
00:14
So for sure we have 3 variables in this problem and we are suggested to use x2, x5 and x6 as free variables.
00:27
That is, in the general form we have been given to fill the entries of the 4 vectors used in that general form.
00:41
The free variables are x2, x5 and x6 and we are going to do so.
00:48
The first thing we see is that from the second equation, right here, from the second equation we can already put x4 in terms of x5 and x6.
01:10
Indeed, if we are going to use x2, x5 and x6 as free variables, then we have to put the other variables in terms of these 3 variables, that is x1, x3 and x4.
01:23
So x4 can be put in terms of x5 and x6 using the second equation.
01:28
There is no x2 in the expression of x4, that is not a problem.
01:34
So the second equation we get x4 is negative 4x5 plus 4x6 minus 5.
01:53
That is the expression we get when we solve this second equation of the linear system 4x4.
02:03
So we have this.
02:07
Now you see that we can put x1 in terms of x2, x5 and x6 using the third equation directly.
02:18
Now from the third equation of the linear system, x1 is negative 3x2 minus 5x5 plus 7x6 minus 4.
02:40
That is the expression for x1 in terms of x2, x5 and x6.
02:50
So we have these 2 variables, we need x3 only to be put in terms of x2, x5 and x6.
02:59
And that we are going to do with the first equation, but you see we have x3 right here, but we have x1 also.
03:09
That is we have x2, x5 and x6 which are the 3 variables, but we have x1 which is not a 3 variable.
03:17
But x1 has been put in terms of x2, x5 and x6, so we are going to use it in the first equation.
03:27
So plug in the expression of x1 in terms of x2, x5 and x6 into the first equation of the linear system, we get, ok so we get x1 directly, the first term will be this expression right here.
04:13
So negative 3x2 minus 5x5 plus 7x6 minus 4, all that is x1.
04:27
Now we have plus 3x2, then plus 2x3, then the other terms are plus 3x5 plus 5x6 and that is equal to 0.
04:58
So now we simplify similar terms and we see that negative 3x2 plus 3x2 cancel out, then we have 2x3 first, then x5 for example, we have it here and here only.
05:36
So 3x5 minus 5x5 is negative 2x5, and then we have x6 here and x6 here, 7x6 plus 5x6 is 12x6 and we are only left with negative 4 here equals 0.
06:05
And so divided by 2 both sides we get x3 minus x5 plus 6x6 minus 2 equals 0, and so x3 which is the value we want to solve is x5 minus 6x6 plus 2.
06:30
So we have written x3 in terms of x2, x5 and x6.
06:35
In fact there is no x2 in this case as well as in the expression for x4 right here.
06:43
So we are done.
06:46
Let me put them as a summary here.
06:48
So x1 is negative 3x2 minus 5x5 plus 7x6 minus 4.
07:16
X3 is x5 minus 6x6 plus 2 and x4 is minus 4x5 plus 4x6 minus 5.
07:50
These are the three expressions we wanted to find...