00:01
If each of the following matrices is the augmented matrix ab of the linear system, a x equal b, we want to find a solution of the linear system.
00:14
So we have three augmentic matrices.
00:17
So let's start with one.
00:19
And the corresponding linear system is the following.
00:24
We have the first three by three sub -matrics of the augmented matrix, as we say here, is matrix a, coefficient matrix, and the last column is the right -hand side b of the linear system.
00:42
So in this case we have three variables, let's say x, y and z.
00:48
So we get x minus y plus 3z equal 3.
00:55
That's the first equation, the first line of the augmented matrix is corresponding to this linear equation.
01:03
And the second line of the augmented matrix translates to 2x plus 2y plus z equal 2.
01:15
The last line, the last row of the augmented matrix corresponds to x minus y plus 0c, that is 0, equal 1.
01:26
So this is the linear system we got to solve.
01:29
In this case, we can use directly the third equation, and from this third equation, x is equal to y plus 1.
01:44
So we put now that in the first equation.
01:54
Put into the first equation of the linear system, we get x that is y plus 1.
02:11
I'm using this equation, remember the first one, y plus 1 minus y plus 3 z equal 3 and the good thing is that y and negative y cancel out so from here we get 1 plus 3 z equal 3 and so 3 z equal 2 and so z equal 2 3 3 calculated already the value of variables z.
02:49
We need now x and y.
02:51
Let's say we are going to use now the second equation with z equal two thirds x equal y plus one here and the second equation of the linear system we get well we put we have we see here we have 2x, that is 2 times y plus 1, plus 2y plus z, that is 2 thirds, equal 2.
03:44
So we develop a little bit of this equation, so we get 2y plus 2, 2, 2 plus 2, 2, distributing this 2 inside parentheses, plus 2y plus 2 3 3rds equal 2.
04:02
And now we can see this two here and this two here on the right each on each side of the equation cancel out so what we get is 2y plus 2y is 4 y is equal to negative 2 3rds and from here y is negative 1 6 so we have y and z and now x can be calculated with by using the value of y through this equation x equal y plus one.
04:44
Now, x is y plus one, that is negative 1, 6 plus 1, and that is 5 over 6.
04:54
So x is 5 over 6.
05:02
Then the solution, the unique solution, in fact, of the linear system associated to this augmented matrix in part 1 is the vector 5.
05:26
6 for x then y is negative 1 over 6 and z is 2 thirds or what is the same x equal 5 over 6 y equal negative 1 over 6 and z equal 2 3rds and that's the answer to first part go to second part now then in second part, the linear system associated to these automatic matrix here, again, we have three variables.
06:13
The coefficient matrix is three by three.
06:16
So we can say the linear system is negative 2x plus y plus 2z equal 1 and x minus y.
06:33
I think you see something here.
06:38
I think i'm making...
06:40
Sorry, this is not an air system.
06:43
I'm writing the third one.
06:45
This is not the second, sorry.
06:48
This is the third.
06:49
So the second is, let's see, x plus 3y.
06:55
That is, that's it.
06:57
X plus 3y minus 2z equal 1.
07:04
And then 3x plus 2y minus z equal 2.
07:17
And the third equation is negative x plus, sorry, minus 3y plus 2z equal 3.
07:34
And so if this linear system had a solution that is values for x, y and c that sold the 3, then we must have the following.
07:47
The first equation is x plus 3y minus 2 z equal 1.
07:57
And the third equation, this equation i simply wrote it again as a c is here, but the third equation is negative x minus 3 y plus 2c equal to 3.
08:13
We can change the signs both sides of the equation.
08:17
So we obtain a new equation that is valid as the original minus 2z equal negative 3.
08:27
So we can see we have the same expression on the left hand side of both equation.
08:35
That means if we had a solution to the linear system, if we had a solution for this linear system, then we would get 1 equal negative 3, which is obviously false.
09:11
And why i say that? because using the first equation as it is, and using the third equation changing the sign both sides of the equation, we get these two equations and we can see we have the same left -hand side.
09:29
And so if we put the solution, we had a solution, if we put the values of x, y, and z that is a solution of the linear system, we will have the left -hand side, the same value, and that should be then equal to 1 and negative 3.
09:46
Or in other words, 1 and negative 3 should be equal because they are equal to the same expression.
09:54
But that's false.
09:55
One is not equal to negative 3.
09:57
That means we cannot have a solution.
09:59
This contradiction will always arise if we had a numerical solution to this linear system...