Answer:
We have the system
[tex]3x+3y=6\\4x+3y=5[/tex]
The augmented matrix of the system is
[tex]A=\left[\begin{array}{ccc}3&3&6\\4&3&5\end{array}\right][/tex]
Pivot the system means to use row operations for find the echelon form of the augmented matrix.
Then,
1. In the second row of A we subtract 4/3 from the first row. We obtain:
[tex]B=\left[\begin{array}{ccc}3&3&6\\0&-1&-3\end{array}\right][/tex]
Then B is the echelon form of A.
We use backward substitution to find the set of solutions.
1.
[tex]-y=-3\\y=3[/tex]
2.
[tex]3x+3y=6\\3x+3(3)=6\\x=-1[/tex]
The system has unique solution [tex](x,y)=(-1,3)[/tex]
By pivoting, we will get:
[tex]\left[\begin{array}{ccc}1&0&|-1\\0&1&|3\end{array}\right][/tex]
So the solutions are:
Here we have the system:
3x + 3y = 6
4x + 3y = 5
We need to solve this by pivoting. This is equivalent to a Gauss-Jordan method.
So first, we want to multiply the first equation by 1/3, so we get:
x + y = 2
4x + 3y = 5
Now we want to remove the x-variable from the second equation, then we subtract 4 times the first equation in both sides:
4x + 3y - 4*(x + y) = 5 - 4*2
-y = -3
Now our system is:
x + y = 2
-y = -3
Now we want to remove the y-variable from the first equation, so we can add the second equation to it:
x + y + (-y) = 2 + (-3)
x = -1
Then our system is:
x = -1
-y = -3
Finally, we want all our coefficients to be equal to 1, so we can multiply the second equation by -1:
-1*-y = -1*-3
y = 3
Then our system can be written as:
x = -1
y = 3
Or, in matrix form:
[tex]\left[\begin{array}{ccc}1&0&|-1\\0&1&|3\end{array}\right][/tex]
This is the pivoted matrix.
If you want to learn more about systems of equations, you can read:
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