I’m just having trouble; can anyone explain this using the same approach that the teacher used?
Hi @Chris.X
Sure! This is the Gaussian elimination process to transform the system of equations into row echelon form.
Start with the system of equations, the left-hand side coefficients are written as a matrix.
Write the augmented matrix, we convert the system to a matrix:
\begin{bmatrix} 1 & 1 & 2 & | 12 \\ 3 & -3 & -1 & | 3 \\ 2 & -1 & 6 & | 24 \end{bmatrix}
The last column represents the right-hand side values.
Elimination to form row echelon form:
1 - Make the first column’s leading coefficient 1 (already done in row 1).
2 - Subtract correct multiples of row 1 from rows 2 and 3 to create zeros below the leading 1.
3 - Further, eliminate coefficients to create an upper triangular form.
The transformed system results in:
\begin{bmatrix} 1 & 1 & 2 & | 12 \\ 0 & -6 & -7 & | -33 \\ 0 & 0 & 6 & | 18 \end{bmatrix}
This equals to:
a + b + 2c = 12
-6b - 7c = -33
6c = 18
which is in row echelon form.
Try doing it yourself with another example to understand it better! Hope it helps and feel free to ask if you need further assistance.
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still dont get
what I got
1, 1,2
0,1, 7/6
0,0, 11/2
reduced:
1, 1,2
0,1, 7/6
0,0, 1
1, 1,0
0,1, 0
0,0, 1
1, 0, 0
0,1, 0
0,0, 1
I just followed the guy’s approach here
Hello, @Chris.X,
What you got is a row echelon form, however, it looks like you are taking it further and solving the unknowns. If you followed the approach, there was still the fourth column which you had missed. Add them back and you complete one way that goes from a row echelon form to solving the unknowns.
Cheers,
Raymond
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You have to use the entire system matrix, including the column to the right of the equal sign.
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thanks, I c