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Gaussian Elimination Explained Simply
Solving a system of 10 equations with 10 unknowns by intuition is impossible. You need an algorithm, and Gaussian elimination is the most powerful tool to solve linear systems of equations by reducing any matrix into a simpler triangular form that makes the answer almost trivial to find.
What is Gaussian elimination and why does it matter?
Gaussian elimination is a systematic process that simplifies a system of equations without changing its solution. The goal is to transform the matrix into row echelon form, a structure that looks like an inverted right triangle, where every row starts further to the right than the one above it.
A matrix is in echelon form when the first nonzero element of each row, called the pivot [00:55], sits to the right of the pivot in the row above, and everything below each pivot is zero. Once your matrix has this triangular shape, solving it becomes straightforward through back substitution.
What is a pivot in a matrix? It is the first nonzero number of a row in echelon form. Pivots guide the elimination process and later reveal the rank of the matrix.
How do you build the augmented matrix from a system?
Let's work with a real three by three example introduced in the lesson [01:30]:
The first move is to create the augmented matrix, written as A | B. On the left you place the coefficient matrix, and on the right you attach the results vector separated by a vertical line.
[ 1 2 1 | 2 ] [ 3 8 1 | 12 ] [ 0 4 1 | 2 ]
This compact form lets you operate on coefficients and results at the same time, which is exactly what the algorithm needs.
How do you eliminate values below each pivot?
The strategy is simple: identify the pivot, then turn every number below it into zero using row operations.
First pivot: clearing the first column
The first pivot is the 1 in position (1,1). Below it sits a 3, so you apply the operation Row 2 minus 3 times Row 1 [03:10]. Multiplying Row 1 by 3 gives (3, 6, 3 | 6). Subtracting from Row 2 produces (0, 2, -2 | 6). The matrix now looks like:
[ 1 2 1 | 2 ] [ 0 2 -2 | 6 ] [ 0 4 1 | 2 ]
The zero we wanted under the first pivot is in place, and the original third row already had a zero there.
Second pivot: clearing the second column
The next pivot is the 2 in the second row. Below it is a 4, so you apply Row 3 minus 2 times Row 2 [05:00]. Multiplying Row 2 by 2 gives (0, 4, -4 | 12). Subtracting from Row 3 yields (0, 0, 5 | -10). The final echelon form is:
[ 1 2 1 | 2 ] [ 0 2 -2 | 6 ] [ 0 0 5 | -10 ]
That staircase shape is exactly what we were chasing.
Why do we want row echelon form? Because the last row isolates a single unknown, letting us solve it directly and propagate the result upward.
How does back substitution finish the job?
With the system in triangular form, you rewrite it as equations and solve from the bottom up [06:30]:
The unique solution is the point (2, 1, -2). Geometrically, each equation represents a plane in three dimensional space, and that point is the only place where all three planes intersect simultaneously [07:45]. Gaussian elimination is, in essence, an algorithm to locate that intersection.
What is back substitution? It is solving a triangular system from the last equation upward, plugging known values into the equations above to isolate each remaining variable.
Practice exercise to lock in the method
Try this small system on your own and share your answer in the comments:
Write the augmented matrix, choose the right row operation to zero out the value under the first pivot, and then back substitute to find x and y. If you want, share your full process the same way we did here.
One last thing worth noticing: during elimination we tracked pivots carefully. The number of pivots you end up with tells you something deep about the matrix itself, a property called the rank, which opens the door to the next topic.
