Thread: Help with solving linear equations using Gaussian elimination...

I have been following notes online to finish my assignment and this is what confuses me:

Gaussian Elimination. Consider a linear system.
1. Construct the augmented matrix for the system;
2. Use elementary row operations to transform the augmented matrix into a triangular one;
3. Write down the new linear system for which the triangular matrix is the associated augmented matrix;
4. Solve the new system. You may need to assign some parametric values to some unknowns, and then apply the method of back substitution to solve the new system. For number 2, it says to use elementary row operations to transform an augmented matrix into a triangular one...

Do each of the leading coefficients have to be a 1?

The notes that my teacher wrote on the board show that she changed each coefficient to a 1 by using the rules.

I am confused

One ERO is to multiply a row by a constant. That will enable you to get leading 1's on each row. This step is technically unnecessary (it's not done on a computer, for instance). What's important is getting the matrix to be upper triangular. Make sense?

Yes it makes sense.

So far i've done the problem again and the leading coefficients are not 1, but I am using the back substitution method to find out the linear equations

Also, is there a way to find out if my answer is correct?

I thought I heard my teacher say you can check your answer by plugging the end result into the original equations...I have a ear condition so it's hard to hear.

Your teacher was correct about checking your work. In fact, I would highly recommend checking your work on every single problem you ever do for the rest of your life. You should always double-check and make sure your answer makes sense. (Right order-of-magnitude? Correct sign? Correct units? etc.) In linear equations, you'll never introduce extraneous solutions by doing an irreversible operation (like squaring, for example), but in nonlinear problems, this can happen quite frequently. Thus, you might get solutions that are spurious. That is why it is important. But it's also good in a linear system to check your answers.