 One of the standard techniques in matrix algebra is known as Gaussian elimination, and this is based on two fundamental ideas. First of all, every row of an augmented coefficient matrix corresponds to an equation in standard form, and that anything you can do to an equation in standard form can be done to a row of the augmented coefficient matrix. And when we talk about Gaussian reduction, what we mean is we want to produce a row echelon matrix from the original matrix. So let's take a look at a system of equations. Here's one, two, three variables, and three equations. Two x plus three y plus equals two, three, and so on. So I can do this, I can solve this system by reducing the augmented coefficient matrix. So first I'll form the augmented coefficient matrix. So I want to rewrite everything, so I have an equation in standard form. I'm adding. I have my coefficients of the different variables. I don't really need to keep track of the variables, and so there's my augmented coefficient matrix, and I'll throw it into a set of parentheses to indicate that it's a matrix. Now I want to perform row operations to get this matrix into row echelon form, which means that I want to make sure that the first non-zero entry of every row is going to be one, and all of the entries below it will be zero. If I don't want to deal with fractions, I'll use the Chinese method and focus on first making all the entries below the first non-zero term. I'm going to clear out the columns, and then I'll worry about getting the coefficients equal to one. So first off I'll copy the first row down, because I'm not going to do anything to that first row, and then I'll try to get zeros below the first entry. So here my first non-zero term in the row is two. I want to get zeros below it, and because I'll be using this number here, this number is called the pivot of this row. Now let's take a look at that. I have my first row, and let's go ahead and work on that second row. Let's work on the second row only, and so the first and second rows correspond to the equations 2x plus 3y plus c equals 2, 3x plus negative y minus y plus 2 equals negative 1. And I'm going to apply our Chinese method if I multiply the first equation by 3, that's my coefficient of x in the second equation, and my second equation by negative 2, that's the negative the coefficient of x. I get this set of equations, and when I add the two equations, because the x terms are equal and opposite, the x terms drop out, my y terms, my z terms, and so on. I get a new equation, and I can use this as my new second row. It's x coefficient 0, y 11, c coefficient negative 1, and constant term is going to be 8, so my terms are going to be 0, 11, negative 1, and 8. Now what I did to get this row, I'll record those row operations, well that was 3 times the first row plus negative 2 times the second row, and then I drop that into the second row, so my row operations 3r1 plus minus 2r2 goes to row 2, and so I record that row operation here. Well now let's take the first and third rows, so again I'll use my first row to clear out the third row, so again that first row corresponds to an equation, that third row corresponds to an equation, and so what I'll do, again I'll take a multiplication, I'll multiply each by the coefficient of the other, what is this going to be negative? So I'll multiply the first equation by 4, I'll multiply the second equation by negative 2, and I'll add them together, and again that clears out my x coefficient, and I get a new equation, and this is my new third row, and again what I did, first row times 4, first equation times 4, second row, sorry third row, third equation times negative 2, add them together, and store the results in the third row. And there's my new coefficient matrix. And the last thing, we do want that first entry in the first row to be 1, so we can get this into row echelon form, so what is that? Well again that's an equation 2x plus 3y plus z equals negative 2, so here's my first row, here's my corresponding equation, and I want that first entry to be 1, that means I want this coefficient to be 1, so I need to multiply through by 1 half, and that gets me the new equation 1x, 3 half y, 1 half z equals negative 1, and I got that by multiplying our first row by 1 half, and then I'm going to store the results in that first row. And so there's my three row operations that get me my new matrix. And so we're getting towards row echelon form, the first entry in the first non-zero entry in the first row is 1, and everything below it is 0, so now I go to the next row, first non-zero entry isn't 1, and more importantly for our purposes, the things below it aren't 0, so we've got to apply a further reduction to this matrix. So we'll copy it down, and we proceed again. So again the first row has the first leading non-zero entry equal to 1, and zero is below it, so it's fine, we'll go ahead and keep it, and now we want to work on the second and third rows, and we'll keep the second row, and we'll use this as a pivot to obtain zeroes below it. So again, second and third rows correspond to the equations, 0x plus 11y minus z equals 8, 0x plus 2y plus 6z equals 20, so here's my two equations, and again I can do multiply the first equation by 2, the first coefficient, the first non-zero coefficient, I'll multiply it by 2, I'll multiply the second equation by negative 11, the coefficient of the other equation, and again that gets me my first variable coefficients equal but opposite, and so those will drop out when I add them, and I'll get a new equation, 0x plus 0y equals minus 68c, and when I add these minus 204, and there's my new third row, and what I did was second row times 2, third row times minus 11, add them, and store the results of the third row. So that's going to be this set of row operations, and finally I want this entry to be 1, so I'm going to go through this second row, corresponds to the equation, 0x plus 11y minus z equals 8, to make the first non-zero entry equal to 1, I need to make this 11 into a 1, I need to multiply by 11, and that's going to give me a new system, a new equation, which I can store as my second row again, and what I did, 11 to the second row, and I've got to store that in row two, and there's my next set of row operations, and I have reduced this matrix still further, and I'll continue with our current coefficient matrix. So again, first row is fine, second row is fine, so we'll keep those, our third row, which will be our working row, we'll copy that over as well, and while we do want to make that, well there's nothing below it, so we don't have to worry about eliminating anything, so now we can just focus on making that first non-zero entry equal to 1, and again remember, this is the equation minus 68, third variable z equals minus 204, if I want to make that 1z equal to something, I'm going to multiply by negative 1 over 68, so I'll record that as my row operation, multiply by 1 over 68, negative, and store that as my value in row three, and that gets me my new third row, and now I can solve five act substitution, every row corresponds to an equation, so this last row corresponds to 0x plus 0y plus 1c equals 3, that's this equation that gives us z equals 3, and my second row corresponds to the equation 0x plus 1y, minus 11z equals 811s, so I have another equation there, I know the value of z, z equals 3, so I can substitute that into this equation, and now I have an equation, don't really have an x value there, but I have an equation where I can solve this equation for y, and I get y equals 1, and finally my first row corresponds to the equation 1x plus 3 halves y plus 1 halves z equals 1, I know the value of y, I know the value of z, so I can substitute those into this equation, and get a nice solvable equation, x equals negative 2, and there's my solution, x equals negative 2, y equals 1, z equals 3.