 Elementary row operations can be used to produce the row echelon form of a matrix using Gaussian elimination. So remember our main steps in Gaussian elimination is to multiply the working row by a constant to make its pivot equal to 1. We'll add multiples of the working row to the rows below to get zeros below the pivot. We'll repeat this process until we've produced the row echelon form of the matrix and then when we're done we'll use back substitution to get the actual solutions. So let's try this out with an actual system of equations. So here we'll try to use Gaussian elimination to reduce the system to row echelon form and once we're done we'll try to express all solutions in vector form, parameterizing as necessary. So first let's rewrite our equations in standard form and set down our augmented coefficient matrix. So first we'll choose an order for the variables and since our variables are x, y and z we might as well use the order x, y, z. We'll rewrite our equations so that all of our variable terms are on one side in order and added using negative coefficients as necessary and the constant is on the other. So this equation x plus 3y equals z plus 4 becomes 1x plus 3y plus negative 1z is equal to 4. And likewise 2x minus 3 equals y minus z getting everything over to the left hand side 2x plus negative 1y plus 1z is equal to 3. And now that our equations are in standard form we can write down the augmented coefficient matrix and looking at our first row the pivot is equal to 1 already so we don't have to do anything with it and if we multiply this first equation by negative 2 and add it to the second equation we'll eliminate the x coefficient. So we'll do that multiplying our equation by negative 2 gives us minus 2x plus negative 6y plus 2z equals negative 8 and then adding this to our second equation 2x plus negative 1y plus 1z equals to 3 gives us a new second equation 0x plus minus 7y plus 3z equals negative 5 and if we want to record this as a row operation we took negative 2 times the first row added it to the second and replaced the second row. And this will change our augmented coefficient matrix. The first equation remains unchanged so the first row remains unchanged the second equation has been altered and so we'll replace that second row with the new second equation. And we can read off the equations corresponding to our matrix. The first row still corresponds to our equation 1x plus 3y plus negative 1z equal to 4 our second row corresponds to the equation negative 7y plus 3z equals negative 5 and again we notice that x and y are leading variables but z is never a leading variable so we'll let it be the free variable and we'll make it equal to a parameter. So we'll let z equal to s and substitute it into our second equation which becomes minus 7y plus 3s equals negative 5. We'll then solve this equation for y and get y equals 5 7s plus 3 7s. Now that we know a value of y and z we can substitute these into our equation for x. So our first row corresponds to the equation x plus 3y plus minus 1z equal to 4 substituting in our values for y and z gives us an equation and then solving this equation for x gives us x equals 13 7s minus 2 7s. Now we can try and put this solution in vector form. So first we'll write our solution x, y, z as a vector. We'll rewrite our components so they all include a constant and an added s term using negative 0 and 1 coefficients as necessary and this allows us to split our vector into two components. The constants will form one vector and the s terms will form the other vector and we can remove the s as a scalar multiple of the second vector and write our solutions as a linear combination.