Row-echelon form is basically a stair-step pattern with leading coefficients of one. When a system is placed into row-echelon form, back substitution is very easy. The bottom row gives the answer for z. That answer is back-substituted into the second equation and y is found. Then both y and z are substituted into the first equation and x is found.
Example: Use back substitution to solve the following system.
x - y + 2z = 5
y - z = -1
z = 3
Gaussian Elimination is named after Carl Friedrich Gauss, the German mathematician who proved the fundamental theorem of algebra.
Two systems of equations are equivalent if they have the same solution set.
There are three basic operations, called elementary operations, that can be performed and that will render an equivalent system.
The addition / elimination technique that we discussed in section 5.2 uses these operations, they just weren't formalized. In the elimination technique, we could switch the two equations around and it wouldn't affect the solution set. We could multiply one or both equations by a non-zero constant and then add them together. There is nothing wrong with multiplying both equations by a non-zero constant, just that what's given here is the elementary or basic operations and multiplying both by a non-zero constant and adding would be a combination of these.
Just like with a 2×2 system of linear equations, a larger system can have one, none, or many solutions.
A consistent and independent system. The solution is given as an ordered triplet.
An inconsistent system. There is no solution, and the answer should be written as the null set or the empty set, but not the set containing the null set. This results if at any point while trying to solve the system, an equivalent form has a contradiction (variables eliminate and the statement is false).
A consistent and dependent system. The solution is given in parametric form. This case results when there are less equations than variables and no contradictions.
Example: Solve the following system of linear equations.
3x - 2y + z = 1
y + 2z = 3
If you have the right number of ordered pairs, you can fit a model by solving a system a system of linear equations. To have an unique solution, you need as many points as constants.
There are two values that need to be found, A and B. Therefore, it takes two points to determine a the equation of a line.
Let's say the line passes through the points (2,3) and (5,7). The resulting system of linear equations, obtained by substituting the values in for x and y is shown. Simply solve the system of linear equations, plug in the values for A and B, and you have the model.
3 = 2A + B 7 = 5A + B
Solving that gives A = 4/3 and B = 1/3. The equation of the line passing through the given points is y = 4/3 x + 1/3.
This time, there are three variables, A, B, and C. Therefore, it takes three non-collinear points to determine a equation of a parabola.
Let's say the parabola passes through the points (2,3), (5,7), and (8,4). Substitute the values in for x and y and the three linear equations that result is the system that needs to be solved to find A, B, and C.
3 = 4A + 2B + C 7 = 25A + 5B + C 4 = 64A + 8B + C
If you're wondering where there 4, 25, and 64 came from, they are the values of x2 when x = 2, x = 5, and x = 8.
Solving the system gives A = -7/18, B = 73/18, and C = -32/9. The equation is y = -7/18 x2 + 73/18 x - 32/9.
This procedure could be extended to find the equation of a cubic passing through four points or a quartic equation passing through five points.
You have three variables, D, E, and F. Therefore, it takes three non-collinear points to determine the equation of a circle.
Let's say the circle passes through the points (2,3), (5,7), and (8,4). Substitute the values in for x and y and the equations that result give you the system of linear equations that can give the coefficients that define the circle. You may find it easier to move the constant to the other side.
4 + 9 + 2D + 3E + F = 0 becomes 2D + 3E + F = -13 25 + 49 + 5D + 7E + F = 0 becomes 5D + 7E + F = -74 64 + 16 + 8D + 4E + F = 0 becomes 8D + 4E + F = -80
If you solve that system, you get D = -69/7, E = -55/7, and F = 212/7.
Substituting into the original model gives x2 + y2 -69/7 x - 55/7 y + 212/7 = 0.
If you were so inclined (this is a pretty ugly problem), you could complete the square to find the center and radius of the circle.
If you have exactly three non-collinear points, then the parabola will pass through the points exactly. If there are more than three points, then you must fit a quadratic model to the data, but it won't necessarily pass through all of the points. Recall the least squares regression line that we discussed in the last section that went with the model y = ax + b.
b∑1 + a∑x = ∑y
b∑x + a∑x2 = ∑xy
Now, we are going to find a system of equations that will fit the model y = ax2 + bx + c. It is an extension of the previous model. Notice it has the same pattern as before. As you work from left to right on the left hand side, each summation has one additional x in it. As you work from top to bottom, each term has one more x than the corresponding term in the previous equation.
c∑1 + b∑x + a∑x2 = ∑y
c∑x + b∑x2 + a∑x3 = ∑xy
c∑x2 + b∑x3 + a∑x4 = ∑x2y
If you wanted to find a least squares regression cubic, then you would need four variables and at least five points (four points would exactly fit the model). Follow the same pattern as before, adding one more variable (for d) and one more row.