Multivariable Linear Systems:

Row-Echelon Form:

Row-echelon form is mainly a stair-step pattern with the leading coefficients of one. Whenever a system is positioned into row-echelon form, back substitution is very simple. The bottom row provides the answer for z. That answer is back-substituted to the second equation and y is found. Then both y and z are replaced to the first equation and x is found.

Illustration: Use back substitution to resolve the given system:

x - y + 2z =  5
y - z = -1
z = 3

a) The bottom equation provides that z = 3.
b) Plugging z = 3 to the second equation provides y - 3 = -1 or y = 2.
c) Plugging y = 2 and z = 3 to the first equation gives x - 2 + 2(3) = 5.
d) Solving for x gives 1, therefore the solution is {(1, 2, 3)}.

Gaussian Elimination:

Gaussian Elimination is introduced by Carl Friedrich Gauss, a German mathematician who proved the fundamental theorem of algebra.

Two systems of equations are equal if they have similar solution set.

Elementary Operations:

There are three fundamental operations, termed as elementary operations, which can be executed and that will render the equivalent system.

a) Interchange two equations.
b) Multiply one equation by the non-zero constant.
c) Multiply an equation by the non-zero constant and add it to the other equation, substituting that equation.

The addition/elimination method uses such operations, they just weren't formalized. In the elimination method, we could switch two equations around and it wouldn't influence the solution set. We could multiply one or both equations by the non-zero constant and then add up them together. There is nothing wrong with multiplying both the equations by a non-zero constant, just that what are given here are the elementary or fundamental operations and multiplying both by a non-zero constant and adding up would be a combination of such.

The number of solutions to a linear system:

Just similar to with a 2×2 system of linear equations, a larger system can contain one, none, or many solutions.

Unique Solution:

It is a consistent and independent system. The solution is given as the ordered triplet.

No Solution:

It is an inconsistent system. There is no solution and the answer must be written as the null set or empty set, however not the set having the null set. This outcomes if at any point while trying to resolve the system, an equivalent form consists of a contradiction (that is, variables eliminate and the statement is false).

Many Solutions:

It is a consistent and dependent system. The solution is provided in parametric form. These case outcomes when there are less equation than variables and no contradictions.

Illustration: Solve the given system of linear equations.

3x - 2y + z = 1
y + 2z = 3

a) Take the second equation and resolve for y to obtain y = 3 - 2z.

b) Replace y = 3 - 2z to the first equation for y. This gives you 3x - 2(3 - 2z) + z = 1, that is an equation with the two variables, x and z. A little bit of simplification provides 3x - 6 + 4z + z = 1 or 3x + 5z = 7.

As you earlier resolved for y in terms of z, we now solve for x in terms of z. It is significant that we solve all the other variables in terms of similar variable. Solving 3x + 5z = 7 for x gives x = (7 - 5z)/3

Give the solution. As x and y is both in terms of a third variable, z, we assume z = t and obtain:

x = 1/3 (7 - 5t)
y = 3 - 2t
z = t

Mathematical Models:

When you have the right number of ordered pairs, you can fit a model by resolving a system a system of linear equations. To encompass a unique solution, you require as many points as constants.

Linear Model: y = Ax+B

There are two values which need to be found, A and B. Thus, it takes two points to find out an equation of a line.

Let us state that the line passes via the points (2, 3) and (5, 7). The resultant system of linear equations, obtained by replacing the values in for x and y is shown below. Just solve the system of linear equations, plug the values for A and B, and you have the model as:

3 = 2A + B
7 = 5A + B

Solving the above gives A = 4/3 and B = 1/3. The equation of the line passing via the given points is y = 4/3 x + 1/3.

Quadratic Model: y = Ax² + Bx + C

This time, there are three variables: A, B, and C. Thus, it takes three non-collinear points to find out an equation of the parabola.

Let us state that the parabola passes via the points (2, 3), (5, 7), and (8, 4). Replace the values for x and y and the three linear equations which result the system which requires to be solved to determine A, B and C.

3 = 4A + 2B + C
7 = 25A + 5B + C
4 = 64A + 8B + C

When we are wondering where there 4, 25 and 64 came from, they are the values of x² if x = 2, x = 5 and x = 8.

On solving the system: A = -7/18, B = 73/18, and C = -32/9. The equation is y = -7/18 x² + 73/18 x - 32/9.

This process could be extended to determine the equation of a cubic passing via four points or the quartic equation passing via five points.

Circle: x² + y² + Dx + Ey + F = 0

We have three variables: D, E and F. Thus, it takes three non-collinear points to find out the equation of a circle.

Let us state the circle passes via the points (2, 3), (5, 7), and (8, 4). Replace the values for x and y and the equations which result give you the system of linear equations that can provide the coefficients which define the circle. You might find it simpler to move 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

When we solve that system, we get D = -69/7, E = -55/7 and F = 212/7.

Replacing to the original model gives x² + y² -69/7 x - 55/7 y + 212/7 = 0.

Least Squares Regression Parabola:

When we have precisely three non-collinear points, then the parabola will pass via the points exactly. When there are more than three points, then you should fit a quadratic model to the data, however it won't essentially pass via all of the points. Remember that the least squares regression line went with the model y = ax + b.

b∑1 + a∑x = ∑y
b∑x + a∑x² = ∑xy

Now, we are going to determine a system of equations which will fit the model y = ax² + bx + c. This is an extension of the previous model. Note that it has similar pattern as before. Since you work from left to right on the left hand side, each and every summation consists of one additional x in it. Since you work from top to bottom, each and every term consists of one more x than the corresponding term in previous equation.

c∑1 + b∑x + a∑x² = ∑y
c∑x + b∑x² + a∑x³ = ∑xy
c∑x² + b∑x³ + a∑x⁴ = ∑x2y

When we wanted to find out a least squares regression cubic, then we would require four variables and at least five points (that is, four points would precisely fit the model). Follow similar pattern as before, adding up one more variable (for d) and one more row.

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