2. It should be clear that matrices provide a concise form for representing simultaneous equations. For example we have the next equations system Where A is the n by n square matrix of coefficients.
3. Small number of equations For solving small sets of simultaneous equations (n≤3) we don’t need a computer, here are some methods to do this things: Graphical method Cranner’s rule Elimination of unknowns.
4. Graphical method Plotting any two equations system on Cartesian coordinates with one axis corresponding to x1 and the other to x2 we can get a graphical solution. If you the system: Let x1 be the abscissa. Solve for x2: solution
5. Graphical solution of a set of two simultaneous linear algebraic equations, the intersections of the lines represents the solution. For three simultaneous equations , each equation would be represent by a plane in a three dimensional coordinate system
6. Cramer’s Rule Given a system of linear equations, Cramer's Rule is a handy way to solve for just one of the variables without having to solve the whole systems of equations Let's use the following system of equations system of equations. 2x + y + z = 3 x – y – z = 0 x + 2y + z = 0
7. We have the left-hand side of the system with the variables (the "coefficient matrix") and the right-hand side with the answer values. Let D be the determinant of the coefficient matrix of the above system, and let Dx be the determinant formed by replacing the x-column values with the answer-column values:
8. Similarly, Dy and Dz would then be Cramer's Rule says that x = Dx ÷ D, y = Dy ÷ D, and z = Dz ÷ D. That is: x = 3/3 = 1, y = –6/3 = –2, and z = 9/3 = 3
9. Elimination of unknowns This method is similar to the method you probably learned for solving simple equations. If you had the equation "x + 6 = 11", you would write "–6" under either side of the equation, and then you'd "add down" to get "x = 5" as the solution. x + 6 = 11 –6 –6x = 5
10. Example Solve the following system using addition. 2x + y = 93x – y = 16 Then I'll draw an "equals" bar under the system, and add down: 2x + y = 93x – y = 165x = 25 Now I can divide through to solve for x = 5, and then back-solve, using either of the original equations, to find the value of y. The first equation has smaller numbers, so I'll back-solve in that one: 2(5) + y = 9 10 + y = 9 y = –1 Then the solution is (x, y) = (5, –1).
