TERMINOLOGY - BASIC THEOREMS, EXAMPLES & EXERCISESLinear Systems and Matrices.Linear equations in the two variables x and y are equations of the type cbyax , where the a, b, and c are constants. Similarly, linear equations in the three variables x, y, and z are equations of the type dczbyax , where the a, b, c, and d are constants.Exercise 1: Classify the following equations as linear or nonlinear:02 ,762 ,1 ,012 ,10322 yxzyxyxyxxyyyx A system of linear equations (also called a linear system) is a finite collection of linear equations.Example 1: 432102 ,762zyxzyxzyxNote: If at least one of the constant terms in the linear system is nonzero, the linear system is called nonhomogeneous, otherwise, the linear system is called homogeneous.By a solution of a system is meant a set of values (normally real numbers) for the variables that satisfy all the equations in the system simultaneously. Example 2: The set of values x = 23, y = -11, and z = 1 constitute a solution of the system432102 ,762zyxzyxzyxbecause 4)1(3)11(22310)1(2)11(23 ,7)1(6)11(223A linear system is said to be consistent if it has at least one solutionExample 3: The system in Example 2. is consistent.Exercise 2: Show that the following system is inconsistent.103 ,762zyzyExercise 3: Give the geometrical interpretations for each of the following:The system222111 ,czbyaczbya(a) has a unique solution;(b) has an infinite solution set;(c) has no solution.Theorem: A linear system of equations either has a unique solution, has no solution, or has infinitely many solutions.Note: Every nonhomogeneous linear system with more variables than equations either has no solution or infinitely many solutions.Exercise 4: Determine whether the following system has a unique solution, no solution, or an infinite solution set. If the system has an infinite solution set, express the solution set in parametric form:682 ,9123yxyxExercise 5: Use the method of elimination to determine if the following linear system isconsistent or inconsistent. If the system is consistent, find the solution if it is unique, otherwise, describe the infinite solution set in terms of arbitrary parameters.432102 ,762zyxzyxzyxExercise 6: Write the augmented coefficient matrix of the linear system in Exercise 5The three Elementary Row Operations (ERO) on a matrix A are:1. Interchange any two rows of matrix A (ijR);2. Multiply any one row of A by a nonzero constant (ikR);3. Add a constant multiple of one row of A to another row (jiRkR ).Exercise 7: Let 6 4 7 16 7 4 31 3 1 2APerform the following sequence of row operations on the matrix A:3121132- ,3 , RRRRR A matrix A is called an echelon matrix if it has the following two properties:1. Any nonzero row of A precedes any zero row (if any);2. In each nonzero row of A, the first nonzero element lies strictly to the right of the firstnonzero element in the preceding nonzero row (if there is a preceding nonzero row).Example 4: The following is a list of some 3 x 3 echelon matrices.0 0 05 0 05 4 1 ,3 0 00 2 00 2 1 ,0 0 05 1 02 3 1 ,0 0 00 0 03 0 2Example 5: The augmented matrix of the system63 23 ,10 262vuvzvuzyxis the echelon matrix6 3 1 0 0 02 3 0 1 0 010 1 2 6 2 1Any matrix can be transformed into an echelon matrix using elementary row operations. This procedure is known as Gaussian elimination.Exercise 8: Use EROs to transform the augmented coefficient matrix of the following system to echelon form. Then solve the system by backward substitution.92217421263321321321xxxxxxxxxA reduced echelon matrix A is an echelon matrix that has the following two additional properties:3. Every leading entry of A is 1;4. Each leading entry of A is the only nonzero element in its column.Example 6: The following is a list some reduced echelon matrices.1 0 0 00 1 0 00 0 4 1 ,0 1 00 0 1 ,0 0 05 1 02 0 1 ,0 0 00 0 03 0 1The process of transforming a matrix into a reduced echelon form by EROs is called Gauss-Jordan elimination.Example 7: Use the Gauss-Jordan elimination to solve the system:174835 33 2zyxzyxzyxReduce the augmented matrix to a reduced echelon form using the following sequence of row operations:12231333231212 ,2 , ,31 ,2 ,3 , RRRRRRRRRRRRR .3/4 1 0 03/2 0 1 0317/ 0 0 13/4 1 0 33/2 0 1 13/31 0 2 13/4 1 0 02 2 1 03 1 2 14 3 0 02 2 1 03 1 2 18 7 2 02 2 1 03 1 2 117 4 8 35 1 3 13 1 2 1The unique solution of the system is 3/4 ,3/2 ,3/17  zyxTheorem: Every matrix is row equivalent to one and only one reduced echelon form.Recall: If the constant terms in the linear system are all zero, then the linear system is called homogeneous,Example 8:04 302 5203 2 zyxzyxzyxNote:1. The trivial solution 0 ..., ,0 ,021nxxxis a solution of every homogeneous linear system (i.e., a homogeneous system is always consistent);2. If a homogeneous linear system has one nontrivial solution, then it must have infinitely many solutions;Theorem: Every homogeneous linear system with less equations than variables has infinitely many solutions.Theorem: A homogeneous system consisting of n equations in n variables has only the trivial solution if, and only if, the coefficient matrix of the system is row equivalent to then x n