1.1 Systems of Linear Equations Recall: a linear equation in x and y has the form y = mx +b or ax + by = c. The standard form is most useful here. Recall: a system of linear equations in two variables is a collection of equations: ax by c dx cy e In reality, most situations depend on several variables, so we write: Definition: A linear equation in x1 , x2 , x3 ...xn has the form a1 x1 a2 x2 a3 x3 an xn b where the ai and b are constants. Note: if an equation can be rearranged algebraically to look like this, it is linear. Example: x2 2 6 x1 x3 is linear since it can be written 2 x1 x2 x3 2 6 Definition: A solution to a system of equations in n variables is a list x1 s1 x2 s 2 x3 s3 xn s n Or an ordered n-tupple s1, s2 , s3 ,..., sn To solve a system of two linear equations, you can use graphing, substitution, or elimination. Example: x1 x2 7 2 x1 x2 8 We can solve graphically on the TI-84. There are 3 possible outcomes when solving a system of equations: Higher Dimensions: Equations in R3 can describe lines or planes. ►The solution set for a linear system is the set of all possible solutions of the linear system. ►Two linear systems with the same solution set are called equivalent systems. Strategy for Solving a System Replace one system with an equivalent system that is easier to solve. x1 2 x2 1 x1 3x2 3 Add equation 1 to equation 2 to get x1 2 x2 1 x2 2 Add 2 times equation 2 to equation 1: x1 3 x2 2 Note: In two dimensions, each equation represents a line. In higher dimensions with more variables, we introduce a way of streamlining the the process. Definition: A matrix is an array of numbers. An m x n matrix has m rows and n columns. We can use these to solve systems that would be difficult or impossible to graph: Example: x1 2 x2 4 x2 x3 0 x1 3x2 2 x3 5 The coefficients of the xi can be put into a matrix called the coefficient matrix: 0 1 2 0 1 1 1 3 2 This is a 3 x 3 matrix where each entry is a coefficient of a variable in the original system. Entries are identified by their row and column number. For example, 2 is in the first row of the second column, so it called the 1, 2-entry. ►We can represent the whole system with an augmented matrix. 1 2 0 4 0 1 1 0 1 3 2 5 Our goal is to create an augmented matrix like the following: 1 0 0 r 0 1 0 s 0 0 1 t This represents the system x1 r x2 s x3 t and shows the solution r , s, t . We can get the first matrix into the form that shows the solution to the system by using row operations. Elementary Row Operationss 1. (Replacement) Replace one row by the sum of itself and another row (or a multiple of another row). 2. (Interchange) interchange two rows. 3. (Scaling) multiply all entries in a row by a non-zero constant. Starting with the augmented matrix of the system above: 1 2 0 4 0 1 1 0 1 3 2 5 Keep 1 in R1 and use it to eliminate the 1 from R3: R3+(-R1) → R3 1 2 0 4 0 1 1 0 0 1 2 1 We say column 1 is cleared. Note: The text shows the systems beside the matrices, but you need only show the matrices. Keep the 1 in the 2,2-entry, and use it to clear the 1 below it. R3+(-R2) → R3 1 2 0 4 0 1 1 0 0 0 1 1 Now R3 is almost perfect. Scale it by -1: 1 2 0 4 0 1 1 0 0 0 1 1 ►The matrix is now in triangular form, all zeroes below the main diagonal. It is usually more efficient to clear C3 before C2. 1 2 0 4 0 1 0 1 0 0 1 1 R1+(-2R2) → R1 1 0 0 6 0 1 0 1 0 0 1 1 This represents the system x1 6 x2 1 x3 1 The unique solution is a point in R3, the ordered triple (6, -1, -1). Check: ►We say two matrices are row equivalent if there is a series of elementary row operations that transforms one matrix into the other. ►Note that each operation is reversible, so reversing the operations turns the second matrix into the first. Existence and Uniqueness A basic question we ask in mathematics is “When does an object exist; and if it exists, is it unique?” ►If a solution exists (one or infinitely many), the system is called consistent. If no solution exists, the system is inconsistent. You do not need to solve a system to figure this out. You can look at the “triangular” form of the matrix. Example: 1 5 2 7 0 1 12 1 0 0 0 1 R3 says that 0 = –1. Since this is false, there is no solution to the system represented by the matrix, and the system is inconsistent. Example: #20 Determine the value(s) of h such that the augmented matrix is the matrix of a consistent system. h 3 1 2 4 6 Start by finding the “triangular” form: R2+2R1→R2 h 3 1 0 2 h 4 0 If the system is consistent, then 2h 4x2 0 This is true if 2h 4 0 or if x2 0 . Thus, the system is consistent for all values of h. Example: 2 x4 3 x1 2 x2 x3 0 x3 3 x4 1 2 x1 3 x2 2 x3 x4 5 has the solution or solution set 1 7 , 0, 0, 3 since 3 7 3 1 3 3 00 0 1 0 3 1 3 7 1 2 0 0 5 3 3 4