Lecture 1: Systems of Linear Equations



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MTH Elementary Matrix Algebra Professor Chao Huang Department of Mathematics and Statistics Wright State University Lecture 1 Systems of Linear Equations ² Systems of two linear equations with two variables ½ ax + by = A cx + dy = B. Recall that a pair of two numbers (x 0, y 0 ) is said to be a solution if x = x 0 and y = y 0 satisfy both equations simultaneously. Graphically, we know that the graph of any linear equation of two variables is a straight line in xy plane. Therefore, a solution (x 0, y 0 ) is the intersection of two lines represented by two linear equations. We know that there are three possibilities (1) two line intersect at exactly one point (the system admits a unique solution), () two lines are identical (there are in nite many solutions), and () two lines are parallel but di erent (there is no solution.) Example 1.1. Consider a system of two equations with two unknowns ½ x y = 1 x + y =. (1) There are two commonly used methods to solve linear systems elimination method and substitution method. We now use the elimination method by adding both equation to obtain y =. Substitute into the rst equation of system (1), we nd x = y 1 =. The system has exact one solution (, ). 1

Thick line x + y = Example 1.. Consider another system ½ x y = 1 x + y =. () Adding both equations leads to 0 =. This contradictory equation indicates that system () cannot possibly have a solution. Thick line x + y = Example 1.. Consider the third system ½ x y = 1 x + y = 1. ()

The second equation is a multiple of the rst equation (by -1). Therefore, this system has only one independent equation. Consequently, there are in nitely many solutions x = 1 + t y = t for any choice of t. These three examples demonstrated three possible situations for systems of linear equations in D one unique solution (good systems), or no solution (the system contradicts to itself), or in nite many solutions (two equation are equivalent, called degenerated systems). This observation and the elimination method (one of the most popular method to solve systems in D) extends to general situations. ² General Linear Systems. A system of m linear equations with n variables, denoted by x 1, x,..., x n, reads as > a 11 x 1 + a 1 x +... + a 1n x n = b 1 a 1 x 1 + a x +... + a n x n = b ()... > a m1 x 1 + a m x +... + a mn x n = b m a ij is called a coe cient. It is in the ith equation and is associated with x j. A solution of system () consists of n ordered numbers (x 1, x,..., x n ) satisfying all m equations. The set of all possible solutions is called a solution set. General systems can be solved using the method of elimination following the same idea for D systems. Example 1.. Solve the following system of equations with unknowns x 1 x + x = 0 x x = x 1 + x + 9x = 9. () Solution We use the same method of elimination. The last equation is replaced by the sum of itself and times the rst equation ( x 1 + x + 9x = 9) + (x 1 x + x = 0) 0 x + 1x = 9 while maintain the other equations. The system reduces to x 1 x + x = 0 x x =. (6) x + 1x = 9

Next, We add Equation # to # in (6) (replace # by the sun of Equation # and #) to arrive at x 1 x + x = 0 x x = x = 1 We now solve x from the last equation to get x Equation #, we nd x µ 1 = =) x = =.. = 1. Substitute it into Finally, substituting x = 1, x = into the rst equation, we obtain 1 µ µ x 1 + 1 = 0 =) x 1 = 6 1 1. The solution is (x 1, x, x ) = µ 6 1,, 1. ² Argumented Matrix This whole process can be simpli ed using symbolic means. To this end, we introduce the coe cient matrix for system () as an array (or table) of m rows and n columns a 11 a 1... a 1n 6 a 1 a... a n............. () a m1 a m... a mn m n We call it a m n matrix. m n is referred as the dimension of the matrix. However, this matrix contains only the left-hand side of system (). The entire information of system () can be found in the following table of m rows by (n + 1) columns 6 a 11 a 1... a 1n j b 1 a 1 a... a n j b............ j... a m1 a m... a mn j b m. () It is called Augmented matrix. This is a m (n + 1) matrix. The vertical dash line is for the purpose of reminding us that the last column represents the righe-hand side of the system. A system of linear equations is represented by its Augmented matrix.

For instance, in Example above, the system x 1 x + x = 0 x x = x 1 + x + 9x = 9 has the augmented matrix 1 1 j 0 0 j 9 j 9 ² Elementary Row Operations The elimination method basically consists of the following row operations for the system 1. Replace one equation by the sum of itself and a multiple of another equation. For instance, in solving Example above, the rst stem is to replaced the third equation by the sum of itself and times the rst equation The resulting system is X ( x 1 + x + 9x = 9) + (x 1 x + x = 0) 0 x + 1x = 9. x 1 x + x = 0 x x = x + 1x = 9.. (9) However, if we perform the same row operation on the augmented matrix replace the third row in its augmented matrix by the sum of itself and times the rst row, i.e., the updated third row is 9 j 9 then we obtain a new matrix + 1 1 j 0 = 0 1 j 9, 1 1 j 0 0 j 0 1 j 9 which is exactly the augmented matrix for (9). Conclusion the row operation for the system is equivalent to the same type of row operation for the augmented matrix. Some times we also need two other row operations X,

. Interchange two equations (equivalent to exchanging two rows). One equation is replaced by a non-zero multiple of itself (equivalent to replacinging one rows by a non-zero multiple of itself). De nition 1.1. The following operations are called elementary row operations 1. Replace one row by the sum of itself and a multiple of another row.. Interchange two rows. One row is replaced by a non-zero multiple of itself. De nition 1.. Two matrices are called row equivalent if one is obtained from the other by a series of elementary row operations. Two linear systems are called equivalent if their augmented matrices are row equivalent. Theorem 1.1 If the augmented matrices of two linear systems are row equivalent, then the solution sets of these two linear systems are identical. In short, elementary row reductions do not alter the solution set. Example 1.. Solve system () using row operations. Solution Augmented matrix 0 j 1 1 j 0 9 j 9 We now perform a series of row operation in the way equivalent to what we did before 1 1 j 0 0 j 9 j 9 R 1 + R! R! 1 1 j 0 0 j 0 1 j 9 R + R! R! 1 1 j 0 0 j. 0 0 j 1 Notice that here, in the symbol the left-hand, R 1 + R! R! R 1 + R 6

stands for the sum of row # and times row #1. The right-hand side, R, means row # is updated by the result of row operation. The corresponding system is x 1 x + x = 0 x x = x = 1 which is exactly the same as before. We can continue row operations 1 1 j 0 1 1 j 0 0 j 6 R /! R 0 j 0 0 j 1! 0 0 1 j 1. 1 1 j 0 1 1 j 0 R + R! R! 6 0 0 j 0 0 1 j 1 R /! R 6 0 1 0 j! 1 0 0 1 j 1 6 6 1 0 1 j 1 0 0 j 1 R + R 1! R 1! 6 0 1 0 j 1 0 0 1 j 1 ( 1) R 1 + 1 + R 1! R 1! 6 0 1 0 j 1 0 0 1 j 1 6 6 1 0 0 j 1 = 6 0 1 0 j 1 (Reduced Echelon Form) =) x 1 1 x = 6 0 0 1 j 1 x 1 1. Example 1.6. Solve x x = x 1 x + x = 1 x 1 x + x = 1 (10) Solution Write down the Matrix form and perform row operations 0 1 j j 1 R! R 1 j 1 0 1 j! j 1 j 1 R 1 /! R 1! 1 / 1 j 1/ 0 1 j j 1 ( 1/) R + R! R! ( ) R 1 + R! R! 1 / 1 j 1/ 0 1 j 0 0 0 j /. 1 / 1 j 1/ 0 1 j 0 1/ j / 1/

We convert back to the system > x 1 x + x = 1 x x = > 0 = =) impossible, means no solution. In general, one can always solve a system of linear equations through a series of elementary row reductions till the augmented matrix reduces to Upper Triangle form, and then solve unknowns from bottom up. A general process will be summarized in the next section. ² Homework 1 1. Solve the following linear systems directly by the method of elimination (a) (b) x 1 x + x = x 1 x + 9x = x 1 + 6x x = x 1 x + x = x 1 x + 9x = x 1 9x + 1x =. For the following linear systems, nd their coe cient matrices and augmented matrices. Then solve the system by performing elementary row operations. (a) (b) x 1 x + x + x = x 1 x + 9x x = x 1 + 6x x + x = x 1 x + x = x 1 x + 9x = x 1 9x + 1x = 0. For each of the following statements, determine whether it is true or false. If your answer is true, state your rationale. If false, provide an counterexample (the example contradicting the statement). (a) Every elementary row operation is reversible. (b) Elementary row operations on an augmented matrix never change the solution set of the associated linear system. (c) Two matrices are row equivalent if they have the same number of rows and columns. (d) Two linear systems are equivalent if they have the same solution set.