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Chapter 1 Systems of Linear Equations and Matrices •Introduction to System of Linear Equations • Gaussian Elimination • Matrices and Matrix Operations • Inverses; Rules of Matrix Arithmetic • Elementary Matrices and a Method for Finding A1 • Further Results on Systems of Equations and Invertibility • Diagonal, Triangular, and Symmetric Matrices 1.1 Introduction to systems of linear equations •Any straight line in xy-plane can be represented algebraically by an equation of the form: a1 x a2 y b •General form: Define a linear equation in the n variables x1 , x2 ,..., xn : a1 x1 a2 x2 ... an xn b where a1, a2, …, an and b are real constants. The variables in a linear equation are sometimes called unknowns. Linear Equations Examples: • Some linear equations: 3x 5 y 6 0,109 z 123x 100 y 98, x1 2 x2 4 x3 5x4 9 • Some NON linear equations: 2 x 6 y 0, 9 xz 3x 2 y, e x y 9, 3ln y x 4 z 8, cos x tan y 6 Observe that •A linear equation does not involve any products or roots of variables •All variables occur only to the first power and do not appear as arguments for trigonometric, logarithmic, or exponential functions. Solutions of Linear Equations •A solution of a linear equation is a sequence of n numbers s1, s2, …, sn such that the equation is satisfied. •The set of all solutions of the equation is called its solution set or general solution of the equation. Example: (a) Find the solution of 3x+4y = 5 Solution: We can assign arbitrary values to any one of the variables and solve for the other variable. In particular, we have x t, y 3t 5 4 where t is an arbitrary value, called a parameter. Example (b): Find the solution of x1 3x2 8 x3 16 Solution: We can assign arbitrary values to any two variables and solve for the third variable. In particular, x1 3s 8t 16, x2 s, x3 t where s, t are parameters Linear Systems A finite set of linear equations in the variables x1, x2, …, xn is called a system of linear equations or a linear system. A sequence of numbers s1, s2, …, sn is called a solution of the system if every equation is satisfied. A system has no solution is said to be inconsistent. If there is at least one solution of the system, it is called consistent. Every system of linear equations has either no solutions, exactly one solution, or infinitely many solutions. A general system of two linear equations: a1 x b1 y c1 (a1 , b1 not both zero) a2 x b2 y c2 (a2 , b2 not both zero) Two line may be parallel – no solution Two line may be intersect at only one point – one solution Two line may coincide – infinitely many solutions An arbitrary system of m linear equations in n unknown can be written as a11 x1 a12 x2 ... a1n xn b1 a21 x1 a22 x2 ... a2 n xn b2 ... am1 x1 am 2 x2 ... amn xn bm Where x1, x2, …, xn are unknowns and the subscripted a’s and b’s denote Constants. Note that aij is in the ith equation and multiplies unknown x j Augmented Matrix A system of m linear equations in n unknowns can be written as a rectangular array of numbers: a11 x1 a12 x2 ... a1n xn b1 a21 x1 a22 x2 ... a2 n xn b2 ... am1 x1 am 2 x2 ... amn xn bm a11 a 21 ... am1 a12 a22 ... ... a1n ... a2 n ... ... am 2 ... amn This is called the augmented matrix for the system. Example: 2 x1 3 x2 x3 5 6 x1 11x2 4 x3 0 5 x1 9 x2 2 xn 1 2 3 1 5 6 11 4 0 5 9 2 1 b1 b2 ... bm Elementary Row Operations Since the rows of an augmented matrix correspond to the equations, we can apply the following three types of operations to solve systems of linear equations. •Multiply an equation through by an nonzero constant • Interchange two equation • Add a multiple of one equation to another These operations are called elementary row operations. We will go to details in the next session. 1.2 Gaussian Elimination A matrix which has the following properties is in reduced row echelon Form. •If a row does not consist entirely of zeros, then the first nonzero number in the row is a 1. We call this a leader 1. •If there are any rows that consist entirely of zeros, then they are grouped together at the bottom of the matrix. •In any two successive rows that do not consist entirely of zeros, the leader 1 in the lower row occurs farther to the right than the leader 1 in the higher row. •Each column that contains a leader 1 has zeros everywhere else. A matrix that has the first three properties is said to be in row echelon form. Note: A matrix in reduced row-echelon form is of necessity in row echelon form, but not conversely Row-Echelon and Reduced Row-Echelon Forms Example: Determine which of the following matrices are in row-echelon form, reduced Row-echelon form, both, or neither. 1 0 0 1 0 0 0 0 1 8 1 0, 0 1 0 1 0 0 0 0 2 0 1 0 3 , 0 0 1 3 0 2 1 0,0 0 0 1 5 0 0 3 0 2 1 4 1 5 4 2 1 0 7 , 0 1 6 1 0 0 1 0 0 0 0 Solution: in row-echelon form: 1 0 0 1 8 2 1 0 0 2 1 5 4 2 , 0 1 6 1 0 1 0 , 0 1 0 , 0 1 0 3 0 0 1 0 0 0 0 0 1 3 0 0 0 0 in reduced row-echelon form: 1 0 0 1 0 0 2 0 1 0 , 0 1 0 3 0 0 1 0 0 1 3 Example: Determine which of the following matrices are in row-echelon form, reduced Row-echelon form, both, or neither. 1 0 0 1 0 0 0 0 1 8 1 0, 0 1 0 1 0 0 0 0 2 0 1 0 3 , 0 0 1 3 0 2 1 0,0 0 0 1 5 0 0 3 0 2 1 4 1 5 4 2 1 0 7 , 0 1 6 1 0 0 1 0 0 0 0 Solution: in both: 1 0 0 1 0 0 2 0 1 0 , 0 1 0 3 0 0 1 0 0 1 3 In neither: 1 0 0 0 1 5 4 0 3 0 , 0 1 0 7 0 2 1 0 0 0 1 More on Row-Echelon and Reduced Row-Echelon Form All matrices of the following types are in row-echelon form (any real numbers substituted for the *’s. ) : 1 0 0 0 * * * 1 1 * * 0 , 0 1 * 0 0 0 1 0 * * * 1 1 * * 0 , 0 1 * 0 0 0 0 0 0 * * * 0 1 * * , 0 0 0 0 0 0 0 0 0 1 * * * * * * * * 0 0 1 * * * * * * 0 0 0 1 * * * * * 0 0 0 0 1 * * * * 0 0 0 0 0 0 0 1 * All matrices of the following types are in reduced row-echelon form (any real numbers substituted for the *’s. ) : 1 0 0 0 0 0 0 1 1 0 0 0 , 0 1 0 0 0 0 1 0 0 0 * 1 1 0 * 0 , 0 1 * 0 0 0 0 0 0 0 * * 0 1 * * , 0 0 0 0 0 0 0 0 0 1 * 0 0 0 * * 0 * 0 0 1 0 0 * * 0 * 0 0 0 1 0 * * 0 * 0 0 0 0 1 * * 0 * 0 0 0 0 0 0 0 1 * Solutions of Linear Equations Example: Suppose that the augmented matrix for a system of linear equations has been reduced by row operations to the given reduced row-echelon form. Solve the system. 1 (a ) 0 0 1 0 (c ) 0 0 0 0 6 1 0 3 , 0 1 8 1 0 0 5 6 (b) 0 1 0 1 3 0 0 1 7 9 8 0 1 0 3 5 , 0 0 1 4 7 0 0 0 0 0 1 0 0 0 ( d ) 0 1 4 0 0 0 0 1 3 0 0 2 Solution (a): The corresponding system of equations is 6 x1 x2 3 x3 8 By inspection, x1 6, x2 3, x3 8 Example: Suppose that the augmented matrix for a system of linear equations has been reduced by row operations to the given reduced row-echelon form.Solve the system. 1 0 0 5 6 (b) 0 1 0 1 3 0 0 1 7 9 Solution (b): The corresponding system of equation is + 5x4 6 x1 Leading variables x4 3 x2 x3 free variables 7 x4 9 Solving for the leading variables in terms of the free variable gives x1 6 5x4 x2 3 x4 x3 9 7 x4 Thus, the general solution is x1 6 5t , x2 3 t , x3 9 7t , x4 t Example: Suppose that the augmented matrix for a system of linear equations has been reduced by row operations to the given reduced row-echelon form.Solve the system. 1 0 (c ) 0 0 3 0 0 0 0 1 0 0 0 0 1 0 2 8 3 5 4 7 0 0 Solution (c): The corresponding system of equation is + 2x5 8 x1 +3x2 +3x5 5 x3 x4 4 x5 7 Solving for the leading variables in terms of the free variable gives x1 8 3x2 2x5 x3 5 3x5 x4 7 4 x5 Thus, the general solution is x1 8 3s 2t , x2 s, x3 5 3t , x4 7 4t , x5 t Example: Suppose that the augmented matrix for a system of linear equations has been reduced by row operations to the given reduced row-echelon form.Solve the system. 1 0 0 0 (d ) 0 1 4 0 0 0 0 1 Solution (d): The corresponding system of equation is x1 0 x2 4 x3 0 0 1 Since 0=1 cannot be satisfied, there is no solution to the system Elimination Method Elimination procedure to reduce a matrix to reduced row-echelon form. 0 0 2 0 7 12 2 4 10 6 12 28 2 4 5 6 5 1 Step 1. Locate the leftmost column that does not consist entirely of zeros. 0 0 2 0 7 12 2 4 10 6 12 28 2 4 5 6 5 1 Step 2. Interchange the top row with another row, if necessary, to bring a nonzero entry to the top of the column found in step 1. R1 R2 2 4 10 6 12 28 0 0 2 0 7 12 2 4 5 6 5 1 Step 3. If the entry that is now at the top of the column found in step 1 is a, multiply The first row by 1/a in order to introduce a leading 1. 1 R1 2 1 2 5 3 6 14 0 0 2 0 7 12 2 4 5 6 5 1 Step 4. Add suitable multiples of the top row to the rows below so that all entries Below the leading 1 become zeros. R3 (2R1 ) 14 1 2 5 3 6 0 0 2 0 7 12 0 0 5 0 17 29 Step 5. Cover the top row in the matrix and begin again with Step 1 applied to the Submatrix that remains. Continue in this way until the entire matrix is in row-echelon form. 1 R2 2 6 14 1 2 5 3 0 0 1 0 7 / 2 6 0 0 5 0 17 29 6 14 1 2 5 3 0 0 1 0 7 / 2 6 0 0 0 0 1/ 2 1 R3 5R1 2R3 6 14 1 2 5 3 0 0 1 0 7 / 2 6 0 0 0 0 1 2 Step 6. Beginning with the last nonzero row and working upward, add suitable multiples of each row to the rows above to introduce zeros above the leading 1’s 6 14 1 2 5 3 0 0 1 0 7 / 2 6 0 0 0 0 1 2 R1 5R2 R2 7 R3 2 1 2 5 3 6 14 0 0 1 0 0 1 0 0 0 0 1 2 R1 6 R3 1 2 5 3 0 2 0 0 1 0 0 1 0 0 0 0 1 2 1 2 0 3 0 7 0 0 1 0 0 1 0 0 0 0 1 2 •Step1~Step5: the above procedure produces a row-echelon form and is called Gaussian elimination •Step1~Step6: the above procedure produces a reduced row-echelon form and is called Gaussian-Jordan elimination Every matrix has a unique reduced row-echelon form but a row-echelon form of a given matrix is not unique Example: Solve by Gauss-Jordan elimination x y 2z 9 2 x 4 y 3z 1 3x 6 y 5 z 0 1 1 2 9 Solution: The augmented matrix for the system is 2 4 3 1 3 6 5 0 1 1 2 9 2 4 3 1 R2 2R1 3 6 5 0 9 1 1 2 7 17 0 1 2 2 0 3 11 27 9 1 1 2 0 2 7 17 3 6 5 0 1 1 2 R3 3R2 0 1 7 2 1 0 0 2 R3 3R1 9 17 2 3 2 9 1 1 2 0 2 7 17 0 3 11 27 2R3 1 1 2 0 1 7 2 0 0 1 9 17 2 3 1 R2 2 1 1 2 0 1 7 2 0 0 1 R1 R2 9 7 17 R2 R3 2 2 3 1 1 2 9 R 2 R 3 0 1 0 2 1 0 0 1 3 1 1 0 3 0 1 0 2 0 0 1 3 1 0 0 1 0 1 0 2 0 0 1 3 x The corresponding system of equations is The solution is x=1, y=2, z=3. y 1 2 z 3 Homogeneous Linear Systems A system of linear equations is said to be homogeneous if the constant terms are all zero; that is, the system has the form: a11 x1 a12 x2 ... a1n xn 0 a21 x1 a22 x2 ... a2 n xn 0 ... am1 x1 am 2 x2 ... amn xn 0 Every homogeneous system of linear equation is consistent, since all such system have x1 = 0, x2 = 0, …, xn = 0 as a solution. This solution is called the trivial solution. If there are another solutions, they are called nontrivial solutions. There are only two possibilities for its solutions: • There is only the trivial solution • There are infinitely many solutions in addition to the trivial solution Example: Solve the homogeneous system of linear equations by GaussJordan elimination 2 x1 +2 x2 x3 + x5 0 x1 x2 2 x3 -3x 4 x5 0 x1 x2 2 x3 x5 0 x3 x4 x5 0 Solution: The augmented matrix is The augmented matrix is 2 2 1 0 1 1 1 2 3 1 1 1 2 0 1 0 0 1 1 1 1 0 0 0 1 0 1 0 0 1 0 0 0 0 1 0 0 0 0 0 1 1 0 0 0 0 0 0 The corresponding system of equations is + x5 0 x1 +x2 x3 x5 0 x4 Solve for the leading variables yields x1 x2 x5 x3 x5 x4 0 The general solution is x1 s t , x2 s, x3 t , x4 0, x5 t Note: the trivial solution is obtained when s = t = 0 0 Theorem 1.2.1 A homogeneous system of linear equations with more unknowns than equations has infinitely many solutions. Remark •This theorem applies only to homogeneous system! • A nonhomogeneous system with more unknowns than equations need not be consistent; however, if the system is consistent, it will have infinitely many solutions.