Chapter 1 Systems of Linear Equations and Matrices System of linear algebraic equations and their solution constitute one of the major topics studied in the course known as linear algebra. In the first section we shall introduce some basic terminology and discuss a method for solving such system. 1.1 Systems of Linear Equations Linear Equations Any straight line in the xy-plane can be represented algebraically by an equation of the form a 1 x+a 2 y = b where a 1, a 2, and b are real constants and a 1 and a 2 are not both zero. An equation of this form is called a linear equation in the variables x and y. More generally, we define a linear equation in the n variables x 1,x 2,...,x n to be one that can be expressed in the form a 1 x 1 +a 2 x 2 + +a n x n = b where a 1,a 2,...,a n, and b are real constants. The variables in a linear equation are sometimes called unknowns. For example, the equations 3x+y = 7, y = 1 5 x+2z +4, and x 1 +3x 2 2x 3 +5x 4 = 7 are linear. The equations x+3y = 2, 3x 2y 5z +yz = 4, and y = cosx are not linear. 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. A solution of a linear equation a 1 x 1 + a 2 x 2 + + a n x n = b is a sequence of n numbers s 1, s 2,..., s n such that the equation satisfied when we substitute x 1 = s 1, x 2 = s 2,..., x n = s n. The set of all solutions of the equation is called its solution set or sometimes the general solution of the equation. 1
MA131 (Section 750002): Prepared by Asst.Prof.Dr. Archara Pacheenburawana 2 Example 1. Find the solution set of (a) 5x 2y = 3 and (b) x 1 3x 2 +4x 3 = 2. Solution Linear Systems A finite set of linear equations in the variables x 1, x 2,..., x n is called system of linear equations or a linear system. A sequence of numbers s 1, s 2,..., s n is called a solution of the system if x 1 = s 1, x 2 = s 2,..., x n = s n is a solution of every equation in the system. For example, the system 4x 1 x 2 +3x 3 = 1 3x 1 +x 2 +9x 3 = 4 the the solution x 1 = 1, x 2 = 2, and x 3 = 1 since these values satisfy both equations. However, x 1 = 1, x 2 = 8, x 3 = 1 is not a solution since these values satisfy only the first equation in the system. Thus, not all system of linear equations have solutions. A system of equations that has no solutions is said to be inconsistent; if there is at least one solution of the system, it is called consistent. To illustrate the possibilities that can occur in solving systems of linear equations, consider a general system of two linear equations in the unknowns x and y: a 1 x+b 1 y = c 1 a 2 x+b 2 y = c 2 (a 1,b 1 not both zero) (a 2,b 2 not both zero) The graphs of these equations are lines; call them l 1 and l 2. Since a point (x,y) lies on a line if and only if the number x and y satisfy the equation of the line, the solutions of the system of equations correspond to points of the intersection of l 1 and l 2. There are three possibilities, illustrated in Figure 1.1
MA131 (Section 750002): Prepared by Asst.Prof.Dr. Archara Pacheenburawana 3 y y y l 1 l 2 l 1 l 2 l 1 and l 2 x x x (a) No solution (b) One solution Figure 1.1: (c) Infinitely many solutions The lines l 1 and l 2 may be parallel, in which case there is no intersection and consequently no solution to the system. The lines l 1 and l 2 may intersect at only one point, in which case the system has exactly one solution. The lines l 1 and l 2 may coincide, in which case there are infinitely many points of intersection and consequently infinitely many solutions to the system. Although we have considered only two equations with two unknowns here, we will show later that the same three possibilities hold for arbitrary linear systems: Every system of linear equations has no solutions, or has exactly one solution, or has infinitely many solutions. An arbitrary system of m linear equations in n unknowns can be written as a 11 x 1 +a 12 x 2 + +a 1n x n = b 1 a 21 x 1 +a 22 x 2 + +a 2n x n = b 2 (1.1).... a m1 x 1 +a m2 x 2 + +a mn x n = b m where x 1, x 2,..., x n are the unknowns and the subscripted a s and b s denote constants. For example, a general system of three linear equations in four unknowns can be written as a 11 x 1 +a 12 x 2 +a 13 x 3 +a 14 x 4 = b 1 a 21 x 1 +a 22 x 2 +a 23 x 3 +a 24 x 4 = b 2 a 31 x 1 +a 32 x 2 +a 33 x 3 +a 34 x 4 = b 3 The double subscripting on the coefficients of the unknowns is a useful device that is used to specify the location of the coefficient in the system. The first subscript on the coefficient a ij indicates the equation in which the coefficient occurs, and the second subscript indicates which unknown it multiplies. Thus, a 12 is in the first equation and multiplies unknown x 2.
MA131 (Section 750002): Prepared by Asst.Prof.Dr. Archara Pacheenburawana 4 Augmented Matrices A system of m linear equations in n unknowns can be abbreviated by writing only the rectangular array of numbers a 11 a 12 a 1n b 1 a 21 a 22 a 2n b 2..... a m1 a m2 a mn b m This is called the augmented matrix for the system. For example, the augmented matrix for the system of equations is x 1 +x 2 +2x 3 = 9 2x 1 +4x 2 3x 3 = 1 3x 1 +6x 2 5x 3 = 0 1 1 2 9 2 4 3 1 3 6 5 0 Remark When constructing an augmented matrix, we must write the unknowns in the same order in each equation, and the constants must be on the right. Exercise 1.1 1. Which of the following are linear equations in x 1, x 2 and x 3? (a) x 1 +5x 2 2x 3 = 1 (b) x 1 +3x 2 +x 1 x 3 = 2 (c) x 1 = 7x 2 +3x 3. (d) x 2 1 +x 2 +8x 3 = 5 (e) x 3/5 1 2x 2 +x 3 = 4 (f) πx 1 2x 2 + 1 3 x 3 = 7 1/3 2. Find the solution set of each of the following linear systems. (a) 7x 5y = 3 (b) 3x 1 5x 2 +4x 3 = 7 (c) 8x 1 +2x 2 5x 3 +6x 4 = 1 (d) 3v 8w+2x y +4z = 0 3. Which of the following are linear systems? (a) x 1 3x 2 = x 3 4 x 4 = 1 x 1 x 1 +x 4 +x 3 2 = 0 (c) 3x xy = 1 x+2xy y = 0 (e) 2x 1 sinx 2 = 3 x 2 = x 1 +x 3 x 1 +x 2 3x 3 = 0 (b) 2x y +3z = 1 x + 2y z = 2 4x y = 1 (d) y = 2x 1 y = x (f) x 1 + x 2 +x 3 = 1 2x 2 1 2x 3 = 1 3x 2 x 3 = 2
MA131 (Section 750002): Prepared by Asst.Prof.Dr. Archara Pacheenburawana 5 4. Find the augmented matrix for each of the following systems of linear equations. (a) 3x 1 2x 2 = 1 4x 1 +5x 2 = 3 7x 1 +3x 2 = 2 (c) x 1 = 1 x 2 = 2 x 3 = 3 (b) 2x 1 +2x 3 = 1 3x 1 x 2 +4x 3 = 7 6x 1 +x 2 x 3 = 0 (d) x 1 + 2x 2 x 4 + x 5 = 1 3x 2 + x 3 x 5 = 2 x 3 + 7x 4 = 1 5. Find a system of linear equations corresponding to the augmented matrix. (a) 2 0 0 3 4 0 0 1 1 (b) 3 0 2 5 7 1 4 3 0 2 1 7 (c) [ 7 2 1 3 5 1 2 4 0 1 ] (d) 1 0 0 0 7 0 1 0 0 2 0 0 1 0 3 0 0 0 1 4 6. (a) Find a linear equation in the variables x and y that has the general solution (b) Show that x = 5+2t, y = t. x = t, y = 1 2 t 5 2 is also the general solution of the equation in part (a). 7. Consider the system of equations ax+by = k cx+dy = l ex+fy = m Indicate what we can say about the relative positions of the line ax + by = k, cx+dy = l, and ex+fy = m when (a) the system has no solutions. (b) the system has exactly one solution. (c) the system has infinitely many solutions. 8. Show that the linear system ax+by = e cx+dy = f where a,b,c,d,e,f are constants, has a unique solution if ad bc 0. Express this solution in terms of a,b,c,d,e, and f.
MA131 (Section 750002): Prepared by Asst.Prof.Dr. Archara Pacheenburawana 6 1. (a), (c), (f) Answer to Exercise 1.1 2. (a) x = 3 7 + 5 7 t, y = t (b) x 1 = 5 3 s 4 3 t 1 8, x 2 = s, x 3 = t (c) x 1 = 1 4 r 5 8 s+ 3 4 t 1 8, x 2 = r, x 3 = s, x 4 = t (d) v = 8 q 2 r+ 1 s 4 t, w = q, x = r, y = s, z = t 3 3 3 3 3. (a), (d) 3 2 1 2 0 2 1 4. (a) 4 5 3 (b) 3 1 4 7 (c) 7 3 2 6 1 1 0 1 2 0 1 1 1 (d) 0 3 1 0 1 2 0 0 1 7 0 1 5. (a) 2x 1 = 0 3x 1 4x 2 = 0 x 2 = 1 (b) 3x 1 2x 3 = 5 7x 1 +x 2 +4x 3 = 3 2x 2 +x 3 = 7 (c) 7x 1 +2x 2 +x 3 3x 4 = 5 x 1 +2x 2 +4x 3 = 1 (d) x 1 = 7 x 2 = 2 x 3 = 3 x 4 = 4 1 0 0 1 0 1 0 2 0 0 1 3 6. (a) x 2y = 5 (b) Let x = t; then t 2y = 5. Solving for y yields y = 1 2 t 5 2. 7. (a) The line have no common point of intersection. (b) The line intersect in exactly one point. (c) The three lines coincide. 1.2 Elementary Operations Elementary Row Operations The basic method for solving a system of linear equations is to replace the given system by a new system that has the same solution set but is easier to solve. This new system is generally obtained in a series of steps by applying the following three types of operations to eliminate unknowns systematically: 1. Multiply any equation, say the ith, by a nonzero constant α. This is denoted by αr i.
MA131 (Section 750002): Prepared by Asst.Prof.Dr. Archara Pacheenburawana 7 2. Interchange of two equations, say the ith and jth. This is denoted by r i r j. 3. Add α times the ith equation to the jth equation. This is denoted by r j +αr i. (This keeps the i th equation unchanged.) Since the rows of an augmented matrix correspond to the equations in the associated system, these three operations correspond to the following operations on the rows of the augmented matrix: 1. Multiply any row by a nonzero constant: αr i. 2. Interchange of two rows: r i r j. 3. Add a multiple of one row to another: r j +αr i. These are called elementary row operations. The following illustrates how these operations can be used to solve system of linear equations. Example 2. Solve the following system by using elementary row operations. x+ y +2z = 9 2x+4y 3z = 1 3x+6y 5z = 0 Solution In the left column we solve a system of linear equations by operating on the equations in the system, and in the right column we solve the same system by operating on the rows of the augmented matrix. x+ y +2z = 9 2x+4y 3z = 1 3x+6y 5z = 0 1 1 2 9 2 4 3 1 3 6 5 0 Add 2 times the first equation to the second to obtain x+ y +2z = 9 2y 7z = 17 3x+6y 5z = 0 Add 3 times the first equation to the third to obtain x+ y + 2z = 9 2y 7z = 17 3y 11z = 27 Add 2 times the first row to the second to obtain 1 1 2 9 0 2 7 17 3 6 5 0 Add 3 times the first row to the third to obtain 1 1 2 9 0 2 7 17 0 3 11 27
MA131 (Section 750002): Prepared by Asst.Prof.Dr. Archara Pacheenburawana 8 Multiply the second equation Multiply by 1 to obtain 2 x+ y + 2z = 9 y 7 z = 2 17 2 3y 11z = 27 the second row by 1 to obtain 2 1 1 2 9 0 1 7 17 2 2 0 3 11 27 Add 3 times the second equation to the third to obtain x+ y +2z = 9 y 7 2 z = 17 2 1 2 z = 3 2 Multiply the third equation by 2 to obtain x+ y +2z = 9 y 7 2 z = 17 2 z = 3 Add 1 times the second equation to the first to obtain x + 11 2 z = 35 2 y 7 2 z = 17 2 z = 3 Add 11 times the third equation 2 to the first and 7 times the third 2 equation to the second to obtain x = 1 y = 2 z = 3 Add 3 times the second row to the third to obtain 1 1 2 9 0 1 7 2 17 2 0 0 1 2 3 2 Multiply the third row by 2 to obtain 1 1 2 9 0 1 7 17 2 2 0 0 1 3 Add 1 times the second row to the first to obtain 11 35 1 0 2 2 0 1 7 17 2 2 0 0 1 3 Add 11 times the third row 2 to the first and 7 times the third 2 row to the second to obtain 1 0 0 1 0 1 0 2 0 0 1 3 The solution x = 1, y = 2, z = 3 is now evident. Echelon Forms In previous example, we solved a linear system in the unknowns x, y, and z by reducing the augmented matrix to the form 1 0 0 1 0 1 0 2. 0 0 1 3 This is an example of a matrix that is in reduced row-echelon form. To be of this form, a matrix must have the following properties: 1. If the row does not consist entirely of zeros, then the first nonzero number in the row is a 1. We call this a leading 1.
MA131 (Section 750002): Prepared by Asst.Prof.Dr. Archara Pacheenburawana 9 2. If there are any rows that consist entirely of zeros, then they are grouped together at the bottom of the matrix. 3. In any two successive rows that do not consist entirely of zeros, the leading 1 in the lower row occurs farther to the right than the leading 1 in the higher row. 4. Each column that contains a leading 1 has zeros everywhere else in that column. A matrix that has the first three properties is said to be in row-echelon matrix. (Thus, a matrix in reduced row-echelon form is of necessity in row-echelon form, but not conversely.) The following matrices are in reduced row-echelon form. 0 1 2 0 0 1 0 0 4 1 0 0 [ ] 0 1 0 7, 0 1 0, 0 0 0 1 3 0 0 0 0 0 0 0, 0 0 0 0 1 1 0 0 1 0 0 0 0 0 The following matrices are in row-echelon form. 1 4 3 7 1 0 0 0 1 6 2, 0 1 0, 0 0 1 1 0 0 0 0 1 2 6 0 0 0 1 1 0 0 0 0 0 1 Example 3. Suppose that the augmented matrix for a linear equations has been reduced by row operations to the given reduced row-echelon form. Solve the system. 1 0 0 2 1 0 0 3 5 (a) 0 1 0 4 (b) 0 1 0 4 1 0 0 1 3 0 0 1 2 6 (c) 1 0 2 3 0 1 1 4 0 0 0 0 (d) 1 0 4 0 0 1 5 0 0 0 0 1 Solution
MA131 (Section 750002): Prepared by Asst.Prof.Dr. Archara Pacheenburawana 10 Elimination Methods We have just seen how easy it is to solve a system of linear equations once its augmented matrix is in reduced row-echelon form. The process of using row operations to transform a linear system into one whose augmented matrix is in row-echelon form is called Gaussian elimination. The process of using row operations to transform a linear system into one whose augmented matrix is in reduced row-echelon form is called Gauss-Jordan elimination. Remark It can be show that every matrix has a unique reduced row-echelon form. Example 4. Solve by Gauss-Jordan elimination. x 1 +3x 2 2x 3 +2x 5 = 0 2x 1 +6x 2 5x 3 2x 4 +4x 5 3x 6 = 1 5x 3 +10x 4 +15x 6 = 5 2x 1 +6x 2 +8x 4 +4x 5 +18x 6 = 6 Solution
MA131 (Section 750002): Prepared by Asst.Prof.Dr. Archara Pacheenburawana 11 Back-Substitution It is sometimes preferable to solve a system of linear equations by using Gaussian elimination to bring the augmented matrix into row-echelon form without continuing all the way to the reduced row-echelon form. When this is done, the corresponding system of equations can be solved by a technique called back-substitution. The next example illustrates the idea. Example 5. From Example 4, solve by Gaussian elimination and back-substitution. Solution
MA131 (Section 750002): Prepared by Asst.Prof.Dr. Archara Pacheenburawana 12 Example 6. Solve x+y +2z = 9 2x+4y 3z = 1 3x+6y 5z = 0 by Gaussian elimination and back-substitution. Solution
MA131 (Section 750002): Prepared by Asst.Prof.Dr. Archara Pacheenburawana 13 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 a 11 x 1 +a 12 x 2 + +a 1n x n = 0 a 21 x 1 +a 22 x 2 + +a 2n x n = 0.. a m1 x 1 +a m2 x 2 + +a mn x n = 0 Every homogeneous system of linear equations is consistent, since all such systems have x 1 = 0, x 2 = 0,..., x n = 0 as a solution. This solution is called the trivial solution; if there are other solutions, they are called nontrivial solution. Because a homogeneous linear system always has the trivial solution, there are only two possibilities for its solutions: The system has only the trivial solution. The system has infinitely many solutions in addition to the trivial solution. In the special case of a homogeneous linear system of two equations in two unknowns, say a 1 x+b 1 y = 0 (a 1,b 1 not both zero) a 2 x+b 2 y = 0 (a 2,b 2 not both zero) the graphs of the equations are lines through the origin, and the trivial solution corresponds to the point of intersection at the origin. y.. y a 1 x+b 1 y = 0 x a 2 x+b 2 y = 0 (a) Only the trivial solution x a 1 x+b 1 y = 0 and a 2 x+b 2 y = 0 (b) Infinitely many solutions There is one case in which a homogeneous system is assured of having nontrivial solution namely, whenever the system involves more unknowns than equations. To see why, consider the following example of four equations in five unknowns.
MA131 (Section 750002): Prepared by Asst.Prof.Dr. Archara Pacheenburawana 14 Example 7. Solve the following homogeneous system of linear equations by using Gauss- Jordan elimination. 2x 1 +2x 2 x 3 +x 5 = 0 x 1 x 2 +2x 3 3x 4 +x 5 = 0 x 1 +x 2 2x 3 x 5 = 0 x 3 +x 4 +x 5 = 0 Solution
MA131 (Section 750002): Prepared by Asst.Prof.Dr. Archara Pacheenburawana 15 Theorem 1.1. A homogeneous system of linear equations with more unknowns than equations has infinitely many solutions. Remark Note that Theorem 1.1 applied only to homogeneous systems. A nonhomogeneous system with more unknowns than equations need not be consistent; however, if the system is consistent, it will have infinitely many solutions. Exercise 1.2 1. Which of the following 3 3 matrices are in reduced row-echelon form? 1 0 0 (a) 0 1 0 0 0 1 1 0 0 (e) 0 0 0 0 0 1 0 1 0 (b) 1 0 0 0 0 0 1 1 0 (f) 0 1 0 0 0 0 0 1 0 (c) 0 0 1 0 0 0 1 0 2 (g) 0 1 3 0 0 0 1 0 0 (d) 0 0 1 0 0 0 0 0 0 (h) 0 0 0 0 0 0 2. Which of the following 3 3 matrices are in row-echelon form? 1 0 0 (a) 0 1 0 0 0 1 1 4 6 (e) 0 0 1 0 1 3 1 3 0 (b) 0 0 1 0 0 0 1 1 0 (f) 0 1 0 0 0 0 1 1 1 (c) 0 1 2 0 0 3 1 3 4 (g) 0 0 1 0 0 0 1 0 0 (d) 0 1 0 0 2 0 1 2 3 (h) 0 0 0 0 0 1 3. In each part determine whether the matrix is in row-echelon form, reduced rowechelon form, both, or neither. 0 1 3 5 7 1 0 0 5 [ ] (a) 0 0 1 2 3 (b) 0 0 1 3 1 0 3 1 (c) 0 1 2 4 0 0 0 0 0 0 1 0 4 (d) [ ] 1 7 5 5 0 1 3 2 1 0 0 1 2 (e) 0 1 0 2 4 0 0 1 3 6 0 1 (f) 0 0 0 0 4. Find the reduced row-echelon form of A when 2 2 1 0 1 0 A = 1 1 2 3 1 0 1 1 2 0 1 0. 0 0 1 1 1 0
MA131 (Section 750002): Prepared by Asst.Prof.Dr. Archara Pacheenburawana 16 5. Find the reduced row-echelon form of B when 0 0 2 0 7 12 B = 2 4 10 6 12 28. 2 4 5 6 5 1 6. In each part 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. (a) (c) (e) 1 0 0 3 0 1 0 0 0 0 1 7 1 3 0 2 0 0 1 2 0 0 0 0 1 6 0 0 3 2 0 0 1 0 4 7 0 0 0 1 5 8 0 0 0 0 0 0 (b) (d) 1 0 0 7 8 0 1 0 3 2 0 0 1 1 5 [ 1 2 0 1 5 0 0 1 3 4 (f) 1 3 0 0 0 0 1 0 0 0 0 1 7. Solve each of the following systems by Gauss-Jordan elimination. ] (a) x 1 + x 2 +2x 3 = 8 x 1 2x 2 +3x 3 = 1 3x 1 7x 2 +4x 3 = 10 (c) 2x 1 3x 2 = 2 2x 1 + x 2 = 1 3x 1 +2x 2 = 1 (e) 4x 1 8x 2 = 12 3x 1 6x 2 = 9 2x 1 +4x 2 = 6 (b) 2x 1 +2x 2 +2x 3 = 0 2x 1 +5x 2 +2x 3 = 1 8x 1 + x 2 +4x 3 = 1 (d) 2b+3c = 1 3a+6b 3c = 2 6a+6b+3c = 5 (f) x 1 +3x 2 + x 3 + x 4 = 3 2x 1 2x 2 + x 3 +2x 4 = 8 3x 1 + x 2 +2x 3 x 4 = 1 (g) x y +2z w = 1 2x+ y 2z 2w = 2 x+2y 4z + w = 1 3x 3w = 3 (h) 3x 1 +2x 2 x 3 = 15 5x 1 +3x 2 +2x 3 = 0 3x 1 + x 2 +3x 3 = 11 6x 1 4x 2 +2x 3 = 30 (i) x 1 + x 2 +x 3 + x 4 = 0 2x 1 + x 2 x 3 +3x 4 = 0 x 1 2x 2 +x 3 + x 4 = 0 (j) 10y 4z + w = 1 x+ 4y z + w = 2 3x+2y + z +2w = 5 2x 8y +2z 2w = 4 x 6y +3z = 1 8. Solve each of the systems in Exercise 2 by Gaussian elimination. 9. Determine which of the following homogeneous systems have nontrivial solutions.
MA131 (Section 750002): Prepared by Asst.Prof.Dr. Archara Pacheenburawana 17 (a) 2x 1 3x 2 +4x 3 x 4 = 0 7x 1 + x 2 8x 3 +9x 4 = 0 2x 1 +8x 2 + x 3 x 4 = 0 (b) x 1 +3x 2 x 3 = 0 x 2 8x 3 = 0 4x 3 = 0 (c) a 11 x 1 +a 12 x 2 +a 13 x 3 = 0 a 21 x 1 +a 22 x 2 +a 23 x 3 = 0 (d) 3x 1 2x 2 = 0 6x 1 4x 2 = 0 10. Solve the following homogeneous systems of linear equations by any method. (a) 2x 1 +x 2 +3x 3 = 0 x 1 +2x 2 = 0 x 2 + x 3 = 0 (c) x 1 +x 2 x 3 +3x 4 = 0 3x 1 +x 2 x 3 x 4 = 0 2x 1 x 2 2x 3 x 4 = 0 (e) 2x+2y +4z = 0 w y 3z = 0 2w+3x+ y + z = 0 2w + x+3y 2z = 0 (b) 3x 1 +x 2 +x 3 +x 4 = 0 5x 1 x 2 +x 3 x 4 = 0 (d) 1x 2 1 + 1x 3 2 + 1x 2 3 = 0 1 x 4 1 2x 3 2 + 1x 4 3 = 0 1 x 4 1 + 1x 3 2 3x 4 3 = 0 (f) 2x y 3z = 0 x+2y 3z = 0 x+ y +4z = 0 11. Consider a linear system whose augmented matrix is of the form 1 2 1 1 1 4 3 2 2 2 α 3 For what values of α will the system have a unique solution? 12. Consider a linear system whose augmented matrix is of the form 1 2 1 0 2 5 3 0 1 1 β 0 (a) Is it possible for the system to be consistent? Explain. (b) For what values of β will the system have infinitely many solutions? 13. Consider a linear system whose augmented matrix is of the form 1 1 3 2 1 2 4 3 1 3 α β (a) For what values of α and β will the system have infinitely many solutions? (b) For what values of α and β will the system be inconsistent?
MA131 (Section 750002): Prepared by Asst.Prof.Dr. Archara Pacheenburawana 18 14. Solve the system 2x 1 x 2 = λx 1 2x 1 x 2 + x 3 = λx 2 2x 1 +2x 2 +x 3 = λx 3 for x 1, x 2, and x 3 in the two cases λ = 1, λ = 2. Answer to Exercise 1.2 1. (a), (b), (d), (g), (h) 2. (a), (b), (c), (f), (g) 3. (a) Row-echelon (b) Neither (c) Both (d) Row-echelon (e) Both (f) Both 1 1 0 0 1 0 4. 0 0 1 0 1 0 1 2 0 3 0 7 0 0 0 1 0 0 5. 0 0 1 0 0 1 0 0 0 0 1 2 0 0 0 0 0 0 6. (a) x 1 = 3, x 2 = 0, x 3 = 7 (b) x 1 = 7t+8, x 2 = 3t+2, x 3 = t 5, x 4 = t (c) x 1 = 2+3t, x 2 = t, x 3 = 2 (d) x 1 = 5 2s t, x 2 = s, x 3 = 4 3t, x 4 = t (e) x 1 = 6s 3t 2, x 2 = s, x 3 = 4t+7, x 4 = 5t+8, x 5 = t (f) Inconsistent 7. (a) x 1 = 3, x 2 = 1, x 3 = 2 (b) x 1 = 1 7 3 7 t, x 2 = 1 7 4 7 t, x 3 = t (c) Inconsistent (d) Inconsistent (e) x 1 = 3+2t, x 2 = t (f) x 1 = 3 4 5 8 t, x 2 = 1 4 1 8 t, x 3 = t, x 4 = 3 (g) x = t 1, y = 2s, z = s, w = t (h) x 1 = 4, x 2 = 2, x 3 = 7 (i) x 1 = 4 3 t, x 2 = 0, x 3 = 1 3 t, x 4 = t (j) x = 14 6 t, y = 7 + 3 t 1 s, z = 2+t, w = t 5 5 10 10 10 9. (a), (c), (d) 10. (a) x 1 = 0, x 2 = 0, x 3 = 0 (b) x 1 = s, x 2 = t s, x 3 = 4s, x 4 = t (c) x 1 = t, x 2 = t, x 3 = t (d) x 1 = 5 3 t, x 2 = t, x 3 = t (e) w = t, x = t, y = t, z = 0 (f) x = 0, y = 0, z = 0 11. α 2 12. β = 2 13. (a) α = 5, β = 4, (b) α = 5, β 4 14. If λ = 1, then x 1 = x 2 = s, x 3 = 0. If λ = 2, then x 1 = 1 2 s, x 2 = 0, x 3 = s 1.3 Matrices and Matrix Operations In this section we begin our study of matrix theory by giving some of the fundamental definitions of the subject. We shall see how matrices can be combined through the arithmetic operations of addition, subtraction, and multiplication.
MA131 (Section 750002): Prepared by Asst.Prof.Dr. Archara Pacheenburawana 19 Matrix Notation and Terminology Definition 1.1. A matrix is a rectangular array of numbers, each of which is called an entry of the matrix. For example, and 2 1 1 2 (1.2) 0 5 ( ) 1 1 0 (1.3) 0 5 0 [ 1 2 π ] (1.4) Each horizontal line of number is called a row of the matrix; each vertical one is called a column. For example, in matrix (1.2) the third row consist of the entries 0 and 5, whereas the entries in the first column are 2, 1, and 0. If the given matrix has m rows and n columns, it is called m n matrix (read, m by n matrix), and the numbers m and n are the sizes of the matrix. The matrices (1.2), (1.3), and (1.4), are respectively, 3 2, 2 3, and 1 3 matrices. A matrix with only one column or m 1 matrix (with m > 1) is called a column matrix (or a column vector), and a matrix with only one row or 1 n matrix (with n > 1) is called a row matrix (or a row vector). For example, the matrix (1.4) is called a row matrix or a row vector. Capital (uppercase) letters, such as A, B, and C, will be used to denote matrices, whereas the entries of a matrix will be denoted by the corresponding lowercase letter with double subscripts. The entry that occurs in row i and column j of a matrix A will be denoted by a ij. Thus, a general 2 4 matrix might be written as [ ] a11 a A = 12 a 13 a 14 a 21 a 22 a 23 a 24 and a general m n matrix as a 11 a 12 a 1n a 21 a 22 a 2n A =.... a m1 a m2 a mn When compactness of notation is desired, the preceding matrix can be written as [a ij ] m n and [a ij ] the first notation being used when it is important in the discussion to know the size, and the second being used when the size need not be emphasized.
MA131 (Section 750002): Prepared by Asst.Prof.Dr. Archara Pacheenburawana 20 The entry in row i and column j of a matrix A is also commonly denoted by the symbol (A) ij. Thus, for the matrix 3 6 1 A = 4 2 0 1 3 5 we have (A) 12 = a 12 = 6, (A) 21 = a 21 = 4, and (A) 33 = a 33 = 5. Row and column matrices are of special importance, and it is common practice to denote them by boldface lowercase letters rather than capital letters. For such matrices, double subscripting of the entries is unnecessary. Thus a general 1 n row matrix a and a general m 1 column matrix b would be written as a = [ a 1 a 2 a n ] b 1 b 2 and b =. A matrix A with n rows and n columns is called a square matrix of order n, and the entries a 11, a 22,..., a nn in (1.5) are said to be on the main diagonal of A. a 11 a 12 a 1n a 21 a 22 a 2n.... (1.5) a n1 a n2 a nn b n Definition 1.2. A square matrix in which all the entries above the main diagonal are zero is called lower triangular, and a square matrix in which all the entries below the main diagonal are zero is called upper triangular. A matrix that is either upper triangular or lower triangular is called triangular. Operations on Matrices Definition 1.3. Given matrices A and B, the entries a ij and b kp are corresponding entries if and only if i = k and j = p. In other words, an entry of A corresponds to an entry of B if and only if they occupy the same position in their respective matrices. For example, if A = [ ] 1 2 3 4 5 6 and B = [ ] 7 8 9 1 2 3
MA131 (Section 750002): Prepared by Asst.Prof.Dr. Archara Pacheenburawana 21 then 1 and 7, 2 and 8, and 4 and 1 are all pairs of corresponding entries. We shall speak of corresponding entries only when the two matrices in question have the same size. Definition 1.4 (Equality of matrices). Two matrices A and B are equal, written A = B, if they have the same size and their corresponding entries are equal. In matrix notation, if A = [a ij ] and B = [b ij ] have the same size, then A = B if and only if a ij = b ij for all i and j. Example 8. Which of the following matrices are equal: [ ] 1 2 1 A =, B = 4 0 1+2 [ ] 1 4 1 C = 2, D = 4 0 3 [ ] 1 2, 4 0 [ ] 1 2 1? 4 0 1 Solution Definition 1.5 (Sum of matrices). Let A and B be matrices with the same size. The sum of A and B, written A+B, is the matrix obtained by adding corresponding entries of A and B Inmatrixnotation, ifa = [a ij ] andb = [b ij ]havethesamesize, thenc = A+B if and only if c ij = a ij +b ij for all i and j. Example 9. [ ] 2 1 5 Let A =, B = 3 0 4 and A+C, if defined [ ] [ ] 3 2 7 3 5, and C =. Find A + B 5 6 2 2 1 Solution
MA131 (Section 750002): Prepared by Asst.Prof.Dr. Archara Pacheenburawana 22 Definition 1.6 (Multiplication by a scalar). Let A be a matrix and c be a scalar. Then the product ca is the matrix obtained by multiplying each entry of A by scalar c. The matrix ca is said to be a scalar multiple of A. In matrix notation, if A = [a ij ], then B = ca if and only if b ij = ca ij for all i and j. Definition 1.7. The negative of a matrix B, written B, is the matrix ( 1)B. It is obtained from B by simply reversing the sign of every entry. Example 10. Let A = [ ] 1 1 0 3. Find 3A and ( 1)A. 2 6 4 8 Solution Definition 1.8 (Difference of matrices). Let A and B be matrices with the same size. The difference of A and B, written A B, is the matrix C given by C = A+( B). Note The matrix A B can be obtained by simply subtracting each entry of B from the corresponding entry of A. Example 11. 5 4 3 0 Find C = 2A B where A = 1 7 and B = 4 2. 9 3 5 7 Solution
MA131 (Section 750002): Prepared by Asst.Prof.Dr. Archara Pacheenburawana 23 If A 1,A 2,...,A n are matrices of the same size and c 1,c 2,...,c n are scalars, then an expression of the form c 1 A 1 +c 2 A 2 + +c n A n is called a linear combination of A 1,A 2,...,A n with coefficients c 1, c 2,..., c n. For example, If [ ] [ ] [ ] 2 3 4 0 2 7 9 6 3 A =, B =, and C =, 1 3 1 1 3 5 3 0 12 then 2A B + 1 [ ] 2 3 4 3 C = 2 1 3 1 [ ] 4 6 8 = + 2 6 2 [ ] 7 2 2 = 4 3 11 [ 0 2 7 +( 1) 1 3 5 ] [ 0 2 7 1 3 5 + ] + 1 [ ] 9 6 3 3 3 0 12 [ ] 3 2 1 1 0 4 is the linear combination of A, B, and C with scalar coefficients 2, 1, and 1 3. Definition 1.9 (Matrix multiplication). If A is an m r matrix and B is an r n matrix, then the product AB is the m n matrix whose entries are determined as follows. To find the entry in row i and column j of AB, single out row i from matrix A and column j from matrix B. Multiply the corresponding entries from the row and column together, and then add up the resulting products. In other words, if C = AB, then or, using summation notation, c ij = a i1 b 1j +a i2 b 2j + +a in b nj c ij = n a ik b kj. k=1 Example 12. Given the pairs of matrices A and B, find the products AB and BA, if defined. [ ] 4 1 4 3 1 2 4 (a) A =, B = 0 1 3 1 2 6 0 2 7 5 2 1 2 3 x (b) A = 4 5 6, B = y 7 8 9 z Solution
MA131 (Section 750002): Prepared by Asst.Prof.Dr. Archara Pacheenburawana 24 Note It should be apparent from Example 12 that multiplication of matrices act differently from multiplication of numbers. The latter has the property of commutativity (for all numbers a and b, ab = ba), but matrix multiplication does not. One situation in which the products AB and BA are both defined (although they still need not be equal) is when A and B have the same size and are square. Partitioned Matrices A matrix can be subdivided or partitioned into small matrices by inserting horizontal and vertical rules between selected rows and columns. For example, the following are three possible partitions of a general 3 4 matrix A: partition of A into four submatrices A 11, A 12, A 21, and A 22 a 11 a 12 a 13 a 14 [ ] A = a 21 a 22 a 23 a 24 A11 A = 12 A a 31 a 32 a 33 a 21 A 22 34 partition of A into its row matrices r 1, r 2, and r 3 a 11 a 12 a 13 a 14 r 1 A = a 21 a 22 a 23 a 24 = r 2 a 31 a 32 a 33 a 34 r 3 partition of A into its column matrices c 1, c 2, c 3, and c 4 a 11 a 12 a 13 a 14 A = a 21 a 22 a 23 a 24 = [ ] c 1 c 2 c 3 c 4 a 31 a 32 a 33 a 34 Matrix Multiplication by Columns and by Rows Sometimes it may be desirable to find a particular row or column of a matrix product AB without computing the entire product. The following results are useful for that purpose: ith row matrix of AB = [ ith row matrix of A ] B (1.6) jth column of AB = A [ jth column of B ] (1.7) Example 13. IfAandB arethematrices inexample 12(a), thenfindthefirstrowandsecond column of AB Solution
MA131 (Section 750002): Prepared by Asst.Prof.Dr. Archara Pacheenburawana 25 If a 1, a 2,..., a m denote the row matrices of A and b 1, b 2,..., b n denote the column matrices of B, then it follows from Formula (1.6) and (1.7) that a 1 a 1 B a 2 AB =. B = a 2 B. a m B a m (AB computed row by row) (1.8) AB = A [ b 1 b 2 b n ] = [ Ab1 Ab 2 Ab n ] (1.9) (AB computed column by column) Matrix Products as Linear Combinations Suppose that Then a 11 a 12 a 1n x 1 a 21 a 22 a 2n A =... and x = x 2. a m1 a m2 a mn Ax = a 11 x 1 +a 12 x 2 + +a 1n x n a 21 x 1 +a 22 x 2 + +a 2n x n... a m1 x 1 +a m2 x 2 + +a mn x n = x 1 a 11 a 21. a m1 +x 2 a 12 a 22. a m2 + +x n x n a 1n a 2n. a mn (1.10) In words, (1.10) tell us that the product Ax of the matrix A with a column matrix x is a linear combination of the column matrices of A with the coefficients coming from the matrix x. Moreover, the product ya of a 1 m matrix y with an m n matrix A is a linear combination of the row matrices of A with scalar coefficients coming from y.
MA131 (Section 750002): Prepared by Asst.Prof.Dr. Archara Pacheenburawana 26 Example 14. The matrix product 2 1 3 1 5 1 3 2 2 = 1 3 2 1 3 10 can be written as the linear combination of the column matrices 2 1 3 5 1 1 +2 3 +3 2 = 1 3 2 1 10 The matrix product [ ] 2 1 3 5 1 10 1 3 2 = [ 21 12 27 ] 3 2 1 can be written as the linear combination of the row matrices 5 [ 2 1 3 ] 1 [ 1 3 2 ] +10 [ 3 2 1 ] = [ 21 12 27 ] It follows from (1.8) and (1.10) that the jth column matrix of a product AB is a linear combination of the column matrices of A with the coefficients coming from the jth column of B.
MA131 (Section 750002): Prepared by Asst.Prof.Dr. Archara Pacheenburawana 27 Example 15. We showed in Example 12(a) that 3 2 [ ] 14 1 3 AB = 2 4 2 1 3 = 12 6 30 4 1 6 1 3 14 2 15 The column matrices of AB can be expressed as linear combinations of the column matrices of A as follows: 14 3 2 12 = 2 2 +4 4 14 1 3 1 3 2 6 = 1 2 +1 4 2 1 3 3 3 2 30 = 3 2 +6 4 15 1 3 Matrix Form of a Linear System Matrix multiplication has an important application to system of linear equations. Consider any system of m linear equations in n unknowns. a 11 x 1 +a 12 x 2 + +a 1n x n = b 1 a 21 x 1 +a 22 x 2 + +a 2n x n = b 2 (1.11).... a m1 x 1 +a m2 x 2 + +a mn x n = b m Since two matrices are equal if and only if their corresponding entries are equal, we can replace the m equations in this system by the single matrix equation a 11 x 1 +a 12 x 2 + +a 1n x n b 1 a 21 x 1 +a 22 x 2 + +a 2n x n... = b 2. a m1 x 1 +a m2 x 2 + +a mn x n b m The m 1 matrix on the left side of this equation can be written as a product to give a 11 a 12 a 1n x 1 b 1 a 21 a 22 a 2n x 2.... = b 2.. a m1 a m2 a mn x m b m
MA131 (Section 750002): Prepared by Asst.Prof.Dr. Archara Pacheenburawana 28 If we designate these matrices A, x, and b, respectively, then the original system of m equations in n unknowns has been replaced by the single matrix equation Ax = b The matrix A in this equation is called the coefficient matrix of the system. The augmented matrix for the system is obtained by adjoining b to A as the last column; thus the augmented matrix is Transpose of a Matrix [ ] A b = Definition 1.10 (Transpose of a matrix). a 11 a 12 a 1n b 1 a 21 a 22 a 2n b 2.... a m1 a m2 a mn b m The transpose of the m n matrix A, written A T, is the n m matrix B whose jth column is the jth row of A. Equivalently, B = A T if and only if b ij = a ji for each i and j.. Example 16. Find the transpose of [ ] 2 7 (a) A = 8 0 (b) B = [ ] 1 2 7 4 5 6 (c) C = [ 1 3 4 ] Solution Definition 1.11. If A is a square matrix, then trace of A, denoted by tr(a), is defined to be the sum of the entries on the main diagonal of A. The trace of A is undefined if A is not a square matrix.
MA131 (Section 750002): Prepared by Asst.Prof.Dr. Archara Pacheenburawana 29 Example 17. Find the trace of matrices A and B. 1 2 7 0 a 11 a 12 a 13 A = a 21 a 22 a 23, B = 3 5 8 4 1 2 7 3 a 31 a 32 a 33 4 2 1 0 Solution Exercise 1.3 1. Suppose that A,B,C,D, and E are matrices with the following dimensions: A B C D E (4 5) (4 5) (5 2) (4 2) (5 4) Determine which of the following matrix expression are defined. For those that are defined, give the dimension of the resulting matrix. (a) BA (b) AC +D (c) AE +B (d) AB +B (e) E(A+B) (f) E(AC) (g) E T A (h) (A T +E)D 2. If a matrix A is 3 5 and the product AB is 3 7, what is the size of B? 3. Solve the following matrix equation for a,b,c, and d [ ] [ ] a b b+c 8 1 = 3d+c 2a 4d 7 6 4. Consider the matrices 3 0 A = 1 2, B = 1 1 1 5 2 D = 1 0 1, E = 3 2 4 Compute the following (where possible). [ ] 4 1, C = 0 2 6 1 3 1 1 2 4 1 3 [ ] 1 4 2, 3 1 5 (a) 2A T +C (b) D T E T (c) (D E) T (d) B T +5C T (e) 1 2 CT 1 4 A (f) B BT (g) 2E T 3D T (h) (2E T 3D T ) T
MA131 (Section 750002): Prepared by Asst.Prof.Dr. Archara Pacheenburawana 30 5. Using the matrices in Exercise 4, compute the following (where possible). (a) AB (b) BA (c) (3E)D (d) (AB)C (e) A(BC) (f) CC T (g) (DA) T (h) (C T B)A T (i) tr(dd T ) (j) tr(4e T D) (k) tr(c T A T +2E T ) [ ] [ ] 1 2 1 2 1 6. Let A = and AB =. Find the first and second columns 2 5 6 9 3 of B. 7. Let Use the method of Example 13 to find 1 2 1 1 0 2 A = 3 3 5 and B = 2 1 1 2 4 1 5 4 1 (a) the first row of AB (c) the second column of AB (e) the third row of AA (b) the third row of AB (d) the first column of BA (f) the third column of AA 8. Let A and B be the matrices Exercise7. Use the method of Example 15 to (a) express each column matrix of AB as a linear combination of the column matrices of A (b) express each column matrix of BA as a linear combination of the column matrices of B 9. In each part, express the given system of linear equations as a single matrix equation Ax = b. (a) 3x 1 +2x 2 = 1 2x 1 3x 2 = 5 (b) x 1 + x 2 = 5 2x 1 + x 2 x 3 = 6 3x 1 2x 2 +2x 3 = 7 (c) 2x 1 + x 2 + x 3 = 4 x 1 x 2 +2x 3 = 2 3x 1 2x 2 x 3 = 0 (d) 4x 1 + 3x 3 + x 4 = 1 5x 1 + x 2 8x 4 = 3 2x 1 5x 2 +9x 3 x 4 = 0 3x 2 x 3 +7x 4 = 2 10. In each part, express the matrix equation as a system of linear equations. 3 2 0 1 w 0 3 1 2 x 1 2 (a) 4 3 7 = 1 (b) 5 0 2 2 x = 0 2 1 5 4 x 2 x 3 3 1 4 7 2 5 1 6 11. In each part, find a 6 6 matrix [a ij ] that satisfies the state condition. Make your answers as general as possible by using letters rather than specific numbers for the y z 0 0
MA131 (Section 750002): Prepared by Asst.Prof.Dr. Archara Pacheenburawana 31 nonzero entries. (a) a ij = 0 if i j (b) a ij = 0 if i > j (c) a ij = 0 if i < j (d) a ij = 0 if i j > 1 12. Find the matrix A such that x x+y A y = x y z 0 for all choices of x, y, and z? 13. Indicate whether the statement is always true or sometimes false. Justify your answer with a counterexample. (a) The expressions tr(aa T ) and tr(a T A) are always defined, regardless of the size of A. (b) tr(aa T ) = tr(a T A) for every matrix A. (c) If the first column of A has all zeros, then so does the first column of every product AB. (d) If the first row of A has all zeros, then so does the first row of every product AB. (e) If A is a square matrix with two identical rows, then AA has two identical rows. (f) If A is a square matrix and AA has a column of zeros, then A must have a column of zeros. (g) If B is an n n matrix whose entries are positive even integers, and if A is an n n matrix whose entries are positive integers, then the entries of AB and BA are positive even integers. (h) If the matrix sum AB +BA is defined, then A and B must be square. Answer to Exercise 1.3 1. (a) Undefined (b) 4 2 (c) Undefined (d) Undefined (e) 5 5 (f) 5 2 (g) Undefined (h) 5 2 2. 5 7 3. a = 5,b = 3,c = 4,d = 1 [ ] 5 0 1 5 0 1 7 2 4 4. (a) (b) 4 1 1 (c) 4 1 1 (d) Undefined 3 5 7 1 1 1 1 1 1 1 3 [ ] 9 1 1 9 13 0 4 2 (e) 9 0 0 1 (f) (g) 13 2 4 (h) 1 2 1 4 1 0 0 1 6 1 4 6 3 4 9 4
MA131 (Section 750002): Prepared by Asst.Prof.Dr. Archara Pacheenburawana 32 12 3 42 108 75 3 45 9 5. (a) 4 5 (b) Undefined (c) 12 3 21 (d) 11 11 17 4 1 36 78 63 7 17 13 3 45 9 [ ] [ ] 12 6 9 (e) 11 11 17 21 17 0 2 11 (f) (g) (h) 48 20 14 17 35 12 1 8 7 17 13 24 8 16 [ ] [ ] 7 8 (i) 61 (j) 35 (k) 28 6. b 1 =, b 4 2 = 5 7. (a) [ 10 6 5 ] (b) [ 15 8 9 ] 6 5 (c) 23 (d) 7 (e) [ 16 20 23 ] 12 (f) 23 8 19 23 10 1 2 1 8. (a) 34 = 1 3 +2 3 +5 5 15 2 4 1 6 1 2 1 23 = 0 3 +1 3 +4 5 8 2 4 1 5 1 2 1 14 = 2 3 +1 3 +1 5 9 2 4 1 5 1 0 2 (b) 7 = 1 2 +3 1 +2 1 19 5 4 1 10 1 0 2 11 = 2 2 +3 1 +4 1 26 5 4 1 3 1 0 2 8 = 1 2 +5 1 +1 1 26 5 4 1 9. (a) [ 3 2 2 3 ][ x1 x 2 ] = 2 1 1 (c) 1 1 2 3 2 1 [ ] 1 5 x 1 x 2 x 3 10. (a) 3x 1 x 2 +2x 3 = 2 4x 1 +3x 2 +7x 3 = 1 2x 1 + x 2 +5x 3 = 4 (b) 3w 2x + z = 0 5w +2y 2z = 0 1 1 0 x 1 5 (b) 2 1 1 x 2 = 6 3 2 2 x 3 7 4 0 3 1 4 = 2 (d) 5 1 0 8 0 2 5 9 1 0 3 1 7 x 1 x 2 x 3 x 4 1 = 3 0 2
MA131 (Section 750002): Prepared by Asst.Prof.Dr. Archara Pacheenburawana 33 3w + x+4y +7z = 0 2w +5x+ y +6z = 0 a 11 0 0 0 0 0 a 11 a 12 a 13 a 14 a 15 a 16 0 a 22 0 0 0 0 0 a 22 a 23 a 24 a 25 a 26 11. (a) 0 0 a 33 0 0 0 0 0 0 a 44 0 0 (b) 0 0 a 33 a 34 a 35 a 36 0 0 0 a 44 a 45 a 46 0 0 0 0 a 55 0 0 0 0 0 a 55 a 56 0 0 0 0 0 a 66 0 0 0 0 0 a 66 a 11 0 0 0 0 0 a 11 a 12 0 0 0 0 a 21 a 22 0 0 0 0 a 21 a 22 a 23 0 0 0 (c) a 31 a 32 a 33 0 0 0 a 41 a 42 a 43 a 44 0 0 (d) 0 a 32 a 33 a 34 0 0 0 0 a 43 a 44 a 45 0 a 51 a 52 a 53 a 54 a 55 0 0 0 0 a 54 a 55 a 56 a 61 a 62 a 63 a 64 a 65 a 66 0 0 0 0 a 65 a 66 1 1 0 12. 1 1 0 0 0 0 [ ] [ ] 0 2 1 2 13. (a) True (b) True (c) False; for example, A = and B = 0 4 3 4 [ ] 1 1 (d) True (e) True (f) False; for example, A = (g) True (h) True 1 1 1.4 Inverses; Rules of Matrix Arithmetic In this section we shall discuss some properties of the arithmetic operations on matrices. We shall see that many of the basic rules of arithmetic for real numbers also hold for matrices, but a few do not. Properties of Matrix Operations For real numbers a and b, we always have ab = ba, which is called the commutative law for multiplication. For matrices, however, AB and BA need not be equal. Example 18. Consider the matrices A = [ ] 1 0, B = 2 3 [ ] 1 2 3 4 Multiplying gives Thus, AB BA AB = [ ] [ ] 1 2 5 6, BA = 11 16 11 16
MA131 (Section 750002): Prepared by Asst.Prof.Dr. Archara Pacheenburawana 34 Theorem 1.2 (Properties of Matrices Arithmetic). Let A, B, and C be matrices and a and b be scalars. Assume that the sizes of the matrices are such that each operation is defined. Then (a) A+B = B +A (Commutative law for addition) (b) A+(B +C) = (A+B)+C (Associative law for addition) (c) A(BC) = (AB) C (d) A(B +C) = AB +AC (e) (B +C)A = BA+CA (f) A(B C) = AB AC (h) a(b +C) = ab +ac (j) (a+b)c = ac +bc (l) (ab)c = a(bc) (Associative law for multiplication) (Left distributive laws) (Right distributive laws) (g) (B C)A = BA CA (i) a(b C) = ab ac (k) (a b)c = ac bc (m) a(bc) = (ab)c = B(aC) Example 19. Let 1 0 1 1 2 3 2 1 1 A = 2 1 1, B = 0 1 0, C = 1 1 0 3 2 0 2 1 0 3 0 3 Find A(BC), (AB)C, A(B +C), and AB +AC. Solution
MA131 (Section 750002): Prepared by Asst.Prof.Dr. Archara Pacheenburawana 35 Zero Matrices Definition 1.12. The matrix analog of the number 0 is called a zero matrix. A zero matrix, denoted by 0, is a matrix of any sizes consisting of only zero entries. For example, the matrices 0 0 0 0 0 0, 0 0 0 [ ] 0 0 0, 0 0 0 0 0, and 0 [ 0 0 0 0 ] are all zero matrices. Since we already know that some of the rules of arithmetic for real number do not carry over to matrix arithmetic. For example, consider the following two standard results in the arithmetic of real numbers. If ab = ac and a 0, then b = c. (This is called the cancellation law.) If ad = 0, then at least one of the factors on the left is 0 As the next example shows, the corresponding results are not generally true in matrix arithmetic. Example 20. Consider the matrices [ ] 0 1 A =, B = 0 2 Then AB = AC = [ ] 1 1, C = 3 4 [ ] 3 4 6 8 [ ] 2 5, D = 3 4 AD = [ ] 0 0 0 0 [ ] 3 7 0 0 Thus, although A 0, it is incorrect to cancel the A from both sides of the equation AB = AC and write B = C. Also, AD = 0, yet A 0 and D 0. Thus, the cancellation law is not valid for matrix multiplication, and it is possible for a product of matrices to be zero without either factor being zero.
MA131 (Section 750002): Prepared by Asst.Prof.Dr. Archara Pacheenburawana 36 Theorem 1.3 (Properties of Zero Matrices). Assuming that the sizes of the matrices are such that the indicated operations can be performed, the following rules of matrix arithmetic are valid. (a) A+0 = 0+A = A (b) A A = 0 (c) 0 A = A (d) A0 = 0; 0A = 0 Identity Matrices Of special interest are square matrices with 1 s on the main diagonal and 0 s off the main diagonal, such as [ ] 1 0 0 0 1 0 0 1 0, 0 1 0, 0 1 0 0 0 1 0 0 1 0. 0 0 1 0 0 0 1 Amatrixofthisformiscalledanidentity matrix andisdenotedbyi. Ifitisimportant to emphasize the size, we shall write I n for the n n identity matrix. The identity matrix is so named because it has the property that AI n = A and I m A = A for any m n matrix A. Thus it is the matrix analog of the number 1, the multiplicative identity for numbers. Example 21. Consider the matrices 3 4 1 b 11 b 12 A = 2 6 3 B = b 21 b 22 0 1 8 b 31 b 32 Then and 1 0 0 3 4 1 3 4 1 I 3 A = 0 1 0 2 6 3 = 2 6 3 = A 0 0 1 0 1 8 0 1 8 b 11 b 12 [ ] b BI 2 = b 21 b 22 1 0 11 b 12 = b 0 1 21 b 22 = B b 31 b 32 b 31 b 32
MA131 (Section 750002): Prepared by Asst.Prof.Dr. Archara Pacheenburawana 37 Theorem 1.4. If R is the reduced row-echelon form of an n n matrix A, then either R has a row of zeros or R is the identity matrix I n. Definition 1.13. Let A be a square matrix. If there exist a matrix B such that AB = BA = I, we say that A is invertible (or nonsingular) and that B is the inverse of A. If A has no inverse, it is said to be noninvertible (or singular). Example 22. Show that the matrix B = [ ] 3 5 is an inverse of A = 1 2 [ ] 2 5. 1 3 Solution Properties of Inverses Theorem 1.5. If B and C are both inverses of the matrix A, then B = C. Asaconsequence oftheorem1.5, wecansaynowspeakof the inverse ofaninvertible matrix. If A is invertible, then its inverse will be denoted by the symbol A 1. Thus, AA 1 = I and A 1 A = I
MA131 (Section 750002): Prepared by Asst.Prof.Dr. Archara Pacheenburawana 38 Theorem 1.6. The matrix A = [ ] a b c d is invertible if ad bc 0, in which case the inverse is given by the formula A 1 = 1 ad bc [ ] d b = c a d ad bc c ad bc b ad bc a. ad bc Theorem 1.7. If A and B are invertible matrices of the same dimension, then AB is invertible and (AB) 1 = B 1 A 1 Theorem 1.8. IfA 1,A 2,...,A n areinvertible matricesofthesamedimension, thena 1 A 2 A n is invertible and (A 1 A 2 A n ) 1 = A 1 n A 1 n 1 A 1 2 A 1 1 Example 23. Let A = [ ] 7 3 and B = 5 2 [ ] 4 1. Show that (AB) 3 1 1 = B 1 A 1. Solution
MA131 (Section 750002): Prepared by Asst.Prof.Dr. Archara Pacheenburawana 39 Powers of a Matrix Definition 1.14. If A is a square matrix, then we define the nonnegative integer powers of A to be A 0 = I A n = AA A }{{} (n > 0) nfactors Moreover, if A is invertible, then we define the negative integer powers to be A n = (A 1 ) n = A 1 1 } A 1 A {{} nfactors Note In general, if A is a square matrix, and r and s are positive integers, we have A r A s = A r+s. It is also true that under the same condition, (A r ) s = A rs. Example 24. Let A = [ ] 3 2 and A 1 1 1 = [ ] 1 2. Find the matrices A 1 3 3 and A 3. Solution Theorem 1.9 (Laws of Exponents). If A is invertible matrix, then: (a) A 1 is invertible and (A 1 ) 1 = A. (b) A n is invertible and (A n ) 1 = (A 1 ) n for n = 0,1,2,... (c) Foranynonzeroscalark,thematrixkAisinvertibleand (ka) 1 = 1 k A 1.
MA131 (Section 750002): Prepared by Asst.Prof.Dr. Archara Pacheenburawana 40 Example 25. If ( ) 1 1 3 A = [ ] 2 1, then find A. 5 3 Solution Properties of the Transpose Theorem 1.10. If the sizes of the matrices are such that the state operations can be performed, then (a) (A T ) T = A (b) (A+B) T = A T +B T and (A B) T = A T B T (c) (ka) T = ka T, where k is any scalar (d) (AB) T = B T A T Invertibility of a Transpose Theorem 1.11. If A is an invertible matrix, then A T is also invertible and (A T ) 1 = (A 1 ) T. Example 26. Let A = [ ] 3 1. Show that (A 5 2 T ) 1 = (A 1 ) T. Solution
MA131 (Section 750002): Prepared by Asst.Prof.Dr. Archara Pacheenburawana 41 Exercise 1.4 1. Let Show that 1 2 0 6 0 2 A = 3 5 1, B = 4 1 1 2 3 4 3 8 5 4 5 3 C = 5 7 2, a = 3, b = 5 3 2 1 (a) A+(B +C) = (A+B)+C (c) (a+b)c = ac +bc (e) a(bc) = (ab)c = B(aC) (g) (B +C)A = BA+CA (i) (A T ) T = A (k) (ac) T = ac T (b) (AB)C = A(BC) (d) a(b C) = ab ac (f) A(B C) = AB AC (h) a(bc) = (ab)c (j) (A+B) T = A T +B T (l) (AB) T = B T A T 2. Let 0 denote a 2 2 matrix, each of whose entries is zero. Find (a) 2 2 matrix A such that A 0 and AA = 0 and (b) 2 2 matrix A such that A 0 and AA = A. 3. Use Theorem 1.6 to compute the inverses of the following matrices. [ ] 1 3 (a) A = 2 4 [ ] 4 5 (c) C = 5 6 [ ] 2 3 (b) B = 1 1 [ ] 3 7 (d) D = 6 13 4. Use the matrices A, B, and C in Exercise3 to verify that (a) (A 1 ) 1 = A (b) (B T ) 1 = (B 1 ) T (c) (AB) 1 = B 1 A 1 (d) (ABC) 1 = C 1 B 1 A 1 [ ] 1 1 5. Let A =. Find the power A 1 2 2 and A 3. 1 0 0 6. Let A = 2 1 0. Find the power A 2 and A 3. 3 2 1 [ ] 1 1 7. Let A =. Find a general formula for A 0 1 n.
MA131 (Section 750002): Prepared by Asst.Prof.Dr. Archara Pacheenburawana 42 8. In each part, use the given information to find A. [ ] [ ] 2 1 3 7 (a) A 1 = (b) (7A) 3 5 1 = 1 2 [ ] [ ] 3 1 1 2 (c) (5A T ) 1 = (d) (I +2A) 5 2 1 = 4 5 [ ] cosθ sinθ 9. Find the inverse of. sinθ cosθ [ 1 ] 10. Find the inverse of 2 (ex +e x 1 ) 2 (ex e x ) 1 2 (ex e x 1 ) 2 (ex +e x. ) 11. Consider the matrix a 11 0 0 0 a 22 0 A =... 0 0 a nn where a 11 a 22 a nn 0. Show that A is invertible and find its inverse. 12. Let A, B, and 0 be 2 2 matrices. Assuming that A is invertible, find a matrix C such that [ ] A 1 0 C A 1 is the inverse of the partitioned matrix [ A 0 B A 13. Use the result in Exercise 12 to find the inverses of the following matrices. ] 1 1 0 0 (a) 1 1 0 0 1 1 1 1 1 1 1 1 1 1 0 0 (b) 0 1 0 0 0 0 1 1 0 0 0 1 14. (a) Find a nonzero 3 3 matrix A such that A is symmetric. (b) Find a nonzero 3 3 matrix A such that A is skew-symmetric. 15. A square matrix A is called symmetric if A T = A and skew-symmetric if A T = A. Show that if B is a square matrix, then (a) BB T and B +B T are symmetric, (b) B B T is skew-symmetric. 16. Indicate whether the statement is always true or sometimes false. Justify your answer with a counterexample. (a) (AB) 2 = A 2 B 2, where A are B are n n matrices.
MA131 (Section 750002): Prepared by Asst.Prof.Dr. Archara Pacheenburawana 43 2. (a) (b) (A B) 2 = (B A) 2, where A are B are n n matrices. (c) (A+B)(A B) = A 2 B 2, where A are B are n n matrices. (d) (AB 1 )(BA 1 ) = I n, where A are B are n n matrices. (e) Let A and B be square matrices such that AB = 0. Show that if A is invertible, then B = 0. (f) If a square matrix A satisfies A 2 3A+I = 0, then A 1 = 3I A. (g) If A and B are m n matrices, then AB T and A T B are defined. (h) If AB = BA and if A is invertible, then A 1 B = BA 1. (i) If A T is singular, then A is singular. (j) A matrix with a row of zeros cannot have an inverse. (k) A matrix with a column of zeros cannot have an inverse. (l) If A is invertible and AB = AC, then B = C. (m) Let A be an n n matrix and let x and y be n 1 matrices. If Ax = Ay and x y, then A is singular. (n) If A is invertible and k is any nonzero scalar, then (ka) n = k n A n for all integer values of n. (o) Sum of two invertible matrices is invertible. (p) If AB is invertible, then B is invertible. [ ] [ ] a b x y (q) + is symmetric. b a y x [ ] 0 1 0 0 3. (a) A 1 = (d) D 1 = [ ] 1 0 (b) 0 0 ] [ 2 3 5 10 1 1 5 10 [ 13 3 7 3 2 1 ] Answer to Exercise 1.4 (b) B 1 = [ ] 1 3 1 2 (c) C 1 = [ ] [ ] 2 3 5 8 5. A 2 =, A 3 5 3 = 8 13 1 0 0 1 0 0 6. A 2 = 4 1 0, A 3 = 6 1 0 7. A n = 10 4 1 21 6 1 [ ] [ ] [ 5 1 2 1 13 13 7 2 8. (a) A = (b) A = (c) A = (d) A = 3 13 [ 9 2 13 1 13 13 2 13 6 13 ] 1 7 3 7 [ ] 1 n 0 1 1 5 1 5 3 5 ] [ ] 6 5 5 4
MA131 (Section 750002): Prepared by Asst.Prof.Dr. Archara Pacheenburawana 44 [ ] [ cosθ sinθ 1 ] 9. 10 2 (ex +e x ) 1 2 (ex e x ) sinθ cosθ 1 2 (ex e x 1 ) 2 (ex +e x ) 1 a 11 0 0 1 0 a 11. A = 22 0... 12. C = A 1 BA 1 1 0 0 a nn 1 1 0 0 2 2 1 1 1 1 0 0 0 0 13. (a) 2 2 1 0 0 1 2 2 (b) 0 1 0 0 0 0 1 1 1 1 0 0 0 1 1 0 2 2 1 2 3 0 1 1 14. (a) 2 1 4 (b) 1 0 1 3 4 5 1 1 0 [ ] [ ] 0 1 1 2 16. (a) False; for example, A = and B = 0 2 3 4 [ ] [ ] 0 1 1 2 (c) False; for example, A = and B = 0 2 3 4 (b) True (d) True (e) True (f) True (g) True (h) True (i) True (j) True (k) True (l) True (m) True [ ] [ ] 1 0 1 0 (n) True (o) False; for example, A = and B = (p) True 0 1 0 1 (q) True 1.5 Elementary Matrices and a Method for Finding A 1 In this section we shall develop an algorithm for finding the inverse of an invertible matrix. We shall also discuss some of the basic properties of invertible matrices. Definition 1.15. An n n matrix is called an elementary matrix if it can be obtained from the n n identity matrix I n by performing a single elementary row operation.