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Linear algebra03fallleturenotes01

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ST ZULAIHA NURHAJARURAHMAH

The document discusses systems of linear equations and Gaussian elimination. It begins with an introduction to systems of linear equations, defining them as a set of linear equations that can have no solution, a unique solution, or infinitely many solutions. It then discusses row-echelon form and reduced row-echelon form, which are special forms that a matrix of a linear system can be put into using elementary row operations. Being in row-echelon or reduced row-echelon form provides information about the solution set of the corresponding linear system. Examples are provided to illustrate these concepts.

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Elementary Linear Algebra Howard Anton ＆ Chris Rorres
Chapter Contents  1.1 Introduction to System of Linear Equations  1.2 Gaussian Elimination  1.3 Matrices and Matrix Operations  1.4 Inverses; Rules of Matrix Arithmetic  1.5 Elementary Matrices and a Method for Finding  1.6 Further Results on Systems of Equations and Invertibility  1.7 Diagonal, Triangular, and Symmetric Matrices 1− A
1.1 Introduction to Systems of Equations
Linear Equations  Any straight line in xy-plane can be represented algebraically by an equation of the form:  General form: define a linear equation in the n variables :  Where and b are real constants.  The variables in a linear equation are sometimes called unknowns. byaxa =+ 21 nxxx ,...,, 21 bxaxaxa nn =+++ ...2211 ,,...,, 21 naaa
Example 1 Linear Equations  The equations and are 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.  The equations are not linear.  A solution of a linear equation is a sequence of n numbers 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 ,13 2 1 ,73 ++==+ zxyyx 732 4321 =+−− xxxx xyxzzyxyx sinand,423,53 ==+−+=+ nsss ,...,, 21
Example 2 Finding a Solution Set (1/2)  Find the solution of  Solution(a) we can assign an arbitrary value to x and solve for y , or choose an arbitrary value for y and solve for x .If we follow the first approach and assign x an arbitrary value ,we obtain  arbitrary numbers are called parameter.  for example 124)a( =− yx 2211 , 4 1 2 1 or 2 1 2, tytxtytx =+=−== 2,1 tt 2 11 as 2 11 ,3solutiontheyields3 21 ==== tyxt
Example 2 Finding a Solution Set (2/2)  Find the solution of  Solution(b) we can assign arbitrary values to any two variables and solve for the third variable.  for example  where s, t are arbitrary values .574(b) 321 =+− xxx txsxtsx ==−+= 321 ,,745
Linear Systems (1/2)  A finite set of linear equations in the variables is called a system of linear equations or a linear system .  A sequence of numbers is called a solution of the system.  A system has no solution is said to be inconsistent ; if there is at least one solution of the system, it is called consistent. nxxx ,...,, 21 nsss ,...,, 21 mnmnmm nn nn bxaxaxa bxaxaxa bxaxaxa =+++ =+++ =+++ ... ... ... 2211 22222121 11212111  An arbitrary system of m linear equations in n unknowns
Linear Systems (2/2)  Every system of linear equations has either no solutions, exactly one solution, or infinitely many solutions.  A general system of two linear equations: (Figure1.1.1)  Two lines may be parallel -> no solution  Two lines may intersect at only one point -> one solution  Two lines may coincide -> infinitely many solution zero)bothnot,( zero)bothnot,( 22222 11111 bacybxa bacybxa =+ =+
Augmented Matrices mnmnmm nn nn bxaxaxa bxaxaxa bxaxaxa =+++ =+++ =+++ ... ... ... 2211 22222121 11212111              mmnmm n n baaa baaa baaa ... ... ... 21 222221 111211   The location of the +’s, the x’s, and the =‘s can be abbreviated by writing only the rectangular array of numbers.  This is called the augmented matrix for the system.  Note: must be written in the same order in each equation as the unknowns and the constants must be on the right. 1th column 1th row
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 which is easier to solve.  Since the rows of an augmented matrix correspond to the equations in the associated system. new systems is generally obtained in a series of steps by applying the following three types of operations to eliminate unknowns systematically. These are called elementary row operations. 1. Multiply an equation through by an nonzero constant. 2. Interchange two equation. 3. Add a multiple of one equation to another.
Example 3 Using Elementary row Operations(1/4) 0563 7172 92 =−+ −=− =++ zyx zy zyx  → secondtheto equationfirstthe times2-add 0563 1342 92 =−+ =−+ =++ zyx zyx zyx           − − 0563 1342 9211           − −− 0563 17720 9211  → thirdtheto equationfirstthe times3-add  → thirdtheto rowfirstthe times3-add  → secondtheto rowfirstthe times2-add
Example 3 Using Elementary row Operations(2/4) 0113 92 2 17 2 7 =− −=− =++ zy zy zyx  → 2 1 byequation secondthemultiply 27113 1772 92 −=− −=− =++ zy zy zyx           −− −− 271130 17720 9211           −− −− 271130 10 9211 2 17 2 7  → thirdtheto equationsecondthe times3-add  → thirdtheto rowsecondthe times3-add  → 2 1 byrow secondemultily th
Example 3 Using Elementary row Operations(3/4) 3 92 2 17 2 7 = −=− =++ z zy zyx  → 2-byequation thirdtheMultiply 2 3 2 1 2 17 2 7 92 −=− −=− =++ z zy zyx           −− −− 2 3 2 1 2 17 2 7 00 10 9211           −− 3100 10 9211 2 17 2 7  → firsttheto equationsecond thetimes1-Add  → firsttheto rowsecond thetimes1-Add  → 2-byrow thirdeMultily th
Example 3 Using Elementary row Operations(4/4) 3 2 1 = = = z y x  → secondtheto equationthirdthe timesandfirsttheto equationthirdthe times-Add 2 7 2 11 3 2 17 2 7 2 35 2 11 = −=− =+ z zy zx           −− 3100 10 01 2 17 2 7 2 35 2 11           3100 2010 1001  → secondtheto rowthirdthetimes andfirsttheto rowthirdthe times-Add 2 7 2 11  The solution x=1,y=2,z=3 is now evident.
1.2 Gaussian Elimination
Echelon Forms  This matrix which have following properties is in reduced row- echelon form (Example 1, 2). 1. 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. 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 leader 1 in the lower row occurs farther to the right than the leader 1 in the higher row. 4. 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 (Example 1, 2).  A matrix in reduced row-echelon form is of necessity in row- echelon form, but not conversely.
Example 1 Row-Echelon & Reduced Row-Echelon form  reduced row-echelon form:                   −                     − 00 00 , 00000 00000 31000 10210 , 100 010 001 , 1100 7010 4001  row-echelon form:           −                     − 10000 01100 06210 , 000 010 011 , 5100 2610 7341
Example 2 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. ) :                                                     *100000000 *0**100000 *0**010000 *0**001000 *0**000*10 , 0000 0000 **10 **01 , 0000 *100 *010 *001 , 1000 0100 0010 0001                                                     *100000000 ****100000 *****10000 ******1000 ********10 , 0000 0000 **10 ***1 , 0000 *100 **10 ***1 , 1000 *100 **10 ***1  All matrices of the following types are in reduced row- echelon form ( any real numbers substituted for the *’s. ) :
Example 3 Solutions of Four Linear Systems (a)           − 4100 2010 5001 (a) 4 2- 5 = = = z y x Solution (a) the corresponding system of equations is : Suppose that the augmented matrix for a system of linear equations have been reduced by row operations to the given reduced row-echelon form. Solve the system.
Example 3 Solutions of Four Linear Systems (b1)           − 23100 62010 14001 (b) Solution (b) 1. The corresponding system of equations is : 23 62 1-4 43 42 41 =+ =+ =+ xx xx xx leading variables free variables
Example 3 Solutions of Four Linear Systems (b2) 43 42 41 3-2 2-6 4-1- xx xx xx = = = tx tx tx tx ,32 ,26 ,41 4 3 2 1 = −= −= −−= 2. We see that the free variable can be assigned an arbitrary value, say t, which then determines values of the leading variables. 3. There are infinitely many solutions, and the general solution is given by the formulas
Example 3 Solutions of Four Linear Systems (c1)             − 000000 251000 130100 240061 (c) 25 13 2-46 54 53 521 =+ =+ =++ xx xx xxx Solution (c) 1. The 4th row of zeros leads to the equation places no restrictions on the solutions (why?). Thus, we can omit this equation.
Example 3 Solutions of Four Linear Systems (c2) Solution (c) 2. Solving for the leading variables in terms of the free variables: 3. The free variable can be assigned an arbitrary value,there are infinitely many solutions, and the general solution is given by the formulas. 54 53 521 5-2 3-1 4-6-2- xx xx xxx = = = tx tx tx sx tsx = = = = = 4 4 3 2 1 ,5-2 3-1 ,4-6-2-
Example 3 Solutions of Four Linear Systems (d)           1000 0210 0001 (d) Solution (d): the last equation in the corresponding system of equation is Since this equation cannot be satisfied, there is no solution to the system. 1000 321 =++ xxx
Elimination Methods (1/7)  We shall give a step-by-step elimination procedure that can be used to reduce any matrix to reduced row-echelon form.           −−− − − 156542 281261042 1270200
Elimination Methods (2/7)  Step1. Locate the leftmost column that does not consist entirely of zeros.  Step2. Interchange the top row with another row, to bring a nonzero entry to top of the column found in Step1.           −−− − − 156542 281261042 1270200 Leftmost nonzero column           −−− − − 156542 1270200 281261042 The 1th and 2th rows in the preceding matrix were interchanged.
Elimination Methods (3/7)  Step3. If the entry that is now at the top of the column found in Step1 is a, multiply the first row by 1/a in order to introduce a leading 1.  Step4. Add suitable multiples of the top row to the rows below so that all entires below the leading 1 become zeros.           −−− − − 156542 1270200 1463521 The 1st row of the preceding matrix was multiplied by 1/2.           −− − − 29170500 1270200 1463521 -2 times the 1st row of the preceding matrix was added to the 3rd row.
Elimination Methods (4/7)  Step5. Now cover the top row in the matrix and begin again with Step1 applied to the submatrix that remains. Continue in this way until the entire matrix is in row-echelon form.           −−− − − 29170500 1270200 1463521 The 1st row in the submatrix was multiplied by -1/2 to introduce a leading 1.          −− −− − 29170500 60100 1463521 2 7 Leftmost nonzero column in the submatrix
Elimination Methods (5/7)  Step5 (cont.)           −− − 210000 60100 1463521 2 7 -5 times the 1st row of the submatrix was added to the 2nd row of the submatrix to introduce a zero below the leading 1.           −− − 10000 60100 1463521 2 1 2 7           −− − 10000 60100 1463521 2 1 2 7 The top row in the submatrix was covered, and we returned again Step1. The first (and only) row in the new submetrix was multiplied by 2 to introduce a leading 1. Leftmost nonzero column in the new submatrix  The entire matrix is now in row-echelon form.
Elimination Methods (6/7)  Step6. Beginning with las nonzero row and working upward, add suitable multiples of each row to the rows above to introduce zeros above the leading 1’s.           210000 100100 703021 7/2 times the 3rd row of the preceding matrix was added to the 2nd row.           − 210000 100100 1463521           − 210000 100100 203521 -6 times the 3rd row was added to the 1st row.  The last matrix is in reduced row-echelon form. 5 times the 2nd row was added to the 1st row.
Elimination Methods (7/7)  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 4 Gauss-Jordan Elimination(1/4)  Solve by Gauss-Jordan Elimination  Solution: The augmented matrix for the system is 6184862 515105 1342562 0x223 65421 643 654321 5321 =−+++ =++ −=−+−−+ =+−+ xxxxx xxx xxxxxx xxx             61848062 515010500 1-3-42-5-62 00202-31
Example 4 Gauss-Jordan Elimination(2/4)  Adding -2 times the 1st row to the 2nd and 4th rows gives  Multiplying the 2nd row by -1 and then adding -5 times the new 2nd row to the 3rd row and -4 times the new 2nd row to the 4th row gives             2600000 0000000 13-02100 00202-31             61808400 515010500 1-3-02-1-00 00202-31
Example 4 Gauss-Jordan Elimination(3/4)  Interchanging the 3rd and 4th rows and then multiplying the 3rd row of the resulting matrix by 1/6 gives the row-echelon form.  Adding -3 times the 3rd row to the 2nd row and then adding 2 times the 2nd row of the resulting matrix to the 1st row yields the reduced row-echelon form.             0000000 100000 0002100 0024031 3 1             0000000 100000 1-3-02-1-00 00202-31 3 1
Example 4 Gauss-Jordan Elimination(4/4)  The corresponding system of equations is  Solution The augmented matrix for the system is  We assign the free variables, and the general solution is given by the formulas: 3 1 6 43 5421 02 0x243 = =+ =+++ x xx xxx 3 1 6 43 5421 2 x243 = −= −−−= x xx xxx 3 1 654321 ,,,2,,243 ===−==−−−= xtxsxsxrxtsrx
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 solved by a technique called back- substitution.  Example 5
Example 5 ex4 solved by Back-substitution(1/2)  From the computations in Example 4, a row-echelon form from the augmented matrix is  To solve the corresponding system of equations  Step1. Solve the equations for the leading variables.             0000000 100000 1-3-02-1-00 00202-31 3 1 3 1 6 43 5421 02 0x243 = =+ =+++ x xx xxx 3 1 6 643 5321 321 x223 = −−= −+−= x xxx xxx
Example5 ex4 solved by Back-substitution(2/2)  Step2. Beginning with the bottom equation and working upward, successively substitute each equation into all the equations above it.  Substituting x6=1/3 into the 2nd equation  Substituting x3=-2 x4 into the 1st equation  Step3. Assign free variables, the general solution is given by the formulas. 3 1 6 43 5321 2 x223 = −= −+−= x xx xxx 3 1 6 43 5321 2 x223 = −= −+−= x xx xxx 3 1 654321 ,,,2,,243 ===−==−−−= xtxsxsxrxtsrx
Example 6 Gaussian elimination(1/2)  Solve by Gaussian elimination and back-substitution. (ex3 of Section1.1)  Solution  We convert the augmented matrix  to the ow-echelon form  The system corresponding to this matrix is 0563 1342 92 =−+ =−+ =++ zyx zyx zyx           − − 0563 1342 9211           −− 3100 10 9211 2 17 2 7 3,,92 2 17 2 7 =−=−=++ zzyzyx
Example 6 Gaussian elimination(2/2)  Solution  Solving for the leading variables  Substituting the bottom equation into those above  Substituting the 2nd equation into the top 3 ,2 ,3 = = −= z y yx 3 , ,29 2 7 2 17 = +−= −−= z zy zyx 3,2,1 === zyx
Homogeneous Linear Systems(1/2)  A system of linear equations is said to be homogeneous if the constant terms are all zero; that is , the system has the form :  Every homogeneous system of linear equation is consistent, since all such system have 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:  The system has only the trivial solution.  The system has infinitely many solutions in addition to the trivial solution. 0... 0... 0... 2211 2222121 1212111 =+++ =+++ =+++ nmnmm nn nn xaxaxa xaxaxa xaxaxa  0,...,0,0 21 === nxxx
Homogeneous Linear Systems(2/2)  In a special case of a homogeneous linear system of two linear equations in two unknowns: (fig1.2.1) zero)bothnot,(0 zero)bothnot,(0 2222 1111 baybxa baybxa =+ =+
Example 7 Gauss-Jordan Elimination(1/3) 0 02 032 022 543 5321 54321 5321 =++ =−−+ =+−+−− =+−+ xxx xxxx xxxxx xxxx             −− −−− − 001000 010211 013211 010122             000000 001000 010100 010011  Solve the following homogeneous system of linear equations by using Gauss-Jordan elimination.  Solution  The augmented matrix  Reducing this matrix to reduced row-echelon form
Example 7 Gauss-Jordan Elimination(2/3) 0 0 0 4 53 521 = =+ =++ x xx xxx 04 53 521 = −= −−= x xx xxx Solution (cont)  The corresponding system of equation  Solving for the leading variables is  Thus the general solution is  Note: the trivial solution is obtained when s=t=0. txxtxsxtsx ==−==−−= 54321 ,0,,,
Example7 Gauss-Jordan Elimination(3/3) (1)0() 0() 0() 2 1 =+ =+ =+ ∑ ∑ ∑ r k k x x x    (2)() () () 2 1 ∑ ∑ ∑ −= −= −= r k k x x x   Two important points:  Non of the three row operations alters the final column of zeros, so the system of equations corresponding to the reduced row-echelon form of the augmented matrix must also be a homogeneous system.  If the given homogeneous system has m equations in n unknowns with m<n, and there are r nonzero rows in reduced row-echelon form of the augmented matrix, we will have r<n. It will have the form:
Theorem 1.2.1 A homogeneous system of linear equations with more unknowns than equations has infinitely many solutions.  Note: theorem 1.2.1 applies only to homogeneous system  Example 7 (3/3)
Computer Solution of Linear System  Most computer algorithms for solving large linear systems are based on Gaussian elimination or Gauss-Jordan elimination.  Issues  Reducing roundoff errors  Minimizing the use of computer memory space  Solving the system with maximum speed
1.3 Matrices and Matrix Operations
Definition A matrix is a rectangular array of numbers. The numbers in the array are called the entries in the matrix.
Example 1 Examples of matrices  Some examples of matrices  Size 3 x 2, 1 x 4, 3 x 3, 2 x 1, 1 x 1 [ ] [ ]4, 3 1 , 000 10 2 ,3-012, 41 03 21 2 1                 −π           −  # rows # columns row matrix or row vector column matrix or column vector entrie s
Matrices Notation and Terminology(1/2)  A general m x n matrix A as  The entry that occurs in row i and column j of matrix A will be denoted . If is real number, it is common to be referred as scalars.             = mnmm n n aaa aaa aaa A ... ... ... 21 22221 11211  ( )ijij Aa or ija
 The preceding matrix can be written as  A matrix A with n rows and n columns is called a square matrix of order n, and the shaded entries are said to be on the main diagonal of A. Matrices Notation and Terminology(2/2) [ ] [ ]ijnmij aa or×             mnmm n n aaa aaa aaa ... ... ... 21 22221 11211  nnaaa ,,, 2211 
Definition Two matrices are defined to be equal if they have the same size and their corresponding entries are equal. [ ] [ ] j.andiallforifonlyandifBAthen size,samethehaveandIf ijij ijij ba bBaA == ==
Example 2 Equality of Matrices  Consider the matrices  If x=5, then A=B.  For all other values of x, the matrices A and B are not equal.  There is no value of x for which A=C since A and C have different sizes.       =      =      = 043 012 , 53 12 , 3 12 CB x A
Operations on Matrices  If A and B are matrices of the same size, then the sum A+B is the matrix obtained by adding the entries of B to the corresponding entries of A.  Vice versa, the difference A-B is the matrix obtained by subtracting the entries of B from the corresponding entries of A.  Note: Matrices of different sizes cannot be added or subtracted. ( ) ( ) ijijijijij ijijijijij baBABA baBABA −=−=− +=+=+ )()( )()(
Example 3 Addition and Subtraction  Consider the matrices  Then  The expressions A+C, B+C, A-C, and B-C are undefined.       =           − − − =           − −= 22 11 , 5423 1022 1534 , 0724 4201 3012 CBA           −− −− −− =−          − =+ 51141 5223 2526 , 5307 3221 4542 BABA
Definition If A is any matrix and c is any scalar, then the product cA is the matrix obtained by multiplying each entry of the matrix A by c. The matrix cA is said to be the scalar multiple of A. [ ] ( ) ( ) ijijij ij caAccA a == = then,Aifnotation,matrixIn
Example 4 Scalar Multiples (1/2)  For the matrices  We have  It common practice to denote (-1)B by –B.       − =      −− =      = 1203 369 , 531 720 , 131 432 CBA ( )       − =      − −− =      = 401 123 , 531 720 1-, 262 864 2 3 1 CBA
Example 4 Scalar Multiples (2/2)
Definition  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 the matrix A and column j from the matrix B .Multiply the corresponding entries from the row and column together and then add up the resulting products.
Example 5 Multiplying Matrices (1/2)  Consider the matrices  Solution  Since A is a 2 ×3 matrix and B is a 3 ×4 matrix, the product AB is a 2 ×4 matrix. And:
Example 5 Multiplying Matrices (2/2)
Examples 6 Determining Whether a Product Is Defined  Suppose that A ,B ,and C are matrices with the following sizes: A B C 3 ×4 4 ×7 7 ×3  Solution:  Then by (3), AB is defined and is a 3 ×7 matrix; BC is defined and is a 4 ×3 matrix; and CA is defined and is a 7 ×4 matrix. The products AC ,CB ,and BA are all undefined.
Partitioned Matrices  A matrix can be subdivided or partitioned into smaller matrices by inserting horizontal and vertical rules between selected rows and columns.  For example, below are three possible partitions of a general 3 ×4 matrix A .  The first is a partition of A into four submatrices A 11 ,A 12, A 21 ,and A 22 .  The second is a partition of A into its row matrices r 1 ,r 2, and r 3 .  The third is a partition of A into its column matrices c 1, c 2 ,c 3 ,and c 4 .
Matrix Multiplication by columns and by Rows  Sometimes it may b desirable to find a particular row or column of a matrix product AB without computing the entire product.  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 Formulas (6)and (7)that
Example 7 Example5 Revisited  This is the special case of a more general procedure for multiplying partitioned matrices.  If A and B are the matrices in Example 5,then from (6)the second column matrix of AB can be obtained by the computation  From (7) the first row matrix of AB can be obtained by the computation
Matrix Products as Linear Combinations (1/2)
Matrix Products as Linear Combinations (2/2)  In words, (10)tells us that the product A x of a 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 .  In the exercises w ask the reader to show that the product y A 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 .
Example 8 Linear Combination
Example 9 Columns of a Product AB as Linear Combinations
Matrix form of a Linear System(1/2) mnmnmm nn nn bxaxaxa bxaxaxa bxaxaxa =+++ =+++ =+++ ... ... ... 2211 22222121 11212111              =             +++ +++ +++ mnmnmm nn nn b b b xaxaxa xaxaxa xaxaxa  2 1 2211 2222121 1212111 ... ... ...             =                         mmmnmm n n b b b x x x aaa aaa aaa  2 1 2 1 21 22221 11211 ... ... ...  Consider any system of m linear equations in n unknowns.  Since two matrices are equal if and only if their corresponding entries are equal.  The m×1 matrix on the left side of this equation can be written as a product to give:
Matrix form of a Linear System(1/2)  If w designate these matrices by A ,x ,and b ,respectively, the original system of m equations in n unknowns has been replaced by the single matrix equation  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
Definition  If A is any m×n matrix, then the transpose of A ,denoted by ,is defined to be the n×m matrix that results from interchanging the rows and columns of A ; that is, the first column of is the first row of A ,the second column of is the second row of A ,and so forth. T A T A T A
Example 10 Some Transposes (1/2)
Example 10 Some Transposes (2/2)  Observe that  In the special case where A is a square matrix, the transpose of A can be obtained by interchanging entries that are symmetrically positioned about the main diagonal.
Definition  If A is a square matrix, then the 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.
Example 11 Trace of Matrix

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Linear algebra03fallleturenotes01

  • 1. Elementary Linear Algebra Howard Anton ＆ Chris Rorres
  • 2. Chapter Contents  1.1 Introduction to System of Linear Equations  1.2 Gaussian Elimination  1.3 Matrices and Matrix Operations  1.4 Inverses; Rules of Matrix Arithmetic  1.5 Elementary Matrices and a Method for Finding  1.6 Further Results on Systems of Equations and Invertibility  1.7 Diagonal, Triangular, and Symmetric Matrices 1− A
  • 4. Linear Equations  Any straight line in xy-plane can be represented algebraically by an equation of the form:  General form: define a linear equation in the n variables :  Where and b are real constants.  The variables in a linear equation are sometimes called unknowns. byaxa =+ 21 nxxx ,...,, 21 bxaxaxa nn =+++ ...2211 ,,...,, 21 naaa
  • 5. Example 1 Linear Equations  The equations and are 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.  The equations are not linear.  A solution of a linear equation is a sequence of n numbers 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 ,13 2 1 ,73 ++==+ zxyyx 732 4321 =+−− xxxx xyxzzyxyx sinand,423,53 ==+−+=+ nsss ,...,, 21
  • 6. Example 2 Finding a Solution Set (1/2)  Find the solution of  Solution(a) we can assign an arbitrary value to x and solve for y , or choose an arbitrary value for y and solve for x .If we follow the first approach and assign x an arbitrary value ,we obtain  arbitrary numbers are called parameter.  for example 124)a( =− yx 2211 , 4 1 2 1 or 2 1 2, tytxtytx =+=−== 2,1 tt 2 11 as 2 11 ,3solutiontheyields3 21 ==== tyxt
  • 7. Example 2 Finding a Solution Set (2/2)  Find the solution of  Solution(b) we can assign arbitrary values to any two variables and solve for the third variable.  for example  where s, t are arbitrary values .574(b) 321 =+− xxx txsxtsx ==−+= 321 ,,745
  • 8. Linear Systems (1/2)  A finite set of linear equations in the variables is called a system of linear equations or a linear system .  A sequence of numbers is called a solution of the system.  A system has no solution is said to be inconsistent ; if there is at least one solution of the system, it is called consistent. nxxx ,...,, 21 nsss ,...,, 21 mnmnmm nn nn bxaxaxa bxaxaxa bxaxaxa =+++ =+++ =+++ ... ... ... 2211 22222121 11212111  An arbitrary system of m linear equations in n unknowns
  • 9. Linear Systems (2/2)  Every system of linear equations has either no solutions, exactly one solution, or infinitely many solutions.  A general system of two linear equations: (Figure1.1.1)  Two lines may be parallel -> no solution  Two lines may intersect at only one point -> one solution  Two lines may coincide -> infinitely many solution zero)bothnot,( zero)bothnot,( 22222 11111 bacybxa bacybxa =+ =+
  • 10. Augmented Matrices mnmnmm nn nn bxaxaxa bxaxaxa bxaxaxa =+++ =+++ =+++ ... ... ... 2211 22222121 11212111              mmnmm n n baaa baaa baaa ... ... ... 21 222221 111211   The location of the +’s, the x’s, and the =‘s can be abbreviated by writing only the rectangular array of numbers.  This is called the augmented matrix for the system.  Note: must be written in the same order in each equation as the unknowns and the constants must be on the right. 1th column 1th row
  • 11. 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 which is easier to solve.  Since the rows of an augmented matrix correspond to the equations in the associated system. new systems is generally obtained in a series of steps by applying the following three types of operations to eliminate unknowns systematically. These are called elementary row operations. 1. Multiply an equation through by an nonzero constant. 2. Interchange two equation. 3. Add a multiple of one equation to another.
  • 12. Example 3 Using Elementary row Operations(1/4) 0563 7172 92 =−+ −=− =++ zyx zy zyx  → secondtheto equationfirstthe times2-add 0563 1342 92 =−+ =−+ =++ zyx zyx zyx           − − 0563 1342 9211           − −− 0563 17720 9211  → thirdtheto equationfirstthe times3-add  → thirdtheto rowfirstthe times3-add  → secondtheto rowfirstthe times2-add
  • 13. Example 3 Using Elementary row Operations(2/4) 0113 92 2 17 2 7 =− −=− =++ zy zy zyx  → 2 1 byequation secondthemultiply 27113 1772 92 −=− −=− =++ zy zy zyx           −− −− 271130 17720 9211           −− −− 271130 10 9211 2 17 2 7  → thirdtheto equationsecondthe times3-add  → thirdtheto rowsecondthe times3-add  → 2 1 byrow secondemultily th
  • 14. Example 3 Using Elementary row Operations(3/4) 3 92 2 17 2 7 = −=− =++ z zy zyx  → 2-byequation thirdtheMultiply 2 3 2 1 2 17 2 7 92 −=− −=− =++ z zy zyx           −− −− 2 3 2 1 2 17 2 7 00 10 9211           −− 3100 10 9211 2 17 2 7  → firsttheto equationsecond thetimes1-Add  → firsttheto rowsecond thetimes1-Add  → 2-byrow thirdeMultily th
  • 15. Example 3 Using Elementary row Operations(4/4) 3 2 1 = = = z y x  → secondtheto equationthirdthe timesandfirsttheto equationthirdthe times-Add 2 7 2 11 3 2 17 2 7 2 35 2 11 = −=− =+ z zy zx           −− 3100 10 01 2 17 2 7 2 35 2 11           3100 2010 1001  → secondtheto rowthirdthetimes andfirsttheto rowthirdthe times-Add 2 7 2 11  The solution x=1,y=2,z=3 is now evident.
  • 17. Echelon Forms  This matrix which have following properties is in reduced row- echelon form (Example 1, 2). 1. 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. 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 leader 1 in the lower row occurs farther to the right than the leader 1 in the higher row. 4. 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 (Example 1, 2).  A matrix in reduced row-echelon form is of necessity in row- echelon form, but not conversely.
  • 18. Example 1 Row-Echelon & Reduced Row-Echelon form  reduced row-echelon form:                   −                     − 00 00 , 00000 00000 31000 10210 , 100 010 001 , 1100 7010 4001  row-echelon form:           −                     − 10000 01100 06210 , 000 010 011 , 5100 2610 7341
  • 19. Example 2 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. ) :                                                     *100000000 *0**100000 *0**010000 *0**001000 *0**000*10 , 0000 0000 **10 **01 , 0000 *100 *010 *001 , 1000 0100 0010 0001                                                     *100000000 ****100000 *****10000 ******1000 ********10 , 0000 0000 **10 ***1 , 0000 *100 **10 ***1 , 1000 *100 **10 ***1  All matrices of the following types are in reduced row- echelon form ( any real numbers substituted for the *’s. ) :
  • 20. Example 3 Solutions of Four Linear Systems (a)           − 4100 2010 5001 (a) 4 2- 5 = = = z y x Solution (a) the corresponding system of equations is : Suppose that the augmented matrix for a system of linear equations have been reduced by row operations to the given reduced row-echelon form. Solve the system.
  • 21. Example 3 Solutions of Four Linear Systems (b1)           − 23100 62010 14001 (b) Solution (b) 1. The corresponding system of equations is : 23 62 1-4 43 42 41 =+ =+ =+ xx xx xx leading variables free variables
  • 22. Example 3 Solutions of Four Linear Systems (b2) 43 42 41 3-2 2-6 4-1- xx xx xx = = = tx tx tx tx ,32 ,26 ,41 4 3 2 1 = −= −= −−= 2. We see that the free variable can be assigned an arbitrary value, say t, which then determines values of the leading variables. 3. There are infinitely many solutions, and the general solution is given by the formulas
  • 23. Example 3 Solutions of Four Linear Systems (c1)             − 000000 251000 130100 240061 (c) 25 13 2-46 54 53 521 =+ =+ =++ xx xx xxx Solution (c) 1. The 4th row of zeros leads to the equation places no restrictions on the solutions (why?). Thus, we can omit this equation.
  • 24. Example 3 Solutions of Four Linear Systems (c2) Solution (c) 2. Solving for the leading variables in terms of the free variables: 3. The free variable can be assigned an arbitrary value,there are infinitely many solutions, and the general solution is given by the formulas. 54 53 521 5-2 3-1 4-6-2- xx xx xxx = = = tx tx tx sx tsx = = = = = 4 4 3 2 1 ,5-2 3-1 ,4-6-2-
  • 25. Example 3 Solutions of Four Linear Systems (d)           1000 0210 0001 (d) Solution (d): the last equation in the corresponding system of equation is Since this equation cannot be satisfied, there is no solution to the system. 1000 321 =++ xxx
  • 26. Elimination Methods (1/7)  We shall give a step-by-step elimination procedure that can be used to reduce any matrix to reduced row-echelon form.           −−− − − 156542 281261042 1270200
  • 27. Elimination Methods (2/7)  Step1. Locate the leftmost column that does not consist entirely of zeros.  Step2. Interchange the top row with another row, to bring a nonzero entry to top of the column found in Step1.           −−− − − 156542 281261042 1270200 Leftmost nonzero column           −−− − − 156542 1270200 281261042 The 1th and 2th rows in the preceding matrix were interchanged.
  • 28. Elimination Methods (3/7)  Step3. If the entry that is now at the top of the column found in Step1 is a, multiply the first row by 1/a in order to introduce a leading 1.  Step4. Add suitable multiples of the top row to the rows below so that all entires below the leading 1 become zeros.           −−− − − 156542 1270200 1463521 The 1st row of the preceding matrix was multiplied by 1/2.           −− − − 29170500 1270200 1463521 -2 times the 1st row of the preceding matrix was added to the 3rd row.
  • 29. Elimination Methods (4/7)  Step5. Now cover the top row in the matrix and begin again with Step1 applied to the submatrix that remains. Continue in this way until the entire matrix is in row-echelon form.           −−− − − 29170500 1270200 1463521 The 1st row in the submatrix was multiplied by -1/2 to introduce a leading 1.          −− −− − 29170500 60100 1463521 2 7 Leftmost nonzero column in the submatrix
  • 30. Elimination Methods (5/7)  Step5 (cont.)           −− − 210000 60100 1463521 2 7 -5 times the 1st row of the submatrix was added to the 2nd row of the submatrix to introduce a zero below the leading 1.           −− − 10000 60100 1463521 2 1 2 7           −− − 10000 60100 1463521 2 1 2 7 The top row in the submatrix was covered, and we returned again Step1. The first (and only) row in the new submetrix was multiplied by 2 to introduce a leading 1. Leftmost nonzero column in the new submatrix  The entire matrix is now in row-echelon form.
  • 31. Elimination Methods (6/7)  Step6. Beginning with las nonzero row and working upward, add suitable multiples of each row to the rows above to introduce zeros above the leading 1’s.           210000 100100 703021 7/2 times the 3rd row of the preceding matrix was added to the 2nd row.           − 210000 100100 1463521           − 210000 100100 203521 -6 times the 3rd row was added to the 1st row.  The last matrix is in reduced row-echelon form. 5 times the 2nd row was added to the 1st row.
  • 32. Elimination Methods (7/7)  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.
  • 33. Example 4 Gauss-Jordan Elimination(1/4)  Solve by Gauss-Jordan Elimination  Solution: The augmented matrix for the system is 6184862 515105 1342562 0x223 65421 643 654321 5321 =−+++ =++ −=−+−−+ =+−+ xxxxx xxx xxxxxx xxx             61848062 515010500 1-3-42-5-62 00202-31
  • 34. Example 4 Gauss-Jordan Elimination(2/4)  Adding -2 times the 1st row to the 2nd and 4th rows gives  Multiplying the 2nd row by -1 and then adding -5 times the new 2nd row to the 3rd row and -4 times the new 2nd row to the 4th row gives             2600000 0000000 13-02100 00202-31             61808400 515010500 1-3-02-1-00 00202-31
  • 35. Example 4 Gauss-Jordan Elimination(3/4)  Interchanging the 3rd and 4th rows and then multiplying the 3rd row of the resulting matrix by 1/6 gives the row-echelon form.  Adding -3 times the 3rd row to the 2nd row and then adding 2 times the 2nd row of the resulting matrix to the 1st row yields the reduced row-echelon form.             0000000 100000 0002100 0024031 3 1             0000000 100000 1-3-02-1-00 00202-31 3 1
  • 36. Example 4 Gauss-Jordan Elimination(4/4)  The corresponding system of equations is  Solution The augmented matrix for the system is  We assign the free variables, and the general solution is given by the formulas: 3 1 6 43 5421 02 0x243 = =+ =+++ x xx xxx 3 1 6 43 5421 2 x243 = −= −−−= x xx xxx 3 1 654321 ,,,2,,243 ===−==−−−= xtxsxsxrxtsrx
  • 37. 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 solved by a technique called back- substitution.  Example 5
  • 38. Example 5 ex4 solved by Back-substitution(1/2)  From the computations in Example 4, a row-echelon form from the augmented matrix is  To solve the corresponding system of equations  Step1. Solve the equations for the leading variables.             0000000 100000 1-3-02-1-00 00202-31 3 1 3 1 6 43 5421 02 0x243 = =+ =+++ x xx xxx 3 1 6 643 5321 321 x223 = −−= −+−= x xxx xxx
  • 39. Example5 ex4 solved by Back-substitution(2/2)  Step2. Beginning with the bottom equation and working upward, successively substitute each equation into all the equations above it.  Substituting x6=1/3 into the 2nd equation  Substituting x3=-2 x4 into the 1st equation  Step3. Assign free variables, the general solution is given by the formulas. 3 1 6 43 5321 2 x223 = −= −+−= x xx xxx 3 1 6 43 5321 2 x223 = −= −+−= x xx xxx 3 1 654321 ,,,2,,243 ===−==−−−= xtxsxsxrxtsrx
  • 40. Example 6 Gaussian elimination(1/2)  Solve by Gaussian elimination and back-substitution. (ex3 of Section1.1)  Solution  We convert the augmented matrix  to the ow-echelon form  The system corresponding to this matrix is 0563 1342 92 =−+ =−+ =++ zyx zyx zyx           − − 0563 1342 9211           −− 3100 10 9211 2 17 2 7 3,,92 2 17 2 7 =−=−=++ zzyzyx
  • 41. Example 6 Gaussian elimination(2/2)  Solution  Solving for the leading variables  Substituting the bottom equation into those above  Substituting the 2nd equation into the top 3 ,2 ,3 = = −= z y yx 3 , ,29 2 7 2 17 = +−= −−= z zy zyx 3,2,1 === zyx
  • 42. Homogeneous Linear Systems(1/2)  A system of linear equations is said to be homogeneous if the constant terms are all zero; that is , the system has the form :  Every homogeneous system of linear equation is consistent, since all such system have 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:  The system has only the trivial solution.  The system has infinitely many solutions in addition to the trivial solution. 0... 0... 0... 2211 2222121 1212111 =+++ =+++ =+++ nmnmm nn nn xaxaxa xaxaxa xaxaxa  0,...,0,0 21 === nxxx
  • 43. Homogeneous Linear Systems(2/2)  In a special case of a homogeneous linear system of two linear equations in two unknowns: (fig1.2.1) zero)bothnot,(0 zero)bothnot,(0 2222 1111 baybxa baybxa =+ =+
  • 45. Example 7 Gauss-Jordan Elimination(2/3) 0 0 0 4 53 521 = =+ =++ x xx xxx 04 53 521 = −= −−= x xx xxx Solution (cont)  The corresponding system of equation  Solving for the leading variables is  Thus the general solution is  Note: the trivial solution is obtained when s=t=0. txxtxsxtsx ==−==−−= 54321 ,0,,,
  • 46. Example7 Gauss-Jordan Elimination(3/3) (1)0() 0() 0() 2 1 =+ =+ =+ ∑ ∑ ∑ r k k x x x    (2)() () () 2 1 ∑ ∑ ∑ −= −= −= r k k x x x   Two important points:  Non of the three row operations alters the final column of zeros, so the system of equations corresponding to the reduced row-echelon form of the augmented matrix must also be a homogeneous system.  If the given homogeneous system has m equations in n unknowns with m<n, and there are r nonzero rows in reduced row-echelon form of the augmented matrix, we will have r<n. It will have the form:
  • 47. Theorem 1.2.1 A homogeneous system of linear equations with more unknowns than equations has infinitely many solutions.  Note: theorem 1.2.1 applies only to homogeneous system  Example 7 (3/3)
  • 48. Computer Solution of Linear System  Most computer algorithms for solving large linear systems are based on Gaussian elimination or Gauss-Jordan elimination.  Issues  Reducing roundoff errors  Minimizing the use of computer memory space  Solving the system with maximum speed
  • 50. Definition A matrix is a rectangular array of numbers. The numbers in the array are called the entries in the matrix.
  • 51. Example 1 Examples of matrices  Some examples of matrices  Size 3 x 2, 1 x 4, 3 x 3, 2 x 1, 1 x 1 [ ] [ ]4, 3 1 , 000 10 2 ,3-012, 41 03 21 2 1                 −π           −  # rows # columns row matrix or row vector column matrix or column vector entrie s
  • 52. Matrices Notation and Terminology(1/2)  A general m x n matrix A as  The entry that occurs in row i and column j of matrix A will be denoted . If is real number, it is common to be referred as scalars.             = mnmm n n aaa aaa aaa A ... ... ... 21 22221 11211  ( )ijij Aa or ija
  • 53.  The preceding matrix can be written as  A matrix A with n rows and n columns is called a square matrix of order n, and the shaded entries are said to be on the main diagonal of A. Matrices Notation and Terminology(2/2) [ ] [ ]ijnmij aa or×             mnmm n n aaa aaa aaa ... ... ... 21 22221 11211  nnaaa ,,, 2211 
  • 54. Definition Two matrices are defined to be equal if they have the same size and their corresponding entries are equal. [ ] [ ] j.andiallforifonlyandifBAthen size,samethehaveandIf ijij ijij ba bBaA == ==
  • 55. Example 2 Equality of Matrices  Consider the matrices  If x=5, then A=B.  For all other values of x, the matrices A and B are not equal.  There is no value of x for which A=C since A and C have different sizes.       =      =      = 043 012 , 53 12 , 3 12 CB x A
  • 56. Operations on Matrices  If A and B are matrices of the same size, then the sum A+B is the matrix obtained by adding the entries of B to the corresponding entries of A.  Vice versa, the difference A-B is the matrix obtained by subtracting the entries of B from the corresponding entries of A.  Note: Matrices of different sizes cannot be added or subtracted. ( ) ( ) ijijijijij ijijijijij baBABA baBABA −=−=− +=+=+ )()( )()(
  • 57. Example 3 Addition and Subtraction  Consider the matrices  Then  The expressions A+C, B+C, A-C, and B-C are undefined.       =           − − − =           − −= 22 11 , 5423 1022 1534 , 0724 4201 3012 CBA           −− −− −− =−          − =+ 51141 5223 2526 , 5307 3221 4542 BABA
  • 58. Definition If A is any matrix and c is any scalar, then the product cA is the matrix obtained by multiplying each entry of the matrix A by c. The matrix cA is said to be the scalar multiple of A. [ ] ( ) ( ) ijijij ij caAccA a == = then,Aifnotation,matrixIn
  • 59. Example 4 Scalar Multiples (1/2)  For the matrices  We have  It common practice to denote (-1)B by –B.       − =      −− =      = 1203 369 , 531 720 , 131 432 CBA ( )       − =      − −− =      = 401 123 , 531 720 1-, 262 864 2 3 1 CBA
  • 61. Definition  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 the matrix A and column j from the matrix B .Multiply the corresponding entries from the row and column together and then add up the resulting products.
  • 62. Example 5 Multiplying Matrices (1/2)  Consider the matrices  Solution  Since A is a 2 ×3 matrix and B is a 3 ×4 matrix, the product AB is a 2 ×4 matrix. And:
  • 64. Examples 6 Determining Whether a Product Is Defined  Suppose that A ,B ,and C are matrices with the following sizes: A B C 3 ×4 4 ×7 7 ×3  Solution:  Then by (3), AB is defined and is a 3 ×7 matrix; BC is defined and is a 4 ×3 matrix; and CA is defined and is a 7 ×4 matrix. The products AC ,CB ,and BA are all undefined.
  • 65. Partitioned Matrices  A matrix can be subdivided or partitioned into smaller matrices by inserting horizontal and vertical rules between selected rows and columns.  For example, below are three possible partitions of a general 3 ×4 matrix A .  The first is a partition of A into four submatrices A 11 ,A 12, A 21 ,and A 22 .  The second is a partition of A into its row matrices r 1 ,r 2, and r 3 .  The third is a partition of A into its column matrices c 1, c 2 ,c 3 ,and c 4 .
  • 66. Matrix Multiplication by columns and by Rows  Sometimes it may b desirable to find a particular row or column of a matrix product AB without computing the entire product.  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 Formulas (6)and (7)that
  • 67. Example 7 Example5 Revisited  This is the special case of a more general procedure for multiplying partitioned matrices.  If A and B are the matrices in Example 5,then from (6)the second column matrix of AB can be obtained by the computation  From (7) the first row matrix of AB can be obtained by the computation
  • 68. Matrix Products as Linear Combinations (1/2)
  • 69. Matrix Products as Linear Combinations (2/2)  In words, (10)tells us that the product A x of a 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 .  In the exercises w ask the reader to show that the product y A 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 .
  • 71. Example 9 Columns of a Product AB as Linear Combinations
  • 72. Matrix form of a Linear System(1/2) mnmnmm nn nn bxaxaxa bxaxaxa bxaxaxa =+++ =+++ =+++ ... ... ... 2211 22222121 11212111              =             +++ +++ +++ mnmnmm nn nn b b b xaxaxa xaxaxa xaxaxa  2 1 2211 2222121 1212111 ... ... ...             =                         mmmnmm n n b b b x x x aaa aaa aaa  2 1 2 1 21 22221 11211 ... ... ...  Consider any system of m linear equations in n unknowns.  Since two matrices are equal if and only if their corresponding entries are equal.  The m×1 matrix on the left side of this equation can be written as a product to give:
  • 73. Matrix form of a Linear System(1/2)  If w designate these matrices by A ,x ,and b ,respectively, the original system of m equations in n unknowns has been replaced by the single matrix equation  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
  • 74. Definition  If A is any m×n matrix, then the transpose of A ,denoted by ,is defined to be the n×m matrix that results from interchanging the rows and columns of A ; that is, the first column of is the first row of A ,the second column of is the second row of A ,and so forth. T A T A T A
  • 76. Example 10 Some Transposes (2/2)  Observe that  In the special case where A is a square matrix, the transpose of A can be obtained by interchanging entries that are symmetrically positioned about the main diagonal.
  • 77. Definition  If A is a square matrix, then the 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.