Chapter 7. Linear Algebra: Matrices, Vectors,

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1 Chapter 7. Linear Algebra: Matrices, Vectors, Determinants. Linear Systems Linear algebra includes the theory and application of linear systems of equations, linear transformations, and eigenvalue problems. Its practical application may arise from electrical networks, optimization problems, linear systems of equations, signal processing, communication theory, etc. Linear algebra makes systematic use of vectors and matrices, and to a lesser extent, determinants. The definitions of matrices and vectors and related concepts are given in Sec. 7.1 and matrix multiplication follows in Sec The topics related to the linear systems of equations are given in the remaining sections: Gauss elimination (Sec. 7.3), the important role of rank (Secs. 7.4, 7.5, 7.7), the basic existence and uniqueness problem for solutions (Sec. 7.5), vector spaces (Sec. 7.9), etc. 7.1 Matrices, Vectors: Addition and Scalar Multiplication A matrix is a rectangular array of numbers enclosed in brackets. These numbers are called entries or elements of the matrix. 6 x a b e,, [ 1 2 3],, 3 x a a a c d 2x 2 e x 1

2 EXAMPLE 1. Coefficient matrix of a linear system of equations 5x 2y+ z = 0 3 x + 4z = 0 ==> A = ; coefficient matrix EXAMPLE 2. Sales figures in matrix form General Concepts and Notations We denote matrices by capital boldface letters, A, B, C,, or A = a jk. By an m n (read m by n matrix ) matrix we mean a matrix with m rows and n columns. In the double-subscript notation for the entries, the first subscript always denotes the row and the second the column. If m= n, we call A an n n square matrix. Then its diagonal containing a11, a22,, a nn is called the main 2

3 diagonal or principal diagonal of matrix A. A matrix that is not square is called a rectangular matrix. Vectors A vector is a matrix that has only one row or one column. Then we call the matrix a row vector or column vector. We denote the vector by a lowercase boldface letter such as a, b,. [ a a a ] a = ==> row vector 1 2 n b1 b 2 b = bm ==> column vector Matrix Addition and Scaler Multiplication Equality of Matrices We say that two matrices have the same size if they are both m n. Two matrices A = a jk and B = bjk are equal, if and only if they have the same size and the corresponding entries are equal, that is, a = b, written as A= B. jk jk EXAMPLE 3. Equality of matrices a11 a A= a21 a = B = if and only if a a = 4, a = 0, = 3, a =

4 Matrix Addition Matrix addition is defined only for matrices and B of the same size. = bjk A = a jk Their sum, written A+ B, is then obtained by adding the corresponding entries. EXAMPLE 4. Addition of Matrices and Vectors A =, B = ==> A+ B = Scalar Multiplication The product of any m n matrix A = a jk and any scalar c, written ca, is obtained by multiplying each entry in A by c. Rules for Matrix Addition and Scalar Multiplication We have some rules for addition and scalar multiplication as follows, ( a) ( b) A+ B= B+ A ( U+ V) + W= U+ ( V+ W) () c ( d) A+ 0= A A+ ( A) = 0 () e c( A+ B) = ca+ cb ( f) ( c+ k) A= ca+ ka ( g) c( ka) = ( ck) A ( h) 1 A= A. 4

5 7.2 Matrix Multiplication The product C= AB of an m n matrix A = a jk and an r p matrix B = bjk is defined if and only if r = n, that is, The product C= Number of rows of 2 nd factor B = Number of columns of 1 st factor A AB has the entries n c = a b = a b + a b + + a b for jk jl lk j1 1k j2 2k jn nk l= 1 j = 1,, m k = 1,, p. EXAMPLE 1. Matrix multiplication 5

6 EXAMPLE 4. CAUTION! AB BA in general. EXAMPLE 5. CAUTION! AB = imply A= 0 or B= 0 or BA= 0. 0 does not necessarily EXAMPLE 6. CAUTION! AC = imply C= D. AD does not necessarily Properties similar to those for numbers are ( a) ( kab ) = k( AB) = A( kb) ( b) ABC ( ) = ( ABC ) () c ( A+ B) C= AC+ BC ( d) CA ( + B) = CA+ CB 6

7 Motivation of Matrix Multiplication by Linear Transformations Consider two linear systems of equations, and y = a x + a x y = a x + a x x = b w + b w = x b w b w By substituting x 1 and x 2 into the first linear system, we have y1 = a11( b11w1 + b12w2 ) + a12 ( b21w1 + b22w2 ) = c11w1 + c12w2 y = a b w + b w + a b w + b w = c w + c w ( ) ( ) ==> c11 = a11b11 + a12b21, c12 = a11b12 + a12b22 c = a b + a b, c = a b + a b Using matrix multiplication, we have Thus, y a a x y = = =, y Ax 2 a21 a 22 x 2 x b b w x= = = x Bw 2 b21 b 22 w 2 y = Ax = A( Bw) = ABw = Cw, where C = AB. 7

8 Transposition EXAMPLE 7. Transposition of a matrices and vectors , then T A= 8 0 = A 1 0 Rules for transposition are T ( a) ( A ) T ( b) ( A+ B) = A + B T () c ( ca) = ca ( d) ( AB) = A T T T T T T T = B A Special Matrices Symmetric matrices and skew-symmetric matrices are square matrices that satisfy 8

9 A A T T = A; Symmetric matrix = A; Skew-symmetric matrix Upper triangular matrices are square matrices that can have nonzero entries only on and above the main diagonal, whereas any entries below the diagonal must be zero. Lower triangular matrices can have nonzero entries only on and below the main diagonal. EXAMPLE 8. Upper and lower triangular matrices Diagonal matrices are square matrices that can have nonzero entries only on the main diagonal. If all the diagonal entries of a diagonal matrix S are equal, we call S a scalar matrix. AS = SA = ca. In particular, a scalar matrix whose entries on the main diagonal are all 1 is called an identity matrix and is 9

10 denoted by I or I n. AI = IA = A. EXAMPLE 10. Diagonal matrix, Scalar matrix, Identity matrix c D= 0 3 0, S= 0 c 0, I = c Linear Systems of Equations: Gauss Elimination The most important practical use of matrices is in the solution of linear systems of equations. We begin in this section with an important solution method, the Gauss elimination. Linear System, Coefficient Matrix, Augmented Matrix A linear system of m equations in n unknowns x,, 1 xn is a set of equations of the form a jk : coefficients of the system, b i : system responses x i : unknowns or solutions of the system x : solution vector If the b i are all zero, then the system is called a homogeneous system. 10

11 If at least one b i is not zero, then it is called a nonhomogeneous system. A linear system is called over-determined if it has more equations than unknowns, and under-determined if it has fewer equations than unknowns. If the system is homogeneous, it has at least the trivial solution x1 = x2 = = x n = 0. A linear system of two equations in three unknowns is a x + a x + a x = b , 23 3 = 2 a x a x a x b for example, 5x + 2 x x = x 4x + 3x = Matrix Form of a Linear System A linear system of equations can be written as Ax = b where The matrix is called the augmented matrix of the system., 11

12 EXAMPLE 1. Geometric interpretation. Existence of solutions Consider the equation a x + a x = b a x a x b = 2 If we interpret x1, x 2 as coordinates in the x1x2 plane, there are three possible cases for solution: (a) No solution if the lines are parallel. (b) Precisely one solution if they intersect. (c) Infinitely many solutions if they coincide. For instance, x + y = 1 x+ y= 0 Case (a) x + y = 1 x y= 0 Case (b) x + y = 1 2x+ 2y= 2 Case (c) From the given example, we may pose the following questions: (a) Does a given system have a solution? (b) Under what conditions does it have precisely one solution? (c) If it has more than one solution, how can we characterize the set of all solutions? 12

13 Gauss Elimination and Back Substitution The Gauss elimination is a standard method for solving linear systems. It is a systematic elimination process, a method of great importance that works in practice and is reasonable with respect to computing time and storage demand. EXAMPLE 2. Gauss elimination. Electrical network Solve the linear system by Gauss elimination. First Step. Elimination of x 1 Second Step. Row Exchange and Elimination of x 2 13

14 Back Substitution. Determination of x 3, x 2, x 1. Elementary Row Operations. Row-Equivalent Systems Elementary Operations for Equations (a) Interchange of two equations (b) Addition of a constant multiple of one equation to another equation (c) Multiplication of an equation by a nonzero constant c. Elementary Row Operations for Matrices (a) Interchange of two rows (b) Addition of a constant multiple of one row to another row (c) Multiplication of a row by a nonzero constant c. 14

15 We call a linear system S 1 row-equivalent to a linear system S 2 if S 1 can be obtained from S 2 by elementary row operations. Gauss Elimination: The Three Possible Cases of Systems EXAMPLE 3. A case of infinitely many solutions Solution. First Step. Elimination of x 1 Second Step. Elimination of x 2 Back Substitution. x2 = 1 x3 + 4x4 and x1 = 2 x4. Since x 3 and x 4 can be selected arbitrarily, we have infinitely many solutions. 15

16 EXAMPLE 4. A case of a unique solution Solve the linear system First Step. Elimination of x 1 Second Step. Elimination of x 2 Back Substitution. Beginning with the last equation, we have x3 = 2, x2 = 1, x1 = 1. EXAMPLE 5. A case of no solution Solve the linear system First Step. Elimination of x 1 Second Step. Elimination of x 2 16

17 ==> Since the last equation cannot be satisfied, there is no solution. Row Echelon Form and Information From It The form of the system and of the matrix in the last step of the Gauss elimination is called the echelon form. At the end of the Gauss elimination, the reduced system will in general have the form There are three possible cases for solutions. (a) No solution if r < m and one of the numbers b,, r+ 1 b m is not zero. See Example 5. (b) Precisely one solution if r = n and b,,, 1 b if present, are zero. See Example 2. (c) Infinitely many solutions if r < n and b,,, 1 b if present, are zero. See Example 3. r+ r+ m m 17

18 7.4 Linear Independence. Rank of a Matrix. Vector Space In the last section we explained the Gauss algorithm, the most important solution method for linear systems of equations. We also saw that such a system may have no solution at all, or one solution, or infinitely many solutions. So we are confronted with the questions of existence and uniqueness of solutions. In order to handle this problem in a systematic way, we introduce the rank of a matrix. Linear Independence and Dependence of Vectors Given any set of m vectors a(1),, a( m), consider the equation If the above equation holds only for c1= c2 = = c m = 0, then we call the vectors a(1),, a( m) linearly independent. Otherwise, if the equation also holds with scalars not all zero, we call the vectors linearly dependent, because then we can express one of them as a linear combination of the others. For instance, if the above equation holds with c1 0, we have a = c c a c c a. (1) 2 1 (2) m 1 ( m) EXAMPLE 1. Linear independence and dependence Check if the three vectors are linearly dependent or not. 18

19 Sol) Solve c1a1+ c2a2+ c3a3 = 0. ==> ==> [ 3 c1 0 2 c1 2c1] [ 6 c2 42c2 24 c2 54c2] [ c c c ] [ ] = c 6c + 21c = c 21c = c + 24 c = 0 2c + 54c 15c = c = 6 c, c = 1/2 c. ==> ==> Infinitely many solutions with not all zeros. ==> Linearly dependent. Rank of a Matrix Def) The maximum number of linearly independent row vectors of a matrix A = a jk is called the rank of A and is denoted by rank A. EXAMPLE 3. Rank The matrix 19

20 has rank 2, because Example 1 shows that the first two row vectors are linearly independent, whereas all three vectors are linearly dependent ==> The third row vector can be expressed as the linear combination of the first two row vectors. THEOREM 1. Row-Equivalent Matrices Row-equivalent matrices have the same rank EXAMPLE 4. Determination of rank ==> rank A = 2. THEOREM 2. Linear independence and Dependence of Vectors p vectors with n components each are linearly independent if the matrix with these vectors as row vectors has rank p, but they are linearly dependent if that rank is less than p. 20

21 THEOREM 3. (Rank in terms of column vectors) The rank of a matrix A equals the maximum number of linearly independent column vectors of A. Hence A and its T transpose A have the same rank. PROOF. Let A be an m n matrix, and let rank A = r. = a jk Then, by definition, A has r linearly independent row vectors and we call them v(1),, v( r). All row vectors a,, a m of A are linear combinations of those (1) ( ) independent vectors, say, ==> Vector equations, each of them being equivalent to n equations for corresponding components. Denoting the components of v (1) by v11,, v1 n, the components of v 2 by v,, v, etc., we have, for k = 1,, n, 21 2n This can be written 21

22 ==> The equation shows that each column vector of A is a linear combination of the r vectors on the right. ==> Hence the maximum number of linearly independent column vectors of A cannot exceed r. ==> The same conclusion applies to the transpose T A of A. T ==> Since the row vectors of A are the column vectors of T A and the column vectors of A are the row vectors of A, the maximum number of linearly independent row vectors of A cannot exceed the maximum number of linearly independent column vectors. ==> Finally the number of linearly independent column vectors must be equal to r. EXAMPLE 5. Illustration of Theorem 3 Consider the matrix A in Example 3. Since rank A = 2, the column vectors should have two linearly independent ones, and the other two should be linear combinations of them. The first two column vectors are linearly independent, and 22

23 ==> not so easy to see. THEOREM 4. Linear Dependence of Vectors p vectors with n < p are always linearly dependent. Vector Space Def) A vector space is a nonempty set V of vectors (or matrices, or functions) satisfying that for any two vectors (or matrices, or functions) a and b in V, all their linear combinations αa+ βb (α, β any real numbers for real vector space) are elements of V. Def) The maximum number of linearly independent vectors in a vector space V is called the dimension of V and is denoted by dim V. Def) A set of a maximum possible number of independent vectors in V is called a basis for V. Thus the number of vectors in a basis equals dim V. EXAMPLE 6. Vector space, dimension, basis 1) R 3 : a set of all real vectors represented by A basis is x y, < x<, < y<, < z<. z 23

24 , 1, and the dimension is 3. 2) A set of all vectors represented by x y, 1 < x <, < y < is not a vector space. Every vector space has its own zero vector. 3) 2 R 2 : a set of all 2 2 real matrices. A basis is ,,, and the dimension is 4. 4) F : a set of all real functions f ( x ). What is the dimension? 5) The set of all polynomials with the degree smaller than 3 is a vector space. A basis is { 1, x, x 2 } and the dimension is 3. Def) The set of all linear combinations of given vectors, a(1), a(2),, a( p), is called the span of these vectors. Obviously, a span is a vector space. Def) By the subspace of a vector space V, we mean a nonempty subset of V that itself forms a vector space. 24

25 EXAMPLE 7. Subspace Consider the vector space M consisting of all 2 2 matrices. 1) A set of all upper triangular matrices U represented by is the subspace of M. a 0 b d 2) A set of all diagonal matrices D given by a 0 is also a subspace of M. 3) Is the identity matrix I a subspace by itself? No. Why? 0 d 4) Zero matrix is a subspace by itself EXAMPLE 8. Vector space, Dimension, Basis The span of the three vectors in Example 1 is a vector space of dimension 2, and a basis is a(1), a (2), or a(1), a (3), etc. Four Subspaces of a Matrix Def) The span of the row vectors of an m n matrix A is n called the row space of A, which is the subspace of R. Def) The span of the columns of an m n matrix A is called the column space of A, which is the subspace of m R. Def) The null space of A is a vector space consisting of all linear combinations of the solution vectors x for Ax = 0 and is denoted by Ν( A ). 25

26 Def) The left null space of A is a vector space consisting of all linear combinations of the solution vectors y for T T A y = 0 and is denoted by Ν( A ). EXAMPLE 9. Find the bases for the column space and row space of R = Ans) The basis for row space is [ ], [ ]. ==> The row space is a subspace of R 4. The basis for column space is , 1, or 0, 4, etc ==> The column space is a subspace of R Solutions of Linear Systems: Existence, Uniqueness Using the concept of rank discussed in the previous section, we can now give necessary and sufficient conditions for the existence and the uniqueness of solutions of linear systems. THEOREM 1. Fundamental Theorem for linear systems (a) Existence. A linear system of m equations in n unknowns x, 1 x, n 26

27 has solutions if and only if rank A= rank A. Here,,. (b) Uniqueness. The system has precisely one solution if and only if rank A= rank A = n. (c) Infinitely many solutions. If rank A= rank A = r < n, the system has infinitely many solutions. All of these are obtained by determining r suitable unknowns in terms of the remaining n r unknowns, to which arbitrary values can be assigned. (d) Gauss elimination. If solutions exist, they can all be obtained by the Gauss elimination. (e) Non-existence. If rank A rank A, there is no solution. PROOF. (a) We can write the linear system in the form Ax = b or in terms of the column vectors of A c c c b (1) x1 + (2) x2 + + ( n) xn =. 27

28 If rank A= rank A, then b must be a linear combination of the column vectors of A, which means that the linear system has a solution. Conversely, if the linear system has a solution x, then b must be a linear combination of the column vectors. Hence we must have rank A= rank A. (b) If rank = rank =n, C = c(1),, c( n) is linearly independent. First, assume that there are two solution vectors such that Then A A then the set { } c c c c b (1) x1 + + ( n) xn = (1) x1 + + ( n) xn =. ( x x ) c + + ( x x ) c = (1) n n ( n) Since the set C = { c(1),, c( n) } is linearly independent, it always follows that x1 x 1 = 0,, xn x n = 0. Hence the solution is unique. Conversely, if the solution is unique, the following equation ( x x ) c + + ( x x ) c = (1) n n ( n) implies that x1 x 1 = 0,, x x = 0. This means the set C = c { (1),, c( n) } n n is linearly independent so that rank A= rank A =n. (c) If rank A= rank A = r < n, there are r linearly independent column vectors of A such that other n r column vectors are linear combinations of those independent vectors. We renumber the columns and unknowns so that the set K = { c (1),, c ( r) } is the linearly independent set. Then we have c c b (1) x1 + + ( n) xn =, 28

29 c where ( r+ 1) ( n),, c are linear combinations of the first r independent column vectors. We can rewrite the equation as c y + + c y = b (1) 1 ( r) with yj = x j + β j, j = 1,, r, where β j results from the terms c ( r+ 1) x r+ 1,, c ( n) x n. Choosing x,, r+ 1 x n fixes the β j and corresponding x j = yj β j. Since x,, r+ 1 x n can be chosen arbitrarily, we have infinitely many solutions. (d) This was proved in Sec (e) If rank A rank A, then b cannot be expressed as the linear combination of the column vectors c(1),, c( n), say, c c c b (1) x1 + (2) x2 + + ( n) xn. This means that there is no solution. The Theorem 1 is illustrated by the examples in Sec In Example 3 we have rank A= rank A = 2 < n = 4 and can choose x 3 and x 4 arbitrarily. Infinitely many solutions! In Example 4 there is a unique solution since rank A= rank A = n = 3. In Example 5 there is no solution, since rank A= 2 < rank A = 3. r 29

30 Homogeneous Linear Systems THEOREM 2. (Homogeneous linear system) (1) A homogeneous linear system always has the trivial solution, x= 0. (2) Nontrivial solutions exist if and only if rank A = r < n. (3) These solutions, together with the trivial solution, form a vector space (see Sec. 6.4) of dimension n r, called the solution space of the linear system or the null space of A. In particular, if x (1) and x (2) are solution vectors, then x= c1 x(1) + c2 x (2) with any scalars c 1 and c 2 is a solution vector. (This does not hold for nonhomogeneous system.) PROOF. (1) The first proposition is obvious. (2) For a homogeneous system, rank A= rank A. By Theorem 1. (a) and (c), nontrivial solutions exist if and only if rank A = r < n. (3) Let C denote the set of all solution vectors. If x (1) and x (2) are solution vectors, then Ax = 0 and (1) Ax(2) = 0. For any scalars c 1 and c, 2 A( c1 x(1) + c2 x(2) ) = c1 Ax(1) + c2 Ax(2) = 0. That is, c1 x(1) + c2 x (2) C. Thus the set of all solution vectors is a vector space. Assume that rank A = r < n and the first r column vectors are independent. By the Fundamental Theorem we can 30

31 choose n r suitable unknowns, x,,, r+ 1 xn in an arbitrary fashion, and every solution is obtained by determining x,,, 1 x r with x,,. r+ 1 xn Let a basis for the null space is y. ( j) Then y ( j) is easily obtained by choosing x r + j = 1 and the other x,, r+ 1 x zero; the corresponding n x,, 1 xr are then determined. Thus the solution space has dimension n r. Theorem 2 states that rank A+ nullity A = n where n is the number of unknowns or columns. THEOREM 3. (Homogeneous linear system with fewer equations than unknowns) A homogeneous linear system with fewer equations than unknowns, m< n, always has nontrivial solutions since rank A = r m< n. Nonhomogeneous Linear Systems THEOREM 4. (Nonhomogeneous system) If a nonhomogeneous linear system of equations has solutions, then all these solutions are of the form x= x0 + x h where x 0 is any fixed solution of the nonhomogeneous system and x h is the general solution of the corresponding homogeneous system. PROOF. Let x be any given solution and x 0 an arbitrarily chosen solution. Then Ax = b, Ax0 = b. Therefore, 31

32 ( ) A x x = Ax Ax = ==> x x 0 is a solution of the corresponding homogeneous equation, say, x h. Hence all solutions can be expressed as x= x + x 0 h. 7.6 Second- and Third-Order Determinants Determinants were originally introduced for solving linear systems by Cramer s rule. They also have important engineering applications in eigenvalue problems, differential equations (Wronskian), vector algebra, and so on. Second-Order Determinants A determinant of second order is defined by EXAMPLE 1 a a D= = = a a a a det A a21 a = = EXAMPLE 2. Cramer s rule for linear systems of 2 equations Derive a solution formula for the system a x + a x = b a x a x b = 2 Solution. (1) a22 (2) a12 ( a11a22 a12a21) x1 = ba 1 22 a12b2 (2) a (1) a a a a a x = a b ba. ( )

33 Assuming that D= a11a22 a12a21 0, we have Cramer s rule for n = 2 equations ==> If the system is homogeneous ( b1 = b2 = 0) and D 0, it has only the trivial solution, and if D = 0, it also has nontrivial solution. Third-Order Determinants A determinant of third order is defined by ==> Expansion by the first column EXAMPLE 3. Cramer s rule for linear systems of 3 equations For linear systems of three equations in three unknowns Cramer s rule is D1 D2 D3 x1 =, x2 =, x3 = ( D 0) D D D with 33

34 7.7 Determinants. Cramer s Rule Determinant of Any Order n A determinant of order n is written and is defined for n 2 by or n n j + k jk jk jk jk k= 1 k= 1 D= a C = a ( 1) M ( j = 1,2,, n) n n j + k jk jk jk jk j= 1 j= 1 D= a C = a ( 1) M ( k = 1,2,, n) where C jk ( ) j+ k = 1 M. jk M jk is called the minor of a jk in D, and cofactor of a jk in D. C jk the 34

35 EXAMPLE 2. Expansions of a third-order determinant ==> Expansion by the first row. The expansions by the third column is EXAMPLE 3. Determinant of a triangular matrix ==> Can you formulate a little theorem on determinants of triangular matrices? Of diagonal matrices? General Properties of Determinants THEOREM 1. (Behavior of determinant under elementary row operations) (a) Interchange of two rows multiplies the value of the determinant by -1. (b) Addition of a multiple of a row to another row does not change the value of the determinant. (c) Multiplication of a row by c multiplies the value of the determinant by c. 35

36 PROOF. (a) For a simple case of n = 2 a b = ad bc, c d = bc ad. c d a b (b) Add c times row i to row j. Then the entries in row j are ajk + caik. Let D be the new determinant. Expanding D by row j, we have D = D + cd = D + cd Note that row j = row i in D. 2 Interchanging the two rows gives D2 by (a), but should also gives D 2 back, which means D 2 = 0. Finally D = D. (c) Expand the determinant by the row that has been multiplied. n Caution! det ( ka) = k det A. EXAMPLE 4. Evaluation of determinants by reduction to triangular form 36

37 THEOREM 2. (Further properties of determinant) (a)-(c) in Theorem 1 hold also for columns. (d) Transposition does not change a determinant. (e) In case of a zero row or column, the determinant becomes zero. (f) Proportional rows or columns render the value of a determinant zero. In particular, a determinant with two identical rows or columns has the value zero. Rank in Terms of Determinants Theorem 3. (Rank in terms of determinants) An n n square matrix A has rank n if and only if det A 0. PROOF. If det A 0, the echelon form  also has nonzero determinant. This means that  has rank n. Since rank A= rank A ˆ, it follows that rank A = n. Conversely, if rank A = n, the echelon form  has nonzero 37

38 determinant. If  has nonzero determinant, then A also has nonzero determinant, that is, det A 0. THEOREM 4. Cramer s Theorem (a) If a linear system of n equations in the same number of unknowns x, 1 x, n has a nonzero coefficient determinant, the system has precisely one solution, which is given by D1 D2 Dn x1 =, x2 =,, xn = D D D where D k is the determinant obtained from D by replacing the k th column by the column with entries b,,. 1 b n (b) Hence if the system is homogeneous and D 0, it has only the trivial solution. If D = 0, the homogeneous system also has nontrivial solutions. 7.8 Inverse of a Matrix: Gauss-Jordan Elimination The inverse of an n n matrix A = a jk is denoted by 1 A and is an n n matrix such that 1 1 AA = A A = I where I is the n n identity matrix., 38

39 If A has an inverse, then A is called a nonsingular matrix. If A has no inverse, then A is called a singular matrix. If A has an inverse, the inverse is unique. Pf) Assume that both B and C are inverses of A. Then, AB = I, AC = I ==> AB AC = A( B C) = 0. ==> ( ). 1 A A B C = B C= 0 B= C. THEOREM 1. (Existence of the inverse) 1 The inverse A of an n n matrix A exists if and only if rank A = n, hence if and only if det A 0. PROOF. Consider the linear system Ax = b. If rank A = n, then by the Fundamental Theorem, the linear system has a unique solution x. The back substitution following the Gauss elimination shows that x j are linear combinations of those of b. Thus we can write x= Bb. Substitution into the given system yields Ax = A( Bb) = ( AB) b = b for any b. ==> AB = I. Similarly, substituting Ax = b into x= Bb, we have Finally x= Bb= B( Ax) = ( BA) x for any x. ==> BA = I. 1 B= A exists. Conversely, if the inverse exists, then 39

40 A Ax x A b 1 = = 1. Since the inverse is unique, the solution vector x is unique. By the Fundamental Theorem in Sec. 6.5, A must have rank n. Determination of the Inverse, Gauss-Jordan Elimination Consider a nonsingular n n matrix A. To obtain the inverse of A, we can form n linear systems e where (1) ( ) Ax = e,, Ax = e, (1) (1) ( n) ( n),, e n are the columns of the n n identity matrix. Combining the above n equations into a single equation, we have AX = I where the unknown matrix have the columns x(1),, x( n). To solve AX = I for X, we can apply the Gauss elimination to A= [ A I], U H where U is a upper triangular resulting [ ] matrix. The Gauss-Jordan elimination reduces [ U H ] to [ I K ] by additional row operations. Since [ I K ] corresponds to the linear system IX = K, we can obtain the 1 inverse matrix X= A by taking K from [ I K ]. EXAMPLE 1. Inverse of a matrix. Gauss-Jordan elimination Find the inverse of Solution. Applying the Gauss elimination to 40

41 we have ==> [ U H ] Now apply the Gauss-Jordan steps to reduce U to I. ==> The last three columns constitute A 1. Check: 41

42 Useful Formulas for Inverses THEOREM 2. (Inverse of a matrix) The inverse of a nonsingular n n matrix A = a jk is given by where A jk is the cofactor of a jk in det A. In particular, the inverse of a11 a12 A = a21 a is 22 A 1 a a = det A a21 a 11 EXAMPLE 2. Inverse of a 2 2 matrix A= 2 4 A = = ,. EXAMPLE 3. Further Illustration of Theorem 2 Find the inverse of

43 Solution. We have det A = 10, and Thus. Diagonal matrices A = ajk, ajk = 0 when j k, have an 1 inverse if and only if all a jj 0. Then A is also diagonal with entries 1 a11,, 1 a nn. EXAMPLE 4. Inverse of a diagonal matrix Inverse of the product of matrices ( AC) = C A ( AC PQ) = Q P C A. PROOF. 1 AC( AC) = I ==> 1 1 CAC ( ) = A ==> ( AC) = C A. 43

44 The inverse of a matrix 1 A is A. That is, 1 1 ( A ) = A. Matrix Multiplication. Cancellation Laws Matrix multiplication differs from the multiplication of numbers. (1) It is not commutative; that is, in general AB BA. (2) AB = 0 does not generally imply A= 0 or B= 0 or BA = 0. (3) AC = AD does not generally imply C= D. THEOREM 3. (Cancellation law) Let A, B, C be n n matrices. Then: (a) If rank A = n and AB = AC, then B= C. (b) If rank A = n, then AB = 0 implies B= 0. (c) If A is singular, so are BA and AB. PROOF. (a) Since rank A = n, there exists both sides by A 1, we have A 1. Thus multiplying 1 1 A ( AB) = A ( AC ) ==> B= C. (b) Multiplying both sides by A 1, we have 1 1 A ( AB) = A 0 ==> B= 0. (c) If A is singular, then Ax = 0 has nontrivial solutions, 44

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