Lecture Notes in Linear Algebra

Lecture Notes in Linear Algebra Dr. Abdullah Al-Azemi Mathematics Department Kuwait University February 4, 2017

Contents 1 Linear Equations and Matrices 1 1.2 Matrices............................................ 1 1.2.1 Special types of matrices:.............................. 2 1.2.2 Matrix Operation:.................................. 4 1.3 Dot Product and Matrix Multiplication........................... 6 1.4 Properties of Matrix Operations............................... 13 1.4.1 Properties of matrix-scalar multiplication...................... 13 1.4.2 Properties of Matrix Addition............................ 14 1.4.3 Properties of Matrix Multiplication......................... 14 1.4.4 Properties of Matrix Transpose........................... 15 1.4.5 Symmetric and Skew-Symmetric Matrices..................... 16 1.6 Solutions of Linear System of Equations.......................... 20 1.6.1 Solving System of Linear Equations......................... 22 1.7 The Inverse of a Matrix................................... 31 2 Determinants 43 2.1 Determinants......................................... 43 2.2 Cofactor Expansion and Applications............................ 53 3 Vectors in R n 65 3.2 Vectors in R n......................................... 65 i

ii CONTENTS 3.2.1 Vectors Operations:.................................. 65 3.2.2 The Dot Product of Vectors:............................. 66 3.2.3 Properties of Dot Product:.............................. 68 3.2.4 The Angle Between Two Vectors.......................... 68 3.2.5 Principle of Unit Vectors in R n........................... 73 5 Real Vector Spaces 77 5.1 Real Vector Spaces...................................... 77 5.2 Real Vector Subspaces.................................... 81 5.3 Linear Independence..................................... 88 5.4 Basis and Dimension..................................... 94 5.5 Homogenous System..................................... 101 5.6 The Rank of a Matrix and its Application......................... 103 6 Diagonalization 111 6.1 Eigenvalues and Eigenvectors................................ 111 6.2 Diagonalization........................................ 116 6.2.1 Case 1: Diagonalization of a Matrix with Distinct Eigenvalues.......... 118 6.2.2 Case 2: Diagonalization of a Matrix with Repeated Eigenvalues......... 119 4 Vectors in R n 125 4.1 The Cross Product in R n................................... 125 4.2 Planes and Lines in R 3.................................... 132

Chapter 1 Linear Equations and Matrices 1.2 Matrices Definition 1.2.1 An m n matrix A is a rectangular array of m n real numbers arranged in m horizontal rows and n vertical columns. That is, a 11 a 12... a 1j... a 1n a 21 a 22... a 2j... a 2n...... A = a i1 a i2... a ij... a in...... a m1 a m2... a mj... a mn In the matrix A above, we have: a 11, a 12,..., a mn are called elements (or entries) of the matrix. The entry a ij lies in the intersection of row i and column j. The size (or order) of A is m by n, written as m n. The matrix is called a square matrix if m = n. We write M m n for the class of all real matrices of size m n. If a matrix A is of size 1 n, then we say that A is a vector of size (length) n and we write A R n. In addition, if A is of size n 1, we say that A R n. Example 1.2.1 A matrix A M 2 3 and another matrix B M 3 3. A = 2 3 2 4 1 1 2 3 B = 3 0 1 4 2 0 1 4 1 3 3 1

2 Chapter 1. Linear Equations and Matrices An m n matrix can be represented by the following ways: A = (a ij ) for 1 i m and 1 j n, row 1 A = row 2, where row i = [a i1 a i2... a in ], for 1 i m, or. row m A = [col 1 col 2... col n], where col j = a 1j a 2j. a mj for 1 j n. Example 1.2.2 List all of the entries of the matrix A = (a ij ) R 3, for 1 i 3 and 1 j 3, where i + j, if i > j a ij = 0, if i = j i j, if i < j We list the entries of A in the following way: a 11 a 12 a 13 0 1 2 A = a 21 a 22 a 23 = 3 0 1 a 31 a 32 a 33 4 5 0 1.2.1 Special types of matrices: 1. The zero matrix, say O, is an m n matrix whose entries are all equal to 0. 0 0 O =..... 0 0

1.2. Matrices 3 2. The diagonal matrix, D, is a square matrix in R n, where D = (d ij ) so that d ij = 0 for all i j. d 11 0 D =... 0 d nn 3. The scalar matrix is a diagonal matrix S whose all diagonal entries are equal to some a scalar c. c 0 S =... 0 c 4. The identity matrix (or unit matrix) is a scalar matrix I n whose all diagonal entries are equal to 1. 1 0 I n =... 0 1 5. The lower triangular matrix, L, is a square matrix in R n, where L = (l ij ) so that l ij = 0 for all i < j. l 11 0 L =.... l n1... l nn 6. The upper triangular matrix, U, is a square matrix in R n, where U = (u ij ) so that u ij = 0 for all i > j. u 11... u 1n U =.... 0 u nn Remark 1.2.1. Two m n matrices A = (a ij ) and B = (b ij ) are said to be equal if a ij = b ij for all 1 i m and 1 j n. Example 1.2.3 Find the values of a, b, c, and d if a + b c d c + d = 1 2. a b 2 1

4 Chapter 1. Linear Equations and Matrices Since the two matrices are equal. Then a + b = 1 a b = 1 c + d = 2 c d = 2 + = 2a = 2 = a = 1 and hence b = 0. + = 2c = 0 = c = 0 and hence d = 2. 1.2.2 Matrix Operation: Matrix transpose Definition 1.2.2 If A = (a ij ) is an m n matrix, then the n m matrix A T 1 i m, is called the transpose of A. = (a ji ), where 1 j n and Example 1.2.1. If A = a b c d e f 2 3 a, then A T = b c d e f 3 2. Example 1.2.4 ( ) Q#6 pg. 20 If A is an n n lower triangular matrix, then A T is an n n upper triangular matrix. Since A is a lower triangular, then a ij = 0 for all i < j. Therefore, the transpose of A, A T = (a ji ) has entries a ji = 0 if j > i and thus it is an upper triangular. Matrix addition and substraction To add two m n matrices A = (a ij ) and B = (b ij ), we must have the size(a) = size(b). Then, A ± B = C, where C = (c ij ) with c ij = a ij ± b ij, for all 1 i m and 1 j n.

1.2. Matrices 5 Example 1.2.5 If possible, find A + B and A T B, where 2 1 A = 3 0 1 1 3 2, and B = 1 2 4 5 1 1 Clearly, A+B is not possible as they have different sizes. On the other hand, size(a T ) = size(b), and A T B = 2 3 1 1 0 1 2 3 1 2 4 5 1 1 2 3 2 3 = 1 1 5 4 1 0. 2 3. Example 1.2.6 ( ) Q#5 pg.20 Show that the sum and the difference of two lower triangular matrices is a lower triangular matrix. Let A = (a ij ) and B = (b ij ) be two n n matrices so that both are lower triangular matrices. Thus, a ij = 0 and b ij = 0 for all i < j. If (c ij ) = C = A ± B with c ij = a ij ± b ij, then c ij = 0 for all i < j. Therefore, C is again a lower triangular matrix. Remark 1.2.1 The sum and the difference of two upper triangular matrices is an upper triangular matrix. Scalar multiplication If A = (a ij ) is an m n matrix and c is a real number, then c A = (c a ij ) for all 1 i m and 1 j n.

6 Chapter 1. Linear Equations and Matrices 1.3 Dot Product and Matrix Multiplication Definition 1.3.1 The dot (inner) product of the n-vectors x = (x 1 x 2 x n ) and y = (y 1 y 2 y n ) is defined by x y = x 1 y 1 + x 2 y 2 + + x n y n = n x i y i. (1.3.1) i=1 An n-vector x = (x 1 x 2 x n ) can be written as x = (x 1, x 2,, x n ). Example 1.3.1 Find x y if x = (1, 0, 2, 1) and y = (3, 5, 1, 4). Clearly, x y = 3 + 0 + ( 2) + ( 4) = 3. Definition 1.3.2 Let A = (a ij ) be an m p matrix and B = (b ij ) be a p n matrix. Then, the product of A and B, (AB), is the m n matrix C = (c ij ) = row i (A) col j (B) where p c ij = a i1 b 1j + a i2 b 2j + + a ip b pj = a ik b kj, for 1 i m and 1 j n. k=1 Note that the entry (i, j) of the product AB equals to the dot product of row i (A) by the col j (B). Remark 1.3.1 To get the product AB, we must have # of columns in A = # of rows in B.

1.3. Dot Product and Matrix Multiplication 7 Example 1.3.2 If A = a 11 a 12 a 13 a 21 a 22 a 23 2 3 b 11 b 12, and B = b 21 b 22 b 31 b 32 3 2, then AB = a 11b 11 + a 12 b 21 + a 13 b 31 a 11 b 12 + a 12 b 22 + a 13 b 32 a 21 b 11 + a 22 b 21 + a 23 b 31 a 21 b 12 + a 22 b 22 + a 23 b 32 2 2 Example 1.3.3 Find AB, AB T, and A T B if possible, where A = 1 0 2, and B = 4 1 1. 1 2 0 2 5 1 Clearly, AB is not possible to compute since # columns in A is not the same as # rows of B. AB T : excercise. 1 1 A T B = 0 2 4 1 1 4 + 2 1 + 5 1 + 1 = 4 10 2 2 5 1 = 6 6 0 4 10 2 2 0 8 2 2 8 2 2 Remark 1.3.2 In general, AB BA. For instance, consider A = 1 0 and B = 0 0. Check it yourself!! 1 0 1 1 It is always true that AB = BA if A = I n or that A = B.

8 Chapter 1. Linear Equations and Matrices Example 1.3.4 Let A = [ ] 1 2 0. Find all values of c so that c A A T = 15. [ ] Note that ca A T = c 1 2 0 1 2 = 5c. Therefore, 5c = 15 and hence c = 3. 0 Example 1.3.5 1 2 3 Let A = 4 2 1, and B = 1 4 3 1. Compute the (3, 2)-entry of AB. 0 1 2 2 2 The (3, 2)-entry = row 3 (A) col 2 (B) = [0 1 2] [4 1 2] = 1 4 = 5. Definition 1.3.3 Let A be an m n matrix and x R n (be an n-vector), then a 11 a 12 a 1n x 1 a 11 x 1 + a 12 x 2 + + a 1n x n Ax = a 21 a 22 a 2n x 2 = a 21 x 1 + a 22 x 2 + + a 2n x n........ a m1 a m2 a mn x n a m1 x 1 + a m2 x 2 + + a mn x n row 1 (A) x a 11 a 12 a 1n = row 2 (A) x = x 1 a 21 + x 2 a 22 + + x n a 2n.... row m (A) x a m1 a m2 = x 1 col 1 (A) + x 2 col 2 (A) + + x n col n (A). a mn That is, Ax is a linear combination of columns of A and the entries of x are the coefficients.

1.3. Dot Product and Matrix Multiplication 9 Example 1.3.6 Write the following product as a linear combination of the columns of the first matrix. 1 3 2 1 3 1 = 3 2 2 + ( 2) 3 1. 4 2 4 2 Example 1.3.7 ( ) Q#9 pg.38 If A is an m p matrix and B is a p n matrix, then (a) Show that col j (AB) = A col j (B). (b) Show that row i (AB) = ( row i (A) ) B. (a) Let C = AB. Clearly for 1 j n, c 1j row 1 (A) col j (B) row 1 (A) col j (AB) = col j (C) = c 2j = row 2 (A) col j (B) = row 2 (A) col j (B) = A col j (B).... c mj row m (A) col j (B) row m (A) (b) Let C = AB. Clearly for 1 i m, i1 c i2 row i (AB) = row i (C) = [c ] c in [ ] = row i (A) col 1 (B) row i (A) col 2 (B) row i (A) col n (B) = row i (A) [ col 1 (B) col 2 (B) ] col n (B) = ( row i (A) ) B. TRUE or FALSE: If a matrix B has a column of zeros, then the product AB has a column of zeros as well. reason: Assume that column j of B is a column of zero. Then, col j (AB) = A col j (B) = 0. (TRUE).

10 Chapter 1. Linear Equations and Matrices Example 1.3.8 1 2 Let A = 3 4 1 5 of columns of A. 3 2 and B = 2 3 4 3 2 1 Clearly, 1 2 col 1 (AB) = A col 1 (B) = 3 4 2 1 = 2 3 3 + 3 2 4 1 5 1 5 1 2 col 2 (AB) = A col 2 (B) = 3 4 3 1 = 3 3 2 + 2 2 4 1 5 1 5 1 2 col 3 (AB) = A col 3 (B) = 3 4 4 1 2 = 4 3 1 + 1 4 1 5 1 5 2 3. Find the columns of AB as a linear combination Example 1.3.9 Compute A 2017, where A = 1 1. 0 1 Here, we will compute some powers of the matrix A to look for a pattern. That is, A 2 = AA = 1 1 1 1 = 1 2, A 3 = A 2 A = 1 2 1 1 = 1 3 0 1 0 1 0 1 0 1 0 1 0 1 A 4 = A 3 A = 1 3 1 1 = 1 4, A n = A n 1 A = 1 (n 1) 1 1 = 1 n. 0 1 0 1 0 1 0 1 0 1 0 1 Therefore, A 2017 = A 2016 A = 1 2017. 0 1

1.3. Dot Product and Matrix Multiplication 11 Matrix Form of Linear System of Equations: Given a linear system of equation on n variables as in (1.3.2) a 11 x 1 + a 12 x 2 + a 13 x 3 + + a 1n x n = b 1 a 21 x 1 + a 22 x 2 + a 23 x 3 +. + a 2n x n = b 2 a m1 x 1 + a m2 x 2 + a m3 x 3 + + a mn x n = b m (1.3.2) we transform this system into the matrix form as in (1.3.3) a 11 a 12 a 1n a 21. a 22.. a 2n. a m1 a m2 a mn x 1 x 2. x n = b 1 b 2. b m (1.3.3) We say that the form of (1.3.3) is the matrix form A x = B, where x is called the vector of unknowns. Moreover, we define the augmented form as [ A B ]. (1.3.4) Example 1.3.10 Write the following system of linear equations in the matrix form and in the augmented form. 2x z + 3w = 1 x + 5y = 1 x + y + z = 0 The matrix form is: 2 0 1 3 1 5 0 0 1 1 1 0 x y z w = while the augmented form is given by 2 0 1 3 1 1 5 0 0 1 1 1 1 0 0. 1 1 0

12 Chapter 1. Linear Equations and Matrices Exercise 1.3.1 1. Let A = 1 2. Compute A 2 + I 2. 0 1 2. ( ) Q#6 pg.31 : The product of two upper triangular matrices is an upper triangular matrix. The proof is NOT required. 1 2 3. Let A = 4 5 and B = 0 1 3. Express the third row of AB as a linear 3 1 4 3 6 combination of the rows of B. 1 1 4. Let A = 0 2 and B = 4 1 1. Find AB and express the second column of AB 2 5 1 2 0 as a linear combination of the columns of A.

1.4. Properties of Matrix Operations 13 1.4 Properties of Matrix Operations In what follows, let A, B, and C be matrices of appropriate sizes, and let r, s R. 1.4.1 Properties of matrix-scalar multiplication 1. r (s A) = (rs) A, 2. (r + s) A = r A + s A, 3. r (A + B) = r A + r B, 4. A(r B) = (r A)B = r (AB). Example 1.4.1 ( ) Q#32 pg.51 If the m n linear system A x = B has more than one solution, then it has infinitely many solutions. Let y and z be two solutions to the system A x = B. Then, for r, s R, we have A y = B r A y = r B A(r y) = r B. (1.4.1) Similarly, we have A(s z) = s B. (1.4.2) Adding up (1.4.1) and (1.4.2), we got A(r y + s z) = (r + s)b. Therefore, (r y + s z) is a solution to the system whenever, r + s = 1. The result follows.

14 Chapter 1. Linear Equations and Matrices TRUE or FALSE: If x 1 and x 2 are two solutions for Ax = B, then 3 2 x 1 2x 2 is also a solution. reason: Since 3 2 2 = 1 2 1. (FALSE). 1.4.2 Properties of Matrix Addition 1. A + B = B + A, 2. (A + B) + C = A + (B + C), 3. there is a unique matrix O such that O + A = A + O = A, the matrix O is called the additive identity, 4. there is a unique matrix D such that A + D = D + A = O. In particular, D = A is called the additive inverse of A. 1.4.3 Properties of Matrix Multiplication 1. A(BC) = (AB)C, 2. A(B + C) = AB + AC, 3. (A + B)C = AC + BC, 4. there is a unique (identity) matrix I such that AI = IA = A. Remark 1.4.1 Let p, q, and r be rational numbers and let A, B M m n, then 1. A p A q = A p+q, 2. (A p ) q = A pq, 3. AB BA,

1.4. Properties of Matrix Operations 15 4. (AB) p A p B p, 5. for a, b R: ab = 0 only if a = 0 or b = 0. This is not true ingeneral for matrices. For instance, if A = [ 0 1 0 1 ], and B = [ 1 1 0 0 ], then AB = O where A and B are both non-zero matrices. 1.4.4 Properties of Matrix Transpose 1. (A T ) T = A, 2. (A + B) T = A T + B T, 3. (AB) T = B T A T, 4. for any real number r R, (ra) T = ra T, Note That: Property 3 can be generalized to (ABC XY Z) T = Z T Y T X T C T B T A T. Example 1.4.2 Let A = 2 0 and B = 1 4. Find A T, B T, AB, (AB) T, and B T A T. 1 1 3 5 Clearly, A T = 2 1 and B T = 1 3. 0 1 4 5 AB = 2 + 0 8 + 0 = 2 8 (AB) T = 2 2 1 3 4 5 2 1 8 1 Moreover, B T A T = 2 + 0 1 3 = 2 2 8 + 0 4 5 8 1

16 Chapter 1. Linear Equations and Matrices Example 1.4.3 ( ) T #9 pg.51 Find a 2 2 matrix B O and B I2 so that AB = BA if A = 1 2 0 1 Let B = a c b. Therefore, d AB = BA a + 2c c b + 2d = a d c 2a + b, 2c + d thus a = a + 2c c = 0 and b + 2d = 2a + b a = d. Let s, t R so that a = s and b = t to get B = s t. Note that s and t can not be both 0 since B O, and the case of t = 0 and 0 s s = 1 can NOT be hold since B I. 1.4.5 Symmetric and Skew-Symmetric Matrices Definition 1.4.1 Let A M n n. A is called symmetric matrix whenever A T = A, and it is called skewsymmetric matrix whenever A T = A. For instance, A = matrix. 1 2 3 2 4 5 3 5 0 is a symmetric matrix; where B = 0 2 is a skew-symmetric 2 0 Example 1.4.4 ( ) T #23 pg.51 Let A and B be two symetric matrices. Show that (a) A + B is symmetric. (b) AB is symmetric ( ) if and only if AB = BA.

1.4. Properties of Matrix Operations 17 Since A and B are both symmetirc, then A T = A and B T = B. (a): Clearly, (A + B) T = A T + B T = A + B. Therefore, A + B is symmetric. (b): We first prove ( ). Let AB be symmetirc. Thus AB = (AB) T = B T A T = BA which shows that AB = BA. To show the other direction ( ), assume that AB = BA. Thus, (AB) T = B T A T = BA = AB which shows that AB is symmetric. Example 1.4.5 If A and B are skew-symmetric matrices and AB = BA, then AB is symmetric. It is given that A T = A and B T = B; thus, (AB) T = B T A T = ( B)( A) = BA = AB. Therefore, AB is symmetric. Example 1.4.6 ( ) T #26 pg.51 Show that A T A and AA T are both symmetric matrices. It is clear that (A T A) T = A T (A T ) T = A T A which shows that A T A is symmetric. In addition, (AA T ) T = (A T ) T A T = AA T shows that AA T is also symmetric. Example 1.4.7 ( ) T #27 pg.51 Let A Mn n. Show that (a) A T + A is symmetirc. (b) A A T is skew-symmetric. [(a)]: (A T + A) T = (A T ) T + A T = A + A T = A T + A. Therefore, A T + A is symmetric. [(b)]: (A A T ) T = A T (A T ) T = A T A = (A A T ). Therefore, A A T is skew-symmetric.

18 Chapter 1. Linear Equations and Matrices Example 1.4.8 ( ) T #28 pg.51 Let A Mn n. Then A can be written as A = S + K, where S is symmetric matrix and K is skew-symmetric matrix. From Example 1.4.7, we know that A T + A is symmetric and that A A T is skew-symmetric. If S = 1 2 (AT + A) which is symmetric and K = 1 2 (A AT ) which is skew-symmetric, then S + K = 1 2 (AT + A) + 1 2 (A AT ) = 1 2 AT + 1 2 A + 1 2 A 1 2 AT = A.

1.4. Properties of Matrix Operations 19 Exercise 1.4.1 1. Let A = 1 1. 0 1 (a) Find A 1977. (b) Find all matrices B such that AB = BA. 2. Let A = 1 2 1 2 1 2 1 2. Find A 100. 3. Show that if A and B are symmetric matrices, then AB BA is a skew-symmetric matrix. 4. Let A be 2 2 skew-symmetric matrix. If A 2 = A, then A = 0. 1 5. Let A = 3. Find all constants c R such that (ca) T (ca) = 5. 1 3 2 1 3 6. Let A = 0 4 1. Find a symmetric matrix S and a skew symmetric matrix K 1 2 3 such that A = S + K.

20 Chapter 1. Linear Equations and Matrices 1.6 Solutions of Linear System of Equations Definition 1.6.1 A matrix A M m n is said to be in the reduced row echelon form (r.r.e.f. for short) if it satisfies the following conditions: 1. the row of zeros (if any) should be at the bottom of A, 2. the leading (first) entry of a non-zero row should be 1, 3. the leading entry of row i + 1 should be on the right of the leading entry of row i, 4. the columns that contain a leading entry 1, all its other entries are zeros. The matrix A is said to be in the row echelon form (r.e.f.) if it does not satisfy the last Condition. Example 1.6.1 Here aresome examples of some matrices to clarify the previous conditions: (1) A = 1 0 0 3 0 0 0 1 3 0 is r.r.e.f. 0 0 0 0 0 1 3 0 4 (2) B = 0 0 0 0 is not r.r.e.f. since it has a row of zeros in the second row. 0 0 1 2 1 0 4 5 (3) C = 0 2 2 3 is not r.r.e.f. since the leading entry in the second row is not 1. 0 0 1 2 1 0 0 2 (4) D = 0 0 1 3 is not r.r.e.f. since the leading entry of the third row is on the left 0 1 0 0 of the leading entry of the second row. Switch 2 nd and 3 rd rows to get the r.r.e.f. form.

1.6. Solutions of Linear System of Equations 21 Given a matrix A M m n, we define the following elementary row operations: 1. interchanging a row by another row, 2. multiplying a row by a non-zero scalar, 3. adding a multiple of a row to another row. That is, replace a i1, a i2,, a in by a i1 + ca k1, a i2 + ca k2,, a in + ca kn, for 1 i, k n. Definition 1.6.2 An m n matrix A is called row equivalent to an m n matrix B if B can be obtained by applying a finite sequence of elementary row operation to A. In this case, we write A B. Properties of matrix equivalence: For any m n matrices A, B, and C we have 1. A A, 2. A B B A, 3. A B and B C A C. Theorem 1.6.1 Every non-zero m n matrix is row equivalent to a unique matrix in the r.r.e.f. Example 1.6.2 1 2 3 9 Find the matrix B which is in r.r.e.f. and is row equivalent to A = 2 1 1 8. 3 0 1 3 A r 2 2r 1 r 2 r 3 3r 1 r 3 1 2 3 9 0 5 5 10 0 6 10 24 1 5 r 2 r 2 1 2 r 3 r 3 1 2 3 9 0 1 1 2 0 3 5 12 r 1 2r 2 r 1 r 3 3r 2 r 3 1 0 1 5 0 1 1 2 0 0 2 6

22 Chapter 1. Linear Equations and Matrices 1 2 r 3 r 3 1 0 1 5 0 1 1 2 0 0 1 3 r 1 r 3 r 1 r 2 r 3 r 2 1 0 0 2 0 1 0 1 = B (in r.r.e.f.) A. 0 0 1 3 Example 1.6.3 2 4 6 0 Find the matrix B which is in r.r.e.f. and is row equivalent to A = 1 2 4 0 1 3 3 1 A 1 2 r 1 r 1 1 2 3 0 1 2 4 0 1 3 3 1 1 0 3 2 r 1 3r 3 r 0 1 0 1 1 0 0 1 0 1 2 3 0 1 2 3 0 r 2 r 1 r 2 r 2 r 0 0 1 0 3 0 1 0 1 r 3 r 1 r 3 0 1 0 1 0 0 1 0 1 0 0 2 0 1 0 1 = B (in r.r.e.f.) A. 0 0 1 0 r 1 2r 2 r 1 1.6.1 Solving System of Linear Equations Solving Ax = B Ax = B where B O (non-homogenous) Ax = O (homogenous) unique sol no sol infinite sol trivial sol non-trivial sol Remark 1.6.1 Guass-Jordan method for solving Ax = B: 1. Find the r.r.e.f. of augmented matrix [ A B ], 2. Solve the reduced system. Note that, the solution of the reduced system is the solution of the original one.

1.6. Solutions of Linear System of Equations 23 Theorem 1.6.2 Let Ax = B and Cx = D be two linear systems of equations. If [ A B ] [ C D ], then the two system have the same solutions. (α) Solving Non-Homogenous System Ax = B with B O Example 1.6.4 Solve the following system using the Guass-Jordan method. x + 2y + 3z = 9 2x y + z = 8 3x z = 3 1 2 3 9 1 2 3 9 r 2 2r 1 r 2 2 1 1 8 1 5 0 5 5 10 r 2 r 2 1 2 3 9 r 1 2r 2 r 0 1 1 2 1 r 3 3r 1 r 3 1 2 3 0 1 3 0 6 10 24 r 3 r 3 r 3 3r 2 r 3 0 3 5 12 1 0 1 5 1 2 0 1 1 2 r 3 r 3 1 0 1 5 1 0 0 2 r 1 r 3 r 1 0 1 1 2 0 1 0 1 r 2 r 3 r 2 0 0 2 6 0 0 1 3 0 0 1 3 2 Therefore, the reduced system is: x = 2, y = 1, and z = 3. Thus, x = 1 is a unique 3 solution to the system. Remark 1.6.2 The system Ax = B has a unique solution if and only if A I.

24 Chapter 1. Linear Equations and Matrices Example 1.6.5 Solve the following system using the Guass-Jordan method. x 1 + x 2 x 3 + 4x 4 = 1 + x 2 3x 3 + 4x 4 = 0 2x 1 + 2x 2 2x 3 + 8x 4 = 2 1 1 1 4 1 1 1 1 4 1 r 3 2r 1 r 3 0 1 3 4 0 0 1 3 4 0 2 2 2 8 2 0 0 0 0 0 Therefor, the reduced system is: r 1 r 2 r 1 1 0 2 0 1 0 1 3 4 0 0 0 0 0 0 x 1 + 2x 3 = 1 x 2 3x 3 + 4x 4 = 0 x 1 = 1 2x 3 x 2 = 3x 3 4x 4 We now fix x 3 = r and x 4 = t for r, t R to get the following infinite many solutions: x 1 = 1 2r x 2 = 3r 4t 1 2r x = 3r 4t for all r, t R. r t Remark 1.6.3 The system Ax = B has infinity many solutions if the number of unknowns is more than the number of equations. Example 1.6.6 Solve the following system using the Guass-Jordan method. x 1 + x 2 x 3 + 4x 4 = 1 + x 2 3x 3 + 4x 4 = 0 2x 1 + 2x 2 2x 3 + 8x 4 = 3

1.6. Solutions of Linear System of Equations 25 1 1 1 4 1 1 1 1 4 1 r 3 2r 1 r 3 0 1 3 4 0 0 1 3 4 0 2 2 2 8 2 0 0 0 0 1 At this point, it can be seen that the third row suggests that 0 = 1 which is not possible. Therefore, this system has no solution. Remark 1.6.4 The system Ax = B has no solution whenever A a matrix with a row of zeros while [ A B ] a matrix with no rows of zeros. Example 1.6.7 Consider the system: Find all values of a so that the system has: (a) a unique solution. (b) infinite many solutions. (c) no solution. x + y + z = 2 x + 2y z = 2 x + 2y + (a 2 5)z = a In this kind of questions, it is not necessary to get the r.r.e.f. So, we will try to focus on the last row which contains the term a as follows: 1 1 1 2 1 1 1 2 r 2 r 1 r 2 1 2 1 2 0 1 2 0 1 2 a 2 r 3 r 1 r 3 5 a 0 1 a 2 6 a 2 At this point, we have: r 3 r 2 r 3 (a) a unique solution if a 2 4 0 a ±2 a R\{ 2, +2}. 1 1 1 2 0 1 2 0 0 0 a 2 4 a 2 (b) infinite solutions if { a 2 4 = 0 and a 2 = 0 } { a = ±2 and a = +2 } a = +2. (c) no solution if { a 2 4 = 0 and a 2 0 } { a = ±2 and a +2 } a = 2.

26 Chapter 1. Linear Equations and Matrices (β) Solving Homogenous System Ax = O Example 1.6.8 Solve the following system using the Guass-Jordan method. x + 2y + 3z = 0 x + 3y + 2z = 0 2x + y 2z = 0 1 2 3 0 1 2 3 0 1 0 5 0 1 r 2 r 1 r 2 1 3 2 0 r 1 2r 2 r 0 1 1 0 1 11 0 1 1 0 r 3 r 3 r 3 2r 1 r 3 r 3 +3r 2 r 3 2 1 2 0 0 3 8 0 0 0 11 0 1 0 5 0 1 0 0 0 r 1 5r 3 r 1 0 1 1 0 0 1 0 0 r 2 +r 3 r 2 0 0 1 0 0 0 1 0 0 Therefore, the reduced system is: x = 0, y = 0, and z = 0. Thus, x = 0 is a trivial solution. 0 Remark 1.6.5 The system Ax = O has a trivial solution whenever A I. Example 1.6.9 Solve the following system using the Guass-Jordan method. x 1 + x 2 x 3 + 4x 4 = 0 + x 2 3x 3 + 4x 4 = 0 2x 1 + 2x 2 2x 3 + 8x 4 = 0

1.6. Solutions of Linear System of Equations 27 1 1 1 4 0 1 1 1 4 0 r 3 2r 1 r 3 0 1 3 4 0 0 1 3 4 0 2 2 2 8 0 0 0 0 0 0 Therefor, the reduced system is: r 1 r 2 r 1 1 0 2 0 0 0 1 3 4 0 0 0 0 0 0 x 1 + 2x 3 = 0 x 2 3x 3 + 4x 4 = 0 x 1 = 2x 3 x 2 = 3x 3 4x 4 We now fix x 3 = r and x 4 = t for r, t R to get the following non-trivial solution: x 1 = 2r x 2 = 3r 4t 2r x = 3r 4t for all r, t R. r t Remark 1.6.6 The system Ax = O has a non-trivial solution if the number of unknowns is greater than the number of equations in the reduced system. Definition 1.6.3 Linear systems with at least one solution are called consistent, and linear system with no solutions are called inconsistent. Example 1.6.10 Discuss the consistency of the following homogenous system: 1 1 1 0 1 1 3 0 1 1 a 2 5 0 The system has: r 2 r 1 r 2 r 3 r 1 r 3 x + y z = 0 x y + 3z = 0 x + y + (a 2 5)z = 0 1 1 1 0 0 2 4 0. 0 0 a 2 4 0

28 Chapter 1. Linear Equations and Matrices (a) trivial solution when a 2 4 0 a ±2 a R\{±2}, (b) non-trivial solution when a 2 4 = 0 a = ±2. Example 1.6.11 Discuss the consistency of the following non-homogenous system: x + 3y + kz = 4 2x + ky + 12z = 6 1 3 k 4 r 2 2r 1 r 2 1 3 k 4. 2 k 12 6 0 k 6 12 2k 2 The system: (a) can not have a unique solution (not equivalent to I because it is not square matrix!!), (b) has infinite solutions when k 6, (c) has no solutions when k = 6. Example 1.6.12 ( ) T #11 pg.89 Let u and v be two solutions of the homogenous system Ax = O. Show that ru + sv (for r, s R) is a solution to the same system. Since u and v are two solution of Ax = O, then Au = O and Av = O (respectively). We need to show that A(ru + sv) = O. Clearly, A(ru + sv) = A(ru) + A(sv) = r(au) + s(av) = O + O = O. Therefore, ru + sv is also a solution to the system.

1.6. Solutions of Linear System of Equations 29 Example 1.6.13 ( ) T #12 pg.89 Let u and v be two solutions of the non-homogenous system Ax = B. Show that u v is a solution to the homogenous system Ax = O. Since u and v are two solution of Ax = B, then Au = B and Av = B (respectively). We need to show that A(u v) = O. Clearly, A(u v) = Au Av = B B = O. Therefore, u v is a solution to the system Ax = O.

30 Chapter 1. Linear Equations and Matrices Exercise 1.6.1 1 1. Let A = 3. Find all constants c R such that (ca) T (ca) = 5. 1 3 2. Find the reduced row echelon form (r.r.e.f.) of the following matrix: 2 4 6 0 1 2 4 0 1 3 3 1 1 0 3 3 3. Let A = 0 1 1 1 and B = 1 2 3 1 0 2 0 a 2 + 1 system Ax = B has at least one solution. 1 0. Find all value(s) of a such that the 0 a + 2 4. Solve the system x + y + z = 0 x + 2y + 3z = 0 x + 3y + 4z = 0 x + 4y + 5z = 0. 5. Find all value(s) of a for which the system x y + (a + 3)z = a 3 a 7 x + ay az = a 2(a 1)y + (a 2 + 2)z = 8a 14. has (a) no solution, (b) unique solution, and (c) infinite many solutions. 6. Show that if C 1 and C 2 are solutions of the system Ax = B, then 4C 1 3C 2 is also a solution of this system.

1.7. The Inverse of a Matrix 31 1.7 The Inverse of a Matrix Unless otherwise specified, all matrices in this section are considered to be square matrices. Definition 1.7.1 A matrix A M n n is called non-singular or invertable if there exists a matrix B M n n such that A B = B A = I n. In particular, B is called the inverse of A and is denoted by A 1. If there is no such B, we say that A is singular which means that A has no inverse. Example 1.7.1 Here is an example of a 2 2 matrix along with its inverse. 1 1 3 1 = 3 1 1 1 = 1 0 = I 2. 2 3 2 1 2 1 2 3 0 1 Theorem 1.7.1 If a matrix has an inverse, then its inverse is unique. Proof: Assume that A M n n is a matrix with two inverses B and C, then B = B I n = B (A C) = (B A) C = I n C = C. Therefore, A has a unique inverse. Remark 1.7.1 How to find A 1 for a given A M n n : 1. form the augmented matrix [ A I n ],

32 Chapter 1. Linear Equations and Matrices 2. find the r.r.e.f. of [ A I n ], say [ C B ]: (a) if C = I n, then A is non-singular and A 1 = B, (b) if C I n, then C has a row of zeros and A is singular matrix with no inverse. Example 1.7.2 Find, if possible, A 1 for A = 1 1 2 3 1 1 1 0 r 2 2r 1 r 2 1 1 1 0 r 1 r 2 r 1 1 0 3 1. 2 3 0 1 0 1 2 1 0 1 2 1 Therefore, A is non-singular and A 1 = 3 1. 2 1 Example 1.7.3 1 0 1 Find, if possible, A 1 for A = 1 1 2 2 0 1 1 0 1 1 0 0 1 0 1 1 0 0 r 2 r 1 r 2 1 1 2 0 1 0 r 0 1 1 1 1 0 3 r 3 2r 1 r 3 2 0 1 0 0 1 0 0 1 2 0 1 1 0 1 1 0 0 1 0 0 1 0 1 r 1 r 3 r 1 0 1 1 1 1 0 0 1 0 3 1 1 r 2 r 3 r 2. 0 0 1 2 0 1 0 0 1 2 0 1 1 0 1 Therefore, A is non-singular and A 1 = 3 1 1. 2 0 1

1.7. The Inverse of a Matrix 33 Example 1.7.4 1 2 3 Find, if possible, A 1 for A = 1 2 1 5 2 3 1 2 3 1 0 0 1 2 3 1 0 0 r 2 r 1 r 2 1 2 1 0 1 0 r 3 3r 2 r 0 4 4 1 1 0 3 r 3 5r 1 r 3 5 2 3 0 0 1 0 12 12 5 0 1 1 2 3 1 0 0 0 4 4 1 1 0. 0 0 0 2 3 1 At this point, we can conclude that this matrix is singular with no inverse because of the fact that the third row in the first part is a zero row. In particular, A 1 does not exist. Remark 1.7.2 Let A be an n n matrix: 1. if A is row equivalent to I n, then it is non-singular, Example 1.7.5 A = 3 0 is row equivalent to I 2. 0 5 2. if A is row equivalent to a matrix with a row of zeros, then it is singular. Example 1.7.6 A = 1 1 is row equivalent to a matrix with a row of zeros. 0 0 TRUE or FALSE:

34 Chapter 1. Linear Equations and Matrices (FALSE). 1 0 is a non- 0 1 If A and B are singular matrices, then so is A + B. reason: A = 1 0 and B = 0 0 are two singular matrices, while A + B = 0 0 0 1 singular. If A and B are non-singular matrices, then so is A + B. (FALSE). reason: A = 1 0 and B = 1 0 are two non-singular matrices, while A + B = 0 0 is a 0 1 0 1 0 0 singular. Theorem 1.7.2 If A is non-singular, then A 1 is non-singular and (A 1 ) 1 = A. Proof: If A is non-singular, then A A 1 = A 1 A = I, or A 1 A = A A 1 = I. Therefore, A 1 is non-singular and (A 1 ) 1 = A. Theorem 1.7.3 If A and B are non-singular matrices, then A B is non-singular and (A B) 1 = B 1 A 1. Proof: Since both A and B are non-singular, then both A 1 and B 1 exist. Thus A B (B 1 A 1 ) = A (B B 1 ) A 1 = A I A 1 = A A 1 = I. A B (B 1 A 1 ) = A (B B 1 ) A 1 = A I A 1 = A A 1 = I. Thus, A B is non-singular and (A B) 1 = B 1 A 1. Theorem 1.7.4 If A is non-singular, then A T is non-singular and (A T ) 1 = (A 1 ) T. Proof: A is non-singular, and thus A A 1 = I = A 1 A. Taking the transpose for the whole thing, we get

1.7. The Inverse of a Matrix 35 (A A 1 ) T = I = (A 1 A) T } (A 1 ) T A T = I = A T (A 1 ) T A T is non-singular and (A T ) 1 = (A 1 ) T. Inverses and System of Linear Equations Remark 1.7.3 Assume that A is a non-singular n n matrix. Then the non-homogenous system Ax = B has a unique solution since A 1 A x = A 1 B I n x = A 1 B x = A 1 B Remark 1.7.4 Assume that A is a non-singular n n matrix. Then the homogenous system Ax = O has a trivial solution since A 1 A x = A 1 O x = O Remark 1.7.5 Assume that A is a singular n n matrix. Then the homogenous system Ax = O has a non-trivial solution. Example 1.7.7 Solve the following system A x = B 1 0 1 x 1 1 2 y = 0 1 2 0 1 z 1

36 Chapter 1. Linear Equations and Matrices We can find the inverse of A as in Example 1.7.3, to get A 1 = 1 0 1 3 1 1 2 0 1 Then, using Remark 1.7.3, x = x y z = A 1 B = 1 0 1 3 1 1 2 0 1 0 1 1 = 1 0 1 which is a unique solution. Example 1.7.8 Solve the following system A x = O 1 0 1 1 1 2 2 0 1 x y z = 0 0 0 We can find the inverse of A as in Example 1.7.3, to get A 1 = 1 0 1 3 1 1 2 0 1 Then, using Remark 1.7.4, x = x y z = A 1 O = 1 0 1 3 1 1 2 0 1 0 0 0 = 0 0 0 which is a trivial solution.

1.7. The Inverse of a Matrix 37 Example 1.7.9 Solve the following system A x = O 1 2 3 x 1 2 1 y = 0 0 5 2 3 z 0 Since A is singular as we have seen in Example 1.7.4, we can not follow Remark 1.7.4. Therefore, we use the usual Guass-Jordan method to solve this system, knowing that this system has nontrivial solution as in Remark 1.7.5. 1 2 3 0 1 2 1 0 5 2 3 0 1 0 1 0 0 1 1 0 0 0 0 0 t x z = 0 Thus, the reduced system is. Let z = t R. Therefore, x = y z = 0 t t Remark 1.7.6 For any A M n n, we have 1. A is non-singular A I n A 1 exits A x = B has a unique solution A x = O has a trivial solution. 2. A is singular A a matrix with a row of zeros A 1 Does Not Exist A x = O has a non-trivial solution.

38 Chapter 1. Linear Equations and Matrices Exercise 1.7.1 1 0 1 1. Let A T = 1 1 0. Find (2A) 1. 2 0 1 [ Hint: If c is a nonzero constant, then what is (ca) 1?] 2. Let A = 1 2, and B = 1 1 2. Find C if (B T + C)A 1 = B T. 1 3 3 2 0 1 2 0 1 1 0 3. Let A 1 = 0 1 1 and B = 1 0 0. Find C if A C = BT. 2 0 1 0 1 1 [ Hint: Consider multiplying both sides from the left with A 1.] 1 2 3 [ ] 4. Let A 1 = 0 1 2. Find all x, y, z R such that x y z A = 0 0 1 0 1 1 1 2 1 5. Let A = 1 1 0 and B = 1 3 0. 0 1 2 1 1 1 (a) find B 1. (b) Find C if A = B C. [ ] 1 2 3. 6. Let A, B, and C be n n matrices such that D = AB + AC is non-singular. (a) Find A 1 if possible. (b) Find (B + C) 1 if possible. 7. Let A be an n n matrix such that A 3 = O. Prove or disprove that (I n A) 1 = A 2 + A + I n.

1.7. The Inverse of a Matrix 39 CHAPTER 1 REVISION Revision of Chapter 1... Have Fun :-) 1. Let A = 5 a 10. Find all values of a such that A 2 = A. 4 2. Find the matrix A satisfying 2A T 3 1 2 1 1 T = 2 3. 1 2 3. Let A = 1 3 and X = 5. Compute, if possible, X T A T, A T X T, X T X, and XX T. 2 4 3 1 1 4. Let A = 0 2 and B = 4 1 1. 2 5 1 2 0 (a) Find, if possible, AB and A T B. (b) Express the second column of AB as a linear combination of the columns of A. (c) Express the second row of AB as a linear combination of the rows of B. 5. Let A be a square matrix. Show that if A = 2A T, then A = O. 1 1 2 6. Let A = 2 1 1. 1 1 0 x 1 (a) Find all solution, if any, to the system A y = 8. z 7 x (b) Find all solution to the homogenous system A y = 0 0. z 0

40 Chapter 1. Linear Equations and Matrices 7. Find all values of a such that the following system has at least one solution. x + y + z = 1 x + 3y + 2z = 2 x + y + (a 2 3)z = a 1. 1 2 1 8. Let A = 1 3 0, B = 1 3 2 0, and C = 2 1 1 1. 1 1 1 1 2 0 3 (a) Find A 1. (b) Find a matrix X such that AX + B = C. (c) Is it possible to find a matrix Y such that Y A + B = C? Explain. 9. Let A and B be skew symmetric matrices. Show that AB is symmetric iff AB = BA. 10. Let A and B be two square matrices such that AB = O. If B is non-singular, find A. 11. Find a 2 2 matrix C such that A T CB = I 2, where A 1 = 1 1 and B 1 = 2 1. 1 0 0 1 1 0 2 12. Let P = 1 1 0 and Q = 1 2 2 1 1 2. 1 1 1 0 1 1 (a) Compute PQ. (b) Is P singular or non-singular matrix? Explain. 13. Consider the system 3x +2y z w = 0 x y +3z 5w = 0 5y +2z 4w = 0 x y z +w = 0, (a) Determine whether (1, 0, 2, 1) is a solution to the above system. (b) Does the above system has infinitely many solutions? Explain. (c) Let A be the coefficient matrix of the above system. Is A singular? Explain.

1.7. The Inverse of a Matrix 41 1 2 3 [ ] 14. Let A 1 = 0 1 2. Find all x, y, z R such that x y z A = 0 0 1 [ ] 1 2 3. 15. Let A, B, and C be n n matrices such that D = AB + AC is non-singular. (a) Find A 1 if possible. (b) Find (B + C) 1 if possible. 16. Let A be an n n matrix such that A 3 = O. Prove or disprove that (I n A) 1 = A 2 + A + I n. 17. Let A be an n n matrix and let U and V be two n 1 column matrices. Show that if AU = AV and U V, then A is singular. 18. Solve the following linear system x +3y z +w = 1 2x y 2z +2w = 2 3x +y z +w = 1. 1 2 4 19. Let A = 3 2 0 and C = 2 5. 4 3 9 1 (a) If possible, compute (AC + I 3 ) T. (b) Write AC as a linear combination of the columns of A. (c) Find a 3 3 matrix B, if any, such that A 1 B + I 3 = A. 20. Let A = 1 2 and B = 1 1 2. 1 3 3 2 0 (a) If possible, find A 2 (3A) T. (b) Find C, if any, such that (B T + C)A 1 = B T.

42 Chapter 1. Linear Equations and Matrices

Chapter 2 Determinants 2.1 Determinants Definition 2.1.1 Let S = {1, 2,, n} be a set of integers from 1 to n. A rearrangement of j 1, j 2,, j n of elements of S is called permutation of S. Example 2.1.1 For S = {1, 2}, we have 2 permutations: 1 2, and 2 1. While, for S = {1, 2, 3}, we have the following 6 permutations: 1 2 3, 1 3 2, 2 1 3, 2 3 1, 3 1 2, and 3 2 1. In general, for S = {1, 2,, n}, we have n! = n(n 1)(n 2) 3 2 1 permutations. Definition 2.1.2 A permutation j 1 j 2 j n of the set S = {1, 2,, n} is said to have an inversion if a larger integer j r preceedes a smaller one j s for r, s S. A permutation is called even with a positive + sign or odd with a negative sign according to whether the total number of inversions in it is even or odd, respectively. 43

44 Chapter 2. Determinants Example 2.1.2 If S = {1, 2, 3}, then permutation #inversions even-odd sign inversions 1 2 3 0 even + 1 2 1 3 2 3 1 3 2 1 odd 1 3 1 2 3 2 2 1 3 1 odd 2 1 2 3 1 3 2 3 1 2 even + 2 3 2 1 3 1 3 1 2 2 even + 3 1 3 2 1 2 3 2 1 3 odd 3 2 3 1 2 1 TRUE or FALSE: The permutation 5 2 1 3 4 has a positive sign. (FALSE). reason: The number of inversions is 5 which are 5 2, 5 1, 5 3, 5 4, and 2 1. So, this permutation has an odd number of inversions and a negative sign. Definition 2.1.3 Let A = (a ij ) M n n. Then, the determinant of A, denoted by det(a) or A, is det(a) = (±)a 1j1 a 2j2 a 3j3 a njn where the summation ranges over all permutations j 1 j 2 j n of the set S = {1, 2,, n}. The sign is taken as + or according to the sign of the permutation. Example 2.1.3 Compute the determinant of A = a 11 a 12. a 21 a 22 Using the definition, we have det(a) = (±)a 1j1 a 2j2, where j 1 j 2 is a permutation of S = {1, 2}. Thus, j 1 j 2 {1 2, 2 1} and det(a) = + a 11 a 22 a 12 a 21.

2.1. Determinants 45 Example 2.1.4 a 11 a 12 a 13 Compute the determinant of A = a 21 a 22 a 23. a 31 a 32 a 33 Using the definition, we have det(a) = (±)a 1j1 a 2j2 a 3j3, where j 1 j 2 j 3 is a permutation of S = {1, 2, 3}. Thus, j 1 j 2 j 3 {1 2 3, 1 3 2, 2 1 3, 2 3 1, 3 1 2, 3 2 1} and det(a) = + a 11 a 22 a 33 a 11 a 23 a 32 a 12 a 21 a 33 + a 12 a 23 a 31 + a 13 a 21 a 32 a 13 a 22 a 31. Moreover, this formula can be found by taking the sum of the positive product of the diagonal entries and the negative product of the anti-diagonal entries in the following matrix: + + + a 11 a 12 a 13 a 11 a 12 a 21 a 22 a 23 a 21 a 22 a 31 a 32 a 33 a 31 a 32 Example 2.1.5 Find det(a) for A = 2 3. 1 5 A = (2)(5) (1)(3) = 10 3 = 7.

46 Chapter 2. Determinants Example 2.1.6 Find det(a) for 2 0 1 A = 2 3 4. 5 1 2 We calculate the det(a) using the following diagram: + + + 2 0 1 2 0 2 3 4 2 3 5 1 2 5 1 In particular, A = (2)(3)(2) + (0)(4)(5) + (1)( 2)(1) (1)(3)(2) (2)(4)(1) (0)( 2)(2) = 12 + 0 + ( 2) 15 8 0 = 13. Properties of Determinants: 1. A = A T. 2. If B is obtained from A by interchanging two rows (or two columns), then B = A. 3. If A has two equal rows (or columns), then A = 0. 4. If A has a row (or a column) of zeros, then A = 0. 5. If B is obtained from A by adding a multiple of a row (or a column) to another row (or column, respectively), then A = B. 6. If B is obtained from A by multiplying a row (or a column) by a scalar c, then B = c A.

2.1. Determinants 47 Remark 2.1.1 In general, for A M n n, we have c A = c n A. 7. The determinant of lower triangular, upper triangular, diagonal, and scalar matrices can be computed as follows. Theorem 2.1.1 The determinant of lower triangular, upper triangular, diagonal, and scalar matrices equal the product of elements of the diagonal. 8. For any A, B M n n, we have A B = A B. 9. Theorem 2.1.2 If A is an n n non-singular matrix, then A 0 and A 1 = 1. This statement A suggests that if A = 0, then A is singular matrix. Proof: Since A is non-singular, A 1 exists, then AA 1 = I n (take the determinant for both sides) A A 1 = In A A 1 = 1. Thus, A 0 and A 1 = 1 A. Example 2.1.7 Let A, B M n n. Show that, if A B = I n, then B A = I n.

48 Chapter 2. Determinants If A B = I n, then A 0 and B 0. A 1 A B = A 1 I n I n B = A 1 B A = A 1 A = I n Thus, A B = I n = B A. Example 2.1.8 ( ) T #16 pg.195 Show that if A Mn n is skew-symmetric matrix and n is odd, then A = 0. Since A is skew-symmetric, then A T = A and taking the determinant for both sides A T = A A T = ( 1) n, A where n is odd and ( 1) n = 1. A = A T = A. Therefore, A = A which means that A = 0. TRUE or FALSE: If A, B M n n with A = B, then A = B. reason: I 2 I 2 while I2 = 1 and I2 = ( 1) 2 I2 = 1. (FALSE). Example 2.1.9 ( ) T #5 pg.194 If A B = 0, then either A = 0 or B = 0. Clearly, A B = A B = 0. Then, A = 0 or B = 0.

2.1. Determinants 49 Example 2.1.10 ( ) T #8 pg.194 If A B = In, then A 0 and B 0. A B = I n = A B = 1 which implies that zero. A B = 1 and both A and B can not be Example 2.1.11 ( ) T #9(b) pg.194 Show that if A 1 = A T, then A = 1 or A = 1. Since A 1 = A T, we have A 1 = A T 1 A = A A 2 = 1 A = ±1. Example 2.1.12 ( ) T #10 pg.194 Show that if A is non-singular and A 2 = A, then A = 1. Since A is non-singular, A 1 exists and A 2 = A A = I n A = In A = 1. Example 2.1.13 ( ) T #10 pg.210 Show that for any A, B, C Mn n, if A B = A C and A 0, then B = C. Since A 0, then A 1 exists and thus AB = AC A 1 AB = A 1 AC B = C.

50 Chapter 2. Determinants Example 2.1.14 Let A, B M 3 3 with A = 2 and B = 2. Find 2 A 1 (B 2 ) T. 2 A 1 (B 2 ) T = 2 3 A 1 B 2 1 = 8 B B A = 8 ( 1 ) ( 2)( 2) = 16. 2 Example 2.1.15 Evaluate the determinant of the matrix A via reduction to triangular form, where 2 2 5 A = 0 3 1. 1 4 2 The idea here is to compute the determinant of A by applying the method of Theorem 2.1.1. A 2 2 5 1 4 2 1 4 2 = r 1 r 3 0 3 1 ( 1) r 3 2r 1 r 0 3 1 3 ( 1) c 2 c 0 3 1 3 1 4 2 2 2 5 0 10 1 1 2 4 1 2 4 r ( 1)( 1) 3 r 2 r 3 }{{} 0 1 3 0 1 3 = (1)(1)( 13) = 13. =+1 0 1 10 0 0-13 Example 2.1.16 a 1 b 1 c 1 a 1 a 2 a 3 Let a 2 b 2 c 2 = 3. Find 2b 1 2b 2 2b 3. a 3 b 3 c 3 3a 1 c 1 3a 2 c 2 3a 3 c 3 We solve this question in the following two ways (either one works):

2.1. Determinants 51 solution 1: a 1 b 1 c 1 a 1 a 2 a 3 a 1 a 2 a 3 transpose ( 2)r 2 r 2 a 2 b 2 c 2 = 3 b 1 b 2 b 3 = 3 ( 1)r 3 r 3 2b 1 2b 2 2b 3 = a 3 b 3 c 3 c 1 c 2 c 3 c 1 c 2 c 3 ( 1)( 2)3 r a 1 a 2 a 3 3+3r 1 r 3 2b 1 2b 2 2b 3 = 6. 3a 1 c 1 3a 2 c 2 3a 3 c 3 solution 2: a 1 a 2 a 3 a 1 a 2 a 3 r 2b 1 2b 2 2b 3 = ( 2)( 1) 3 +3r 1 r b 1 b 2 b 3 3 3a 1 c 1 3a 2 c 2 3a 3 c 3 c 1 3a 1 c 2 3a 2 c 3 3a 3 a 1 a 2 a 3 a 1 b 1 c 1 transpose 2 b 1 b 2 b 3 2 a 2 b 2 c 2 = 2(3) = 6. c 1 c 2 c 3 a 3 b 3 c 3

52 Chapter 2. Determinants Exercise 2.1.1 0 4 4 0 1. Let D = 1 1 2 0. Evaluate D. 1 3 5 3 0 1 2 6 [ Hint: D = 12.] 2. Let A = a + e b + f, B = a b, and C = e f. Show that A = B + C. c d c d c d a 1 b 1 c 1 b 1 b 2 b 1 3b 3 3. Given that a 2 b 2 c 2 = 7, evaluate a 1 a 2 a 1 3a 3. a 3 b 3 c 3 c 1 c 2 c 1 3c 3 [ Hint: The result is 21.] a 1 a 2 a 3 4. Let A = b 1 b 2 b 3, and B = 2a 3 2a 2 2a 1 b 3 a 3 b 2 a 2 b 1 a 1. If A = 4, find B. c 1 c 2 c 3 c 3 + 3b 3 c 2 + 3b 2 c 1 + 3b 1 [ Hint: The result is 8.] 7 1 0 3 5. Let A be a matrix with A 1 = 2 0 0 0. Find det (A). 1 3 5 4 6 2 0 5 1 2 4 6. Find all values of α for which the matrix 1 3 9 is singular. 1 α α 2 7. Let A and B be two n n matrices such that A is invertible and B is singular. Prove that A 1 B is singular. 8. If A and B are 2 2 matrices with det (A) = 2 and det (B) = 5, compute 3 A 2 (AB 1 ) T.

2.2. Cofactor Expansion and Applications 53 2.2 Cofactor Expansion and Applications Definition 2.2.1 Let A = (a ij ) be an n n matrix. Let M ij be the (n 1) (n 1) submatrix of A obtained by deleting the i th row and the j th column of A. The determinant, det(m ij ), is called the minor of a ij. The cofactor denoted by A ij of a ij is defined by A ij = ( 1) i+j det(m ij ). Moreover, the cofactor matrix denoted by coef(a) = (A ij ) where 1 i, j n. Example 2.2.1 1 2 1 Compute the minors, cofactors, and the coef(a) for A = 0 1 1. 3 1 2 det(m 11 ) = det(m 12 ) = det(m 13 ) = det(m 21 ) = det(m 22 ) = det(m 23 ) = minors: 1 1 1 2 0 1 3 2 0 1 3 1 2 1 1 2 1 1 3 2 1 2 3 1 cofactors: = -1 = A 11 = ( 1) 2 ( 1) = -1 = 3 = A 12 = ( 1) 3 (3) = -3 = -3 = A 13 = ( 1) 4 ( 3) = -3 = -5 = A 21 = ( 1) 3 ( 5) = 5 = -5 = A 22 = ( 1) 4 ( 5) = -5 = -5 = A 23 = ( 1) 5 ( 5) = 5

54 Chapter 2. Determinants det(m 31 ) = 2 1 = -3 = A 31 = ( 1) 4 ( 3) = -3 1 1 det(m 32 ) = 1 1 = -1 = A 32 = ( 1) 5 ( 1) = 1 0 1 det(m 33 ) = 1 2 = 1 = A 33 = ( 1) 6 (1) = 1 0 1 1 3 3 Therefore, coef(a) = 5 5 5. 3 1 1 Theorem 2.2.1 Let A = (a ij ) be an n n matrix with cofactors A ij for all 1 i, j n, then for each 1 i n we have A if i = k, a i1 A k1 + a i2 A k2 + + a in A kn = 0 if i k. and for each 1 j n A if j = k, a 1j A 1k + a 2j A 2k + + a nj A nk = 0 if j k. Example 2.2.2 Compute the determinant of A by using the cofactor expansion method, where A is given in Example 2.2.1. Computing the determinant of A using the first row of A, we have A = a 11 A 11 + a 12 A 12 + a 13 A 13 = (1)( 1) + (2)( 3) + (1)( 3) = 10. Moreover, we can get the same result using (for instance) the second column of A, A = a 12 A 12 + a 22 A 22 + a 32 A 32 = (2)( 3) + (1)( 5) + (1)(1) = 10.

2.2. Cofactor Expansion and Applications 55 Note that, if we choose the first row of A and the cofactors of (for instance) the second row of A, we get a 11 A 21 + a 12 A 22 + a 13 A 23 = (1)(5) + (2)( 5) + (1)(5) = 0. Example 2.2.3 0 3 0 1 Compute 2 1 1 2 using the cofactor expansion method. 0 0 1 2 0 1 0 1 We solve this problem by choosing the row or the column with the maximum number of zeros as this will ease the computations: Choosing the first column, we have A = a11 A 11 + a 21 A 21 + a 31 A 31 + a 41 A 41 0 3 0 1 = (0) A 11 + (0) A 21 + 2 1 1 2 0 0 1 2 0 1 0 1 =O =O {}}{{}}{ 3 0 1 3 0 1 3 0 1 + = ( 2) 0 1 2 + 0 1 2 + 0 1 2 1 0 1 1 0 1 1 0 1 = ( 2) (+1) 3 1 = ( 2) [3 1] = ( 2)(2) = 4. 1 1 + (0) A 41 = ( 2) choose 2 nd column {}}{ 3 0 1 0 1 2 1 0 1

56 Chapter 2. Determinants Definition 2.2.2 Let A = (a ij ) M n n. The adjoint matrix of A, denoted by adj(a), is given by adj(a) = (coef(a)) T. Example 2.2.4 Compute the adjoint of A, where A as given in Example 2.2.1: 1 2 1 A = 0 1 1. 3 1 2 Here, we use the result of Example 2.2.1, where 1 3 3 coef(a) = 5 5 5. 3 1 1 Therefore, adj(a) = (coef(a)) T = 1 5 3 3 5 1. 3 5 1 Theorem 2.2.2 Let A = (a ij ) M n n. Then, A adj(a) = adj(a) A = A In. Proof:

2.2. Cofactor Expansion and Applications 57 We first prove that A adj(a) = A In, A adj(a) = a 11 a 12 a 1n A 11 A 21 A n1 a 21 a 22 a 2n A 12 A 22 A n2............ = a n1 a n2 a nn A 1n A 2n A nn A 0 0 0 0 A 1 0 0 0 0....... =... 0 A 1........ 0. 0.. 0 0 0 A 1 = A In. The proof of adj(a) A = A I n is similar. Example 2.2.5 Let adj(a) = 1 2. Find A. 4 6 Assume that A = x z y. Then coef(a) = w w y z. Therefore, x adj(a) = (coef(a)) T = w z y = 1 2. x 4 6 Thus, x = 6, y = 2, z = 4, and w = 1, and A = 6 2 4 1

58 Chapter 2. Determinants Theorem 2.2.3 If A is non-singular, then A 1 = 1 A adj(a). Proof: A is non-singular so that A 0 and A 1 exists. Then, by Theorem 2.2.2, A adj(a) = A In multiply from left by A 1 A 1 A adj(a) = A 1 A In adj(a) = A A 1 A 1 = 1 A adj(a). Remark 2.2.1 Let A = a c b be any 2 2 non-singular matrix. Thus, ad bc 0, and d A 1 = 1 A adj(a) = 1 d ad bc c b. a Confirmation: A A 1 = 1 ad bc ad bc cd cd ab + ab = 1 0 = I 2. cb + ad 0 1 Example 2.2.6 Find the inverse of A = 1 2. 4 6 By Remark 2.2.1, we have A 1 = 1 6 2. The reader can check the solution by confirming that either (which leads to both) A A 1 = I 2 or A 1 A = I 2 2 4 1.

2.2. Cofactor Expansion and Applications 59 Example 2.2.7 ( ) T #8 pg.210 Show that if A is a non-singular n n matrix, then adj(a) = A n 1. Given that A 0, we use Theorem 2.2.2 to get A A adj(a) = A I n A adj(a) = A In adj(a) = A n I n adj(a) = A n 1 Example 2.2.8 ( ) T #12 pg.210 Show that if A is non-singular, then adj(a) is non-singular and ( adj(a) ) 1 = 1 A A = adj(a 1 ). Since A is non-singular, A 0. We use Theorem 2.2.2 to get adj(a) A = A adj(a) = A I n ( ) 1 adj(a) A A = 1 A A ( adj(a) ) = I n which shows that adj(a) is non-singular and Now, if we replace A by A 1 in Theorem 2.2.2, we get ( ) 1 1 adj(a) = A. (2.2.1) A adj(a 1 ) A 1 = A 1 adj(a 1 ) = A 1 In.

60 Chapter 2. Determinants Thus, adj(a 1 ) A 1 = adj(a 1 ) A 1 A = adj(a 1 ) = 1 A I n 1 A I n A 1 A A (2.2.2) Adding Equation 2.2.1 and Equation 2.2.2, we get ( adj(a) ) 1 = 1 A A = adj(a 1 ). Example 2.2.9 Let A be a non-singular 4 4 matrix with A 1 = 2. Find (a) adj(a), and (b) 1 2 AT adj(a 1 ). We have A = 1 2 since A 1 = 2. Then (a): adj(a) = A 4 1 = A 3 ( ) 3 = 1 2 = 1. 8 (b): 1 2 AT adj(a 1 ) = ( 1 2 ) 4 A T adj(a 1 ) = 1 2 4 A A 1 3 = 1 2 4 1 2 ( 2)3 = 1 4.

2.2. Cofactor Expansion and Applications 61 Example 2.2.10 Let A be a 4 4 matrix with adj(a) = 8. Find 2A adj(2a). By Theorem 2.2.2, we have (2A) adj(2a) = adj(2a) (2A) = 2A I 4. Since adj(a) = 8, we have A = 2. Thus, 2A adj(2a) = 2A I 4 = (2) 4 A I4 = (16) ( 2) I 4 = 32 I 4.

62 Chapter 2. Determinants Exercise 2.2.1 1. Let adj(a) = 2 8. Find A. 1 4 2. Let A be a nonsingular n n matrix. (a) Show that adj(a) is nonsingular. (b) Prove that adj(a 1 ) = 1 A. det (A) 1 4 2 3. Let A = 5 3 6. Compute the cofactors A 11, A 12, and A 13, and show that 5A 11 2 3 2 3A 12 + 6A 13 = 0. 1 1 2 4. Let A = 0 3 1. 1 3 1 (a) Compute det (A) using a cofactor expansion along row 2. (b) Find A adj(a). (c) Find det (2A 1 adj(a)). 1 1 2 5. Let A = 0 3 1. 1 3 1 (a) Find A using a cofactor expansion along row 2. (b) Find A adj(a). (c) Find 2A 1 adj(a).

2.2. Cofactor Expansion and Applications 63 Revision of Chapter 2... Have Fun :-) CHAPTER 2 REVISION 1. Find all 3 3 matrices A, if any, such that adj(a) = 3. 2. Let A and B be two 4 4 nonsingular matrices such that det (2A 1 B 2 ) = 32 and det (B 1 adj(a)) = 4. Find A and B. 3. Let A be a nonsingular n n matrix. Prove that det (adj(a)) = [det (A)] n 1. 4. Show that if A is a nonsingular matrix, then adj(a) is nonsingular and (adj(a)) 1 = 1 2 2 3 5. Let A = 0 1 3 1. Find det(a) by: 0 2 6 1 2 4 3 0 (a) reducing A to triangular form. (b) using a cofactor expansion of det(a). 1 det (A) A = adj(a 1 ). (c) using the cofactor expansion on the fourth row of A. 0 0 6 0 6. Let A be a matrix with adj(a) = 1 3 0 7 0 2. Find A 1. 1 4 8 0 7 1 9 6 1 1 1 1 7. Find adj(a), given that A 1 = 1 7 6 4. 4 4 4 0 0 1 1 1 1 4 2 8. Let A = 5 3 6. Compute the cofactors A 11, A 12, and A 13, and show that 5A 11 2 3 2 3A 12 + 6A 13 = 0.