= j = 1 1 1 2 0 EXAMPLE: Compute the determinant of A = 3 1 2 - PDF document

3.1 Introduction to Determinants Notation: A ij is the matrix obtained from matrix A by deleting the i th row and j th column of A . EXAMPLE: 1 2 3 4 5 6 7 8 A = A 23 = 9 10 11 12 13 14 15 16 a b Recall that det = ad − bc and we let det  a  = a . c d For n ≥ 2 , the determinant of an n × n matrix A =  a ij  is given by det A = a 11 det A 11 − a 12 det A 12 + ⋯ +  − 1  1 + n a 1 n det A 1 n n  − 1  1 + j a 1 j det A 1 j = ∑ j = 1 1

1 2 0 EXAMPLE: Compute the determinant of A = 3 − 1 2 2 0 1 Solution − 1 2 3 2 3 − 1 det A = 1det − 2det + 0det 0 1 2 1 2 0 = ______________________________ = ______ 3 2 3 2 Common notation: det . = 2 1 2 1 So 1 2 0 − 1 2 3 2 3 − 1 = 1 − 2 + 0 3 − 1 2 0 1 2 1 2 0 2 0 1 The  i , j  -cofactor of A is the number C ij where C ij =  − 1  i + j det A ij . 1 2 0 = 1 C 11 + 2 C 12 + 0 C 13 3 − 1 2 2 0 1 (cofactor expansion across row 1) 2

THEOREM 1 The determinant of an n × n matrix A can be computed by a cofactor expansion across any row or down any column: (expansion across row i ) det A = a i 1 C i 1 + a i 2 C i 2 + ⋯ + a in C in (expansion down det A = a 1 j C 1 j + a 2 j C 2 j + ⋯ + a nj C nj column j ) Use a matrix of signs to determine  − 1  i + j ⋯ − + + ⋯ − − + ⋯ − + + ⋮ ⋮ ⋮ ⋱ 1 2 0 EXAMPLE: Compute the determinant of A = 3 − 1 2 2 0 1 using cofactor expansion down column 3. Solution 1 2 0 3 − 1 1 2 1 2 = 1 . = 0 − 2 + 1 3 − 1 2 2 0 2 0 3 − 1 2 0 1 3

1 2 3 4 0 2 1 5 EXAMPLE: Compute the determinant of A = 0 0 2 1 0 0 3 5 Solution 1 2 3 4 0 2 1 5 0 0 2 1 0 0 3 5 2 1 5 2 3 4 2 3 4 2 3 4 = 1 − 0 + 0 − 0 0 2 1 0 2 1 2 1 5 2 1 5 0 3 5 0 3 5 0 3 5 0 2 1 2 1 = 1 ⋅ 2 = 14 3 5 Method of cofactor expansion is not practical for large matrices - see Numerical Note on page 190 . 4

Triangular Matrices: 0 0 0 0 ∗ ∗ ⋯ ∗ ∗ ∗ 0 0 0 0 ∗ ⋯ ∗ ∗ ∗ ∗ 0 0 0 0 ⋱ ∗ ∗ ∗ ∗ ⋱ 0 0 0 0 ∗ ∗ ⋯ ∗ ∗ ∗ 0 0 0 0 ∗ ∗ ⋯ ∗ ∗ ∗ (upper triangular) (lower triangular) THEOREM 2 : If A is a triangular matrix, then det A is the product of the main diagonal entries of A . EXAMPLE: 2 3 4 5 0 1 2 3 = _____________________ = − 24 0 0 − 3 5 0 0 0 4 5

3.2 Properties of Determinants THEOREM 3 Let A be a square matrix. a. If a multiple of one row of A is added to another row of A to produce a matrix B , then det A = det B . b. If two rows of A are interchanged to produce B , then det B = − det A . c. If one row of A is multiplied by k to produce B , then det B = k ⋅ det A . 1 2 3 4 0 5 0 0 EXAMPLE: Compute . 2 7 6 10 2 9 7 11 Solution 1 2 3 4 1 3 4 1 3 4 0 5 0 0 = 5 = 5 2 6 10 0 0 2 2 7 6 10 2 7 11 2 7 11 2 9 7 11 1 3 4 1 3 4 = _________ = ____ . = 5 = − 5 0 0 2 0 1 3 0 1 3 0 0 2 6

∗ ∗ ∗ ∗ ∗ ∗ Theorem 3(c) indicates that . = k − 2 k 5 k 4 k − 2 5 4 ∗ ∗ ∗ ∗ ∗ ∗ 2 4 6 EXAMPLE: Compute 5 6 7 7 6 10 Solution 2 4 6 1 2 3 1 2 3 = 2 = 2 5 6 7 5 6 7 0 − 4 − 8 7 6 10 7 6 10 0 − 8 − 11 1 2 3 1 2 3 = 2  − 4  = 2  − 4  0 1 2 0 1 2 0 − 8 − 11 0 0 5 = 2  − 4  1  1  5  = − 40 7

2 3 0 1 4 7 0 3 EXAMPLE : Compute using a combination of 7 9 − 2 4 1 2 0 4 row reduction and cofactor expansion. 2 3 0 1 2 3 1 2 3 1 4 7 0 3 Solution = − 2 = − 2 4 7 3 0 1 1 7 9 − 2 4 1 2 4 1 2 4 1 2 0 4 2 3 1 1 2 4 1 2 4 = 2 = − 2 = − 2 1 2 4 2 3 1 0 − 1 − 7 0 1 1 0 1 1 0 1 1 1 2 4 = − 2  1  − 1  − 6  = − 12 . = − 2 0 − 1 − 7 0 0 − 6 8

■ ⋯ ∗ ∗ ∗ 0 ■ ⋯ ∗ ∗ Suppose A has been reduced to U = by 0 0 ■ ⋯ ∗ 0 0 0 ⋱ ⋮ 0 0 0 0 ■ row replacements and row interchanges, then product of  − 1  r when A is invertible det A = pivots in U 0 when A is not invertible THEOREM 4 A square matrix is invertible if and only if det A ≠ 0 . If A is an n × n matrix, then det A T = det A . THEOREM 5 Partial proof ( 2 × 2 case) a b and det = ad − bc c d T a b a c det = det = ad − bc c d b d a b a c . ⇒ det = det c d b d 9

( 3 × 3 case) a b c e f d f d e det = a − b + c d e f h i g i g h g h i a d g e h d g d g det = a − b + c b e h f i f i e h c f i a b c a d g . ⇒ det = det d e f b e h g h i c f i Implications of Theorem 5 ? Theorem 3 still holds if the word row is replaced with ___________________. 10

THEOREM 6 (Multiplicative Property) For n × n matrices A and B , det  AB  =  det A  det B  . EXAMPLE: Compute det A 3 if det A = 5 . det A 3 = det  AAA  =  det A  det A  det A  Solution: = ______________ = ________. EXAMPLE: For n × n matrices A and B , show that A is singular if det B ≠ 0 and det AB = 0 . Solution: Since  det A  det B  = det AB = 0 and det B ≠ 0, then det A = 0 . Therefore A is singular. 11