Elementary Linear Algebra Anton & Rorres, 9th Edition Lecture Set 02 Chapter 2: Determinants Chapter Content Determinants by Cofactor Expansion Evaluating Determinants by Row Reduction Properties of the Determinant Function A Combinatorial Approach to Determinants 2 2-1 Minor and Cofactor Definition Let A be mn

The (i,j)-minor of A, denoted Mij is the determinant of the (n-1) (n1) matrix formed by deleting the ith row and jth column from A The (i,j)-cofactor of A, denoted Cij, is (-1)i+j Mij Remark Note that Cij = Mij and the signs (-1)i+j in the definition of cofactor form a checkerboard pattern: ... ... ... ... 3 2-1 Example 1

Let 3 A 2 1 1 4 5 6 4 8 3 2 1 1 4 5 5 6 4 4 8 The minor of entry a11 is M 11 The cofactor of a11 is C11 = (-1)1+1M11 = M11 = 16 3 2 1

6 16 8 1 4 3 5 6 2 4 8 Similarly, the minor of entry a32 is M 32 The cofactor of a32 is C32 = (-1)3+2M32 = -M32 = -26 4 26 6 4 2-1 Cofactor Expansion The definition of a 33 determinant in terms of minors and cofactors det(A) = a M +a (-M )+a M

11 11 12 12 13 13 = a11C11 +a12C12+a13C13 this method is called cofactor expansion along the first row of A Example 2 3 1 0 4 3 2 3 2 -4 det( A) 2 4 3 3 1 0 3( 4) (1)( 11) 0 1 4 2 52 5 4 5 4 2 5

2-1 Cofactor Expansion det(A) =a11C11 +a12C12+a13C13 = a11C11 +a21C21+a31C31 =a21C21 +a22C22+a23C23 = a12C12 +a22C22+a32C32 =a31C31 +a32C32+a33C33 = a13C13 +a23C23+a33C33 Theorem 2.1.1 (Expansions by Cofactors) The determinant of an nn matrix A can be computed by multiplying the entries in any row (or column) by their cofactors and adding the resulting products; that is, for each 1 i, j n det(A) = a1jC1j + a2jC2j + + anjCnj (cofactor expansion along the jth column) and det(A) = ai1Ci1 + ai2Ci2 + + ainCin (cofactor expansion along the ith row) 6 2-1 Example 3 & 4 Example 3 cofactor expansion along the first column of A

3 1 0 4 3 1 0 1 0 det( A) 2 4 3 3 ( 2) 5 3( 4) ( 2)( 2) 5(3) 1 4 2 4 2 4 3 5 4 2 Example 4 smart 0 of 0 row - 1or column 1choice 3 A 1 2

1 2 0 0 -2 0 det(A) = ? 2 1 1 7 2-1 Adjoint of a Matrix If A is any nn matrix and Cij is the cofactor of aij, then the matrix C C C 11 12

C C 21 22 Cn1 Cn 2 1n C2 n Cnn is called the matrix of cofactors from A. The transpose of this matrix is called the adjoint of A and is denoted by adj(A) Remarks If one multiplies the entries in any row by the corresponding cofactors from a different row, the sum of these products is always zero. 8 2-1 Example 5 Let a11

A a21 a31 a12 a22 a32 a13 a23 a33 a11C31 + a12C32 + a13C33 = ? Let a11 A' a21 a11 a12 a22 a12 a13 a23 a13 9

2-1 Example 6 & 7 3 2 1 Let A 1 6 3 2 4 0 The cofactors of A are: C11 = 12, C12 = 6, C13 = -16, C21 = 4, C22 = 2, C23 = 16, C31 = 12, C32 = -10, C33 = 16 The matrix of cofactor and adjoint of A are 12 6 16 4 2 16 12 10 16 12 4 12 adj( A) 6 2 10 16 16 16

The inverse (see below) is A 1 12 12 4 1 1 adj( A) 6 2 10 det( A) 64 16 16 16 10 Theorem 2.1.2 (Inverse of a Matrix using its 1 If A is an invertible matrix, then A adj( A) Adjoint) det( A) 1 Show first that A adj( A) det( A) I

ai1Ci1 ai 2Ci 2 ainCin det( A) ai1C j1 ai 2C j 2 ainC jn 0(i j ) 11 Theorem 2.1.3 If A is an n n triangular matrix (upper triangular, lower triangular, or diagonal), then det(A) is the product of the entries on the main diagonal of the matrix; det(A) = a11a22ann E.g. 2 A a11 0 a21 a22

0 0 a31 a32 a41 a42 a33 a43 7 3 8 3 0 3 7 5 1 0 0 6 7 6 ? 0 0

0 9 8 0 0 0 0 4 0 0 0 a44 12 2-1 Prove Theorem 1.7.1c A triangular matrix is invertible if and only if its diagonal entries are all nonzero 13 2-1 Prove Theorem 1.7.1d

The inverse of an invertible lower triangular matrix is lower triangular, and the inverse of an invertible upper triangular matrix is upper triangular 14 Theorem 2.1.4 (Cramers If Ax = b is a system of n linear equations in n unknowns such Rule) that det(I A) 0 , then the system has a unique solution. This solution is det( An ) det( A1 ) det( A2 ) x1 , x2 , , xn det( A) det( A) det( A) where Aj is the matrix obtained by replacing the entries in the column of A by the entries in the matrix b = [b1 b2 bn]T 15 2-1 Example 9

Use Cramers rule to solve x1 2 x3 6 3 x1 4 x2 6 x3 30 x1 2 x2 3 x3 8 Since 1 0 A 3 4 1 2 Thus, det( A ) x1 1 det( A) 2 6 0 6 , A1 30 4 8 2 3

2 1 6 6 , A2 3 30 1 8 3 2 1 0 6 , A3 3 4 1 2 3 6 30 8 40 10 det( A2 ) 72 18 det( A3 ) 152 38 , x2 , x3 44 11 det( A) 44 11 det( A)

44 11 16 Exercise Set 2.1 Question 13 17 Exercise Set 2.1 Question 23 18 Exercise Set 2.1 Question 2 7 19