Linear Algebra Chapter 4: Determinants Section 4.2. The Determinant - PowerPoint PPT Presentation

Linear Algebra Chapter 4: Determinants Section 4.2. The Determinant of a Square Matrix—Proofs of Theorems March 31, 2019 () Linear Algebra March 31, 2019 1 / 30

Table of contents Example 4.2.A 1 Page 262 Number 12 2 Example 4.2.B 3 Example 4.2.C 4 Page 255 Example 4 5 Theorem 4.2.A. Properties of the Determinant 6 Page 261 Number 8 7 Theorem 4.3. Determinant Criterion for Invertibility 8 Theorem 4.4. The Multiplicative Property 9 10 Page 262 Number 28 11 Page 262 Number 30 12 Page 262 Number 32 () Linear Algebra March 31, 2019 2 / 30

Example 4.2.A Example 4.2.A. Example 4.2.A. Find A 11 , A 12 , and A 13 for   a 11 a 12 a 13  . A = a 21 a 22 a 23  a 31 a 32 a 33 Solution. To find A 11 , we simply eliminate the first row and first column � a 22 � a 21 � � a 23 a 23 of A to get A 11 = . Similarly, A 12 = and a 32 a 33 a 31 a 33 � a 21 � a 22 A 13 = . � a 31 a 32 () Linear Algebra March 31, 2019 3 / 30

Example 4.2.A Example 4.2.A. Example 4.2.A. Find A 11 , A 12 , and A 13 for   a 11 a 12 a 13  . A = a 21 a 22 a 23  a 31 a 32 a 33 Solution. To find A 11 , we simply eliminate the first row and first column � a 22 � a 21 � � a 23 a 23 of A to get A 11 = . Similarly, A 12 = and a 32 a 33 a 31 a 33 � a 21 � a 22 A 13 = . � a 31 a 32 () Linear Algebra March 31, 2019 3 / 30

Page 262 Number 12 Page 262 Number 12   4 − 1 2 Page 262 Number 12. Find the cofactor of 3 in A = 3 1 0  .  − 1 2 1 21 = ( − 1) 2+1 det( A 21 ) where Solution. We have a 21 = 3, so we need a ′ � − 1 � 2 A 21 = is a minor matrix. So 2 1 � � − 1 2 � � a ′ 21 = − det( A 21 ) = − � = − (( − 1)(1) − (2)(2)) = 5 . � � � 2 1 � () Linear Algebra March 31, 2019 4 / 30

Page 262 Number 12 Page 262 Number 12   4 − 1 2 Page 262 Number 12. Find the cofactor of 3 in A = 3 1 0  .  − 1 2 1 21 = ( − 1) 2+1 det( A 21 ) where Solution. We have a 21 = 3, so we need a ′ � − 1 � 2 A 21 = is a minor matrix. So 2 1 � � − 1 2 � � a ′ 21 = − det( A 21 ) = − � = − (( − 1)(1) − (2)(2)) = 5 . � � � 2 1 � () Linear Algebra March 31, 2019 4 / 30

Example 4.2.B Example 4.2.B   2 1 0 1 3 2 1 2   Example 4.2.B. Find the determinant of A =  .   4 0 1 4  1 0 2 1 Solution. We have det( A ) = a 11 a ′ 11 + a 12 a ′ 12 + a 13 a ′ 13 + a 14 a ′ 14 = 2 a ′ 11 + a ′ 12 + a ′ 14 where � � 2 1 2 � � ( − 1) 1+1 det( A 11 ) = � � a ′ = 0 1 4 11 � � � � 0 2 1 � � () Linear Algebra March 31, 2019 5 / 30

Example 4.2.B Example 4.2.B   2 1 0 1 3 2 1 2   Example 4.2.B. Find the determinant of A =  .   4 0 1 4  1 0 2 1 Solution. We have det( A ) = a 11 a ′ 11 + a 12 a ′ 12 + a 13 a ′ 13 + a 14 a ′ 14 = 2 a ′ 11 + a ′ 12 + a ′ 14 where � � 2 1 2 � � ( − 1) 1+1 det( A 11 ) = � � a ′ = 0 1 4 11 � � � � 0 2 1 � � � � � � � � 1 4 0 4 0 1 � � � � � � = (2) � − (1) � + (2) � by the � � � � � � 2 1 0 1 0 2 � � � definition of determinant of a 3 × 3 matrix = (2) ((1)(1) − (4)(2)) − (1) ((0)(1) − (4)(0)) + (2) ((0)(2) − (1)(0)) = 2( − 7) − 0 + 0 = − 14 , () Linear Algebra March 31, 2019 5 / 30

Example 4.2.B Example 4.2.B   2 1 0 1 3 2 1 2   Example 4.2.B. Find the determinant of A =  .   4 0 1 4  1 0 2 1 Solution. We have det( A ) = a 11 a ′ 11 + a 12 a ′ 12 + a 13 a ′ 13 + a 14 a ′ 14 = 2 a ′ 11 + a ′ 12 + a ′ 14 where � � 2 1 2 � � ( − 1) 1+1 det( A 11 ) = � � a ′ = 0 1 4 11 � � � � 0 2 1 � � � � � � � � 1 4 0 4 0 1 � � � � � � = (2) � − (1) � + (2) � by the � � � � � � 2 1 0 1 0 2 � � � definition of determinant of a 3 × 3 matrix = (2) ((1)(1) − (4)(2)) − (1) ((0)(1) − (4)(0)) + (2) ((0)(2) − (1)(0)) = 2( − 7) − 0 + 0 = − 14 , () Linear Algebra March 31, 2019 5 / 30

Example 4.2.B Example 4.2.B (continued 1) Solution (continued). � � 3 1 2 � � ( − 1) 1+2 det( A 12 ) = − � � a ′ = 4 1 4 12 � � � � 1 2 1 � � � � � � � � � � 1 4 4 4 4 1 � � � � � � = − (3) � − (1) � + (2) � � � � � � 2 1 1 1 1 2 � � � � = − (3) ((1)(1) − (4)(2)) + ((4)(1) − (4)(1)) − 2 ((4)(2) − (1)(1)) = − 3( − 7) + (0) − 2(7) = 7 , � � 3 2 1 � � ( − 1) 1+4 det( A 14 ) = − � � a ′ = 4 0 1 14 � � � � 1 0 2 � � � � � � � � � � 0 1 4 1 4 0 � � � � � � = − (3) � − (2) � + (1) � � � � � � 1 2 1 0 0 2 � � � � = − (3) ((0)(2) − (1)(0)) + (2) ((4)(2) − (1)(1)) − ((4)(0) − (0)(1)) = 0 + 2(7) − 0 = 14 . () Linear Algebra March 31, 2019 6 / 30

Example 4.2.B Example 4.2.B (continued 1) Solution (continued). � � 3 1 2 � � ( − 1) 1+2 det( A 12 ) = − � � a ′ = 4 1 4 12 � � � � 1 2 1 � � � � � � � � � � 1 4 4 4 4 1 � � � � � � = − (3) � − (1) � + (2) � � � � � � 2 1 1 1 1 2 � � � � = − (3) ((1)(1) − (4)(2)) + ((4)(1) − (4)(1)) − 2 ((4)(2) − (1)(1)) = − 3( − 7) + (0) − 2(7) = 7 , � � 3 2 1 � � ( − 1) 1+4 det( A 14 ) = − � � a ′ = 4 0 1 14 � � � � 1 0 2 � � � � � � � � � � 0 1 4 1 4 0 � � � � � � = − (3) � − (2) � + (1) � � � � � � 0 2 1 2 1 0 � � � � = − (3) ((0)(2) − (1)(0)) + (2) ((4)(2) − (1)(1)) − ((4)(0) − (0)(1)) = 0 + 2(7) − 0 = 14 . () Linear Algebra March 31, 2019 6 / 30

Example 4.2.B Example 4.2.B (continued 2)  2 1 0 1  3 2 1 2   Example 4.2.B. Find the determinant of A =  .   4 0 1 4  1 0 2 1 Solution (continued). So a 11 a ′ 11 + a 12 a ′ 12 + a 13 a ′ 13 + a 14 a ′ det( A ) = 14 = 2 a ′ 11 + a ′ 12 + a ′ 14 = 2( − 14) + (7) + (14) = − 7 . � () Linear Algebra March 31, 2019 7 / 30

Example 4.2.B Example 4.2.B (continued 2)  2 1 0 1  3 2 1 2   Example 4.2.B. Find the determinant of A =  .   4 0 1 4  1 0 2 1 Solution (continued). So a 11 a ′ 11 + a 12 a ′ 12 + a 13 a ′ 13 + a 14 a ′ det( A ) = 14 = 2 a ′ 11 + a ′ 12 + a ′ 14 = 2( − 14) + (7) + (14) = − 7 . � () Linear Algebra March 31, 2019 7 / 30

Example 4.2.C Example 4.2.C   0 0 0 1 0 1 2 0   Example 4.2.C. Find the determinant of A =  .   0 4 5 9  1 15 6 57 Solution. By Theorem 4.2, “General Expansion by Minors,” we can find the determinant by expanding along any row or column, so we choose to start by expanding along the first column. () Linear Algebra March 31, 2019 8 / 30

Example 4.2.C Example 4.2.C   0 0 0 1 0 1 2 0   Example 4.2.C. Find the determinant of A =  .   0 4 5 9  1 15 6 57 Solution. By Theorem 4.2, “General Expansion by Minors,” we can find the determinant by expanding along any row or column, so we choose to start by expanding along the first column. We then have � � 0 0 1 � � � � det( A ) = (0) − (0) + (0) − (1) 1 2 0 � � � � 4 5 9 � � � � � � 1 2 � � = − (0) − (0) + (1) expanding along the first row � � 4 5 � � = − ((0) − (0) + ((1)(5) − (2)(4))) = 3 . So det( A ) = 3. � () Linear Algebra March 31, 2019 8 / 30

Example 4.2.C Example 4.2.C   0 0 0 1 0 1 2 0   Example 4.2.C. Find the determinant of A =  .   0 4 5 9  1 15 6 57 Solution. By Theorem 4.2, “General Expansion by Minors,” we can find the determinant by expanding along any row or column, so we choose to start by expanding along the first column. We then have � � 0 0 1 � � � � det( A ) = (0) − (0) + (0) − (1) 1 2 0 � � � � 4 5 9 � � � � � � 1 2 � � = − (0) − (0) + (1) expanding along the first row � � 4 5 � � = − ((0) − (0) + ((1)(5) − (2)(4))) = 3 . So det( A ) = 3. � () Linear Algebra March 31, 2019 8 / 30

Page 255 Example 4 Page 255 Example 4 Page 255 Example 4. Show that the determinant of an upper- or lower-triangular square matrix is the product of its diagonal elements. Solution. Let   u 11 u 12 u 13 · · · u 1 , n − 1 u 1 n 0 · · · u 22 u 23 u 2 , n − 1 u 2 n     0 0 u 33 · · · u 3 , n − 1 u 3 n   U =  . . . . .  ... . . . . .   . . . . .     0 0 0 · · · u n − 1 , n − 1 u n − 1 , n   0 0 0 · · · 0 u nn be an upper triangular matrix. () Linear Algebra March 31, 2019 9 / 30