1. Slide 3.1- 1 3 INTRODUCTION TO DETERMINANTS Determinants 3.1

2. Slide 3.1- 2 © 2012 Pearson Education, Inc. INTRODUCTION TO DETERMINANTS  The determinant of a 2×2 matrix, A = [aij ], is the number det A = a 11 a 22 − a 12 a 21  For a 1×1 matrix—say, A=[a 11 ]—we define det A=a 11.  For any square matrix A, let A ij denote the submatrix formed by deleting the ith row and j th column of A.

3. Slide 3.1- 3 © 2012 Pearson Education, Inc. INTRODUCTION TO DETERMINANTS  For instance, if  then A 32 is obtained by crossing out row 3 and column 2,  so that

4. Slide 3.1- 4 © 2012 Pearson Education, Inc. INTRODUCTION TO DETERMINANTS  DEFINITION For n ≥ 2, the determinant of an n×n matrix A = [a ij ] is the sum of n terms of the form ±a 1j det A 1j, with plus and minus signs alternating, where the entries a 11, a 12,..., a 1n are from the first row of A. In symbols,

5. Slide 3.1- 5 © 2012 Pearson Education, Inc. INTRODUCTION TO DETERMINANTS  EXAMPLE 1 Compute the determinant of  Solution Compute det A = a 11 det A 11 − a 12 det A 12 + a 13 det A 13 :

6. Slide 3.1- 6 © 2012 Pearson Education, Inc. INTRODUCTION TO DETERMINANTS  Another common notation for the determinant of a matrix uses a pair of vertical lines in place of brackets.  Thus the calculation in Example 1 can be written as

7. Slide 3.1- 7 © 2012 Pearson Education, Inc. INTRODUCTION TO DETERMINANTS  Given A = [a ij ], the (i, j)-cofactor of A is the number C ij given by C ij = (−1) i+j det A ij (4)  Then det A = a 11 C 11 + a 12 C 12 + · · · + a 1n C 1n  This formula is called a cofactor expansion across the first row of A.

8. Slide 3.1- 8 © 2012 Pearson Education, Inc. INTRODUCTION TO DETERMINANTS  THEOREM 1 The determinant of an n×n matrix A can be computed by a cofactor expansion across any row or down any column. The expansion across the ith row using the cofactors in (4) is det A = a i1 C i1 + a i2 C i2 + · · · + a in C in The cofactor expansion down the j th column is det A = a 1j C 1j + a 2j C 2j + · · · + a nj C nj

9. Slide 3.1- 9 © 2012 Pearson Education, Inc. INTRODUCTION TO DETERMINANTS  The plus or minus sign in the (i, j )-cofactor depends on the position of a ij in the matrix, regardless of the sign of a ij itself. The factor (−1) i+j determines the following checkerboard pattern of signs:

10. Slide 3.1- 10 © 2012 Pearson Education, Inc. INTRODUCTION TO DETERMINANTS  EXAMPLE 2 Use a cofactor expansion across the third row to compute det A, where  Solution Compute det A = a 31 C 31 + a 32 C 32 + a 33 C 33 =(−1) 3+1 a 31 detA 31 + (−1) 3+2 a 32 detA 32 + (−1) 3+3 a 33 detA 33 = =0 + 2(−1) + 0=−2

11. Slide 3.1- 11 © 2012 Pearson Education, Inc. INTRODUCTION TO DETERMINANTS  EXAMPLE 3 Compute det A, where  Solution The cofactor expansion down the first column of A has all terms equal to zero except the first. Thus

12. Slide 3.1- 12 © 2012 Pearson Education, Inc. INTRODUCTION TO DETERMINANTS  Henceforth we will omit the zero terms in the cofactor expansion.  Next, expand this 4×4 determinant down the first column, in order to take advantage of the zeros there.  We have  This 3×3 determinant was computed in Example 1 and found to equal −2. Hence det A = 3·2·(−2)=−12.

13. Slide 3.1- 13 © 2012 Pearson Education, Inc. INTRODUCTION TO DETERMINANTS  THEOREM 2 If A is a triangular matrix, then det A is the product of the entries on the main diagonal of A.

14. Slide 3.2- 14 3 PROPERTIES OF DETERMINANTS Determinants 3.2

15. Slide 3.2- 15 © 2012 Pearson Education, Inc. PROPERTIES OF DETERMINANTS  THEOREM 3 Row Operations Let A be a square matrix. a. If a multiple of one row of A is added to another row to produce a matrix B, then det B = det A. 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.

16. Slide 3.2- 16 © 2012 Pearson Education, Inc. PROPERTIES OF DETERMINANTS  EXAMPLE 1 Compute det A, where A =.  Solution The strategy is to reduce A to echelon form and then to use the fact that the determinant of a triangular matrix is the product of the diagonal entries.  The first two row replacements in column 1 do not change the determinant:

17. Slide 3.2- 17 © 2012 Pearson Education, Inc. PROPERTIES OF DETERMINANTS  An interchange of rows 2 and 3 reverses the sign of the determinant, so

18. Slide 3.2- 18 © 2012 Pearson Education, Inc. PROPERTIES OF DETERMINANTS  EXAMPLE 2 Compute det A, where A =  Solution To simplify the arithmetic, we want a 1 in the upper-left corner. We could interchange rows 1 and 4.  Instead, we factor out 2 from the top row, and then proceed with row replacements in the first column:

19. Slide 3.2- 19 © 2012 Pearson Education, Inc. PROPERTIES OF DETERMINANTS  Next, we could factor out another 2 from row 3 or use the 3 in the second column as a pivot.  We choose the latter operation, adding 4 times row 2 to row 3:

20. Slide 3.2- 20 © 2012 Pearson Education, Inc. PROPERTIES OF DETERMINANTS  Finally, adding −1/2 times row 3 to row 4, and computing the “triangular” determinant, we find that

21. Slide 3.2- 21 © 2012 Pearson Education, Inc. PROPERTIES OF DETERMINANTS  THEOREM 4 A square matrix A is invertible if and only if det A=0.  Theorem 4 adds the statement “det A=0” to the Invertible Matrix Theorem.  A useful corollary is that det A = 0 when the columns of A are linearly dependent.  Also, det A = 0 when the rows of A are linearly dependent.

22. Slide 3.2- 22 © 2012 Pearson Education, Inc. PROPERTIES OF DETERMINANTS  In practice, linear dependence is obvious only when two columns or two rows are the same or a column or a row is zero.  EXAMPLE 3 Compute det A, where A =.  Solution Add 2 times row 1 to row 3 to obtain  because the second and third rows of the second matrix are equal.

23. Slide 3.3- 23 3 CRAMER’S RULE, VOLUME, AND LINEAR TRANSFORMATIONS Determinants 3.3

24. Slide 3.3- 24 © 2012 Pearson Education, Inc. Cramer’s Rule  For any n×n matrix A and any b in R n, let A i (b) be the matrix obtained from A by replacing column i by the vector b.  THEOREM 7 Cramer’s Rule Let A be an invertible n×n matrix. For any b in Rn, the unique solution x of Ax = b has entries given by (1)

25. Slide 3.3- 25 © 2012 Pearson Education, Inc. Cramer’s Rule  PROOF Denote the columns of A by a 1,..., a n and the columns of the n×n identity matrix I by e 1,..., e n. If Ax = b, the definition of matrix multiplication shows that A·I i (x) = A[ e 1 · · · x · · · e n ] = [ Ae 1 · · · Ax · · · Ae n ] = [a 1 · · · b · · · a n ] = A i (b)

26. Slide 3.3- 26 © 2012 Pearson Education, Inc. Cramer’s Rule  By the multiplicative property of determinants, (det A)(det I i (x)) = det A i (b)  The second determinant on the left is simply x i. (Make a cofactor expansion along the ith row.)  Hence (det A)·x i = det A i (b).  This proves (1) because A is invertible and det A ≠ 0.

27. Slide 3.3- 27 © 2012 Pearson Education, Inc. Cramer’s Rule  EXAMPLE 1 Use Cramer’s rule to solve the system 3x 1 − 2x 2 = 6 −5x 1 + 4x 2 = 8  Solution View the system as Ax = b. Using the notation introduced above,

28. Slide 3.3- 28 © 2012 Pearson Education, Inc. Cramer’s Rule  Since det A = 2, the system has a unique solution.  By Cramer’s rule,

29. Slide 3.3- 29 © 2012 Pearson Education, Inc. Application to Engineering  EXAMPLE 2 Consider the following system in which s is an unspecified parameter. Determine the values of s for which the system has a unique solution, and use Cramer’s rule to describe the solution. 3sx 1 − 2x 2 = 4 −6x 1 + sx 2 = 1

30. Slide 3.3- 30 © 2012 Pearson Education, Inc. Application to Engineering  Solution View the system as Ax = b.  Then  Since det A = 3s 2 − 12 = 3(s + 2)(s − 2)  the system has a unique solution precisely when s ≠ ± 2.

31. Slide 3.3- 31 © 2012 Pearson Education, Inc. Application to Engineering  For such an s, the solution is (x 1, x 2 ), where

32. Slide 3.3- 32 © 2012 Pearson Education, Inc. A Formula for A −1  Given A = [a ij ], the (i, j)-cofactor of A is the number C ij given by C ij = (−1) i+j det A ij  Then  The matrix of cofactors on the right side of (4) is called the adjugate (or classical adjoint) of A, denoted by adj A. (4)

33. Slide 3.3- 33 © 2012 Pearson Education, Inc. A Formula for A −1  THEOREM 8 An Inverse Formula Let A be an invertible n×n matrix. Then  EXAMPLE 3 Find the inverse of the matrix A=.

34. Slide 3.3- 34 © 2012 Pearson Education, Inc. A Formula for A −1  Solution The nine cofactors are

35. Slide 3.3- 35 © 2012 Pearson Education, Inc. A Formula for A −1  The adjugate matrix is the transpose of the matrix of cofactors. [For instance, C 12 goes in the (2, 1) position.] Thus  We could compute det A directly, but the following computation provides a check on the calculations above and produces det A:

36. Slide 3.3- 36 © 2012 Pearson Education, Inc. A Formula for A −1  Since (adj A)A = 14I, Theorem 8 shows that det A = 14 and

37. Slide 3.3- 37 © 2012 Pearson Education, Inc. Determinants as Area or Volume  THEOREM 9  If A is a matrix, the area of the parallelogram determined by the columns of A is.  PROOF  The theorem is obviously true for any diagonal matrix:

38. Slide 3.3- 38 © 2012 Pearson Education, Inc. Determinants as Area or Volume

39. Slide 3.3- 39 © 2012 Pearson Education, Inc. Determinants as Area or Volume  Let a 1 and a 2 be nonzero vectors. Then for any scalar c, the area of the parallelogram determined by a 1 and a 2 equals the area of the parallelogram determined by a 1 and a 2 + ca 1.

40. Slide 3.3- 40 © 2012 Pearson Education, Inc. Determinants as Area or Volume  The proof for R 3 is similar. The theorem is obviously true for a diagonal matrix. See Fig 3, and any matrix A can be transformed into a diagonal matrix using column operations that do not change

41. Slide 3.3- 41 © 2012 Pearson Education, Inc. Determinants as Area or Volume A parallelepiped is shown in Fig 4 as a shaded box with two sloping sides. Its volume is the area of the base in the plane Span {a 1, a 3 } times the altitude of a 2 above Span {a 1, a 3 }. Any vector a 2 + ca 1 has the same altitude because a 2 +ca 1 lies in the plane a 2 + Span {a 1, a 3 }, which is parallel to Span {a 1, a 3 }. Hence the volume of the parallelepiped is unchanged when [a 1 a 2 a 3 ] is changed to [a 1 a 2 + ca 1 a 3 ].

42. Slide 3.3- 42 © 2012 Pearson Education, Inc. Determinants as Area or Volume

43. Slide 3.3- 43 © 2012 Pearson Education, Inc. Determinants as Area or Volume EXAMPLE 4 Calculate the area of the parallelogram determined by the points (- 2, -2), (0, 3), (4, -1), and (6, 4). See Fig 5. SOLUTION First translate the parallelogram to one having the origin as a vertex. For example, subtract the vertex (-2, -2) from each of the four vertices. The new parallelogram has the same area, and its vertices are (0, 0), (2, 5), (6, 1), and (8, 6). See Fig 5(b). This parallelogram is determined by the columns of Since, the area of the parallelogram is 28.3

44. Slide 3.3- 44 © 2012 Pearson Education, Inc. Determinants as Area or Volume

45. Slide 3.3- 45 © 2012 Pearson Education, Inc. Linear Transformations THEOREN 10 Let T ： R 2 → R 2 be the linear transformation determined by a matrix A. If S is a parallelogram in R 2, then (5) If T is determined by a matrix A, and if S is a parallelepiped in R 3, then {volume of T(S)}= ‧ {volume of S} (6) PROOF Consider the case, with A = [a 1 a 2 ].

46. Slide 3.3- 46 © 2012 Pearson Education, Inc. Linear Transformations A parallelogram at the origin in R 2 determined by vectors b 1 and b 2 has the form S = {s 1 b 1 + s 2 b 2 ： 0 ≦ s 1 ≦ 1, 0 ≦ s 2 ≦ 1} The image of S under T consists of points of the form T(s 1 b 1 + s 2 b 2 ) = s 1 T(b 1 ) + s 2 T(b 2 ) = s 1 Ab 1 + s 2 Ab 2 Where 0 ≦ s 1 ≦ 1, 0 ≦ s 2 ≦ 1. It follows that T(S) is the parallelogram determined by the columns of the matrix [Ab 1 Ab 2 ]. This matrix can be written as AB, where B = [b 1 b 2 ].

47. Slide 3.3- 47 © 2012 Pearson Education, Inc. Linear Transformations By theorem 9 and the product theorem for determinants, {area of T(S)} = │det AB│=│det A│ ‧ │det B│ = │det A│ ‧ {area of S} An arbitrary parallelogram has the form p + S, where p is vector and S is a parallelogram at the origin, as above. It is easy to see that T transforms p + S into T(p) + T(S). {area of T(p + S)} = {area of T(p) + T(S)} = {area of T(S)} = │det A│ ‧ {area of S} = │det A│ ‧ {area of p + S}

48. Slide 3.3- 48 © 2012 Pearson Education, Inc. Linear Transformations This shows that (5) holds for all parallelograms in R 2. The proof of (6) for the case is analogous.

49. Slide 3.3- 49 © 2012 Pearson Education, Inc. Linear Transformations