3.6 Determinants We said in Section 3.3 that a 2 2 matri a b is invertible if and onl if its c d erminant, ad bc, is nonzero, and we saw the erminant used in the formula for the inverse of a 2 2 matri. In this section we see how to compute the erminant of n n matrices for arbitrar n and also, in the case of 2 2 and 3 3 matrices, how to interpret it geometricall. In the net section we will see one of its important applications: we can use it to write down eplicit formulas for solutions of sstems of linear equations. Determinant of an n n Matri Although it is possible to write down a formula for the erminant of an n n matri in terms of its entries, this formula is rarel actuall used to calculate erminants. From the point of view of calculation, it is better to specif the erminant of an n n matri recursivel we state how to find the erminant of a larger matri in terms of the erminants of smaller matrices. If A is a square matri, we will write its erminant as (A). We alread know from Section 3.3 how to calculate the erminant of a 2 2 matri: If a b A c d then ad bc. The erminant of a matri is even simpler: If A [a], then a. Before generalizing to n n matrices, we introduce a new term: The Minor Matri and Minor of an Entr If A is an n n matri with n 2 and a ij is one of its entries, the associated minor matri m ij is the (n ) (n ) matri obtained b deleting both the row and the column passing through a ij. The erminant of the minor matri m ij is called the minor M ij. Thus, m ij Minor matri (delete row and column through a ij ) M ij Minor (m ij ) Quick Eamples. If A, then 2 2 m [ 2] Delete the row and column through a 2 2.
2 3.6 Determinants M (m ) ([ 2]) 2 The erminant of a matri is its onl entr. m 2 [2] Delete the row and column through a 2 2 2. M 2 (m 2 ) ([2]) 2 m 2 [ ] Delete the row and column through a 2 2 2. M 2 (m 2 ) ([ ]) m 22 [] Delete the row and column through a 2 2 22. M 22 (m 22 ) ([]) 2. If A 3 2 2 0, then 7 5 m 2 3 2 2 0 2 0 Delete the row and column through a 7 2. 7 5 2 0 M 2 (m 2 ) (2)() (0)(7) 2 7 m 3 3 2 2 0 Delete the row and column through a 3. 2 0 7 5 M 3 (m 3 ) ()(0) ( )( 2) 2 2 0 We now give the promised recursive definition of the erminant of a matri. Since we know how to compute the erminants of and 2 2 matrices, we start with the erminants of 3 3 matrices. Computing the Determinant of a Square Matri The erminant of the n n matri A, written (A) or sometimes A, is an associated real number computed for and 2 2 matrices as above, and for larger matrices as follows: 3 3 Matri The erminant of the 3 3 matri A a a 2 a 3 a 2 a 22 a 23 is given b a 3 a 32 a 33 a M a 2 M 2 + a 3 M 3. (The formula involves computing 2 2 minors.)
3.6 Determinants 3 Quick Eample Let A 3 2 2 0. Then 7 5 a M a 2 M 2 + a 3 M 3 [ 2 0 2 0 3 5 7 3 ( 2) 2 + ( ) 24 32. ] + ( ) 2 2 7 5 4 4 Matri a a 2 a 3 a 4 The erminant of the 4 4 matri A a 2 a 22 a 23 a 24 is given b a 3 a 32 a 33 a 34 a 4 a 42 a 43 a 44 (A) a M a 2 M 2 + a 3 M 3 a 4 M 4. Notice the alternating pattern in the signs. (The formula involves computing 3 3 minors.) Quick Eample 0 0 0 Let A 2 4 0 0 2 2 0. Then 2 3 4 (A) a M a 2 M 2 + a 3 M 3 a 4 M 4 4 0 0 2 2 0 0 2 0 0 2 0 2 3 4 3 4 + 0 2 4 0 2 0 0 2 4 0 2 2 2 4 2 3 4 2 4 32. Notice that the erminant of a lower triangular matri like this (no entries above the main diagonal) is just the product of the entries on the main diagonal. n n Matri In general, the erminant of an n n matri is given b the following formula with alternating signs: (A) a M a 2 M 2 + a 3 M 3 ±a n M n The formula involves computing (n ) (n ) minors.
4 3.6 Determinants Technolog: Computing Determinants with the TI-83/84 Plus On a TI-83/84, ou can find the inverse of the square matri [A] b entering ([A]) ENTER is found in the MATRX MATH menu Computing Determinants with Ecel The formula MDETERM can be used to compute the erminant of an square matri. In the following worksheet, the erminant of 3 2 2 0 is computed in cell E3 b entering the formula shown: 7 5 Computing Determinants with the Online Matri Algebra Tool On the Web site, follow On-Line Utilities Matri Algebra Tool. There, enter our matri A and the formula (A) as shown, and press Compute. of A 3 The figure shows how one would compute the erminant 2 2 0 : 7 5 Now we know how to compute the erminant, but what is it good for? Determinants give us a method to compute volumes, to ermine whether a square matri is singular, and to compute the inverse of a nonsingular matri. In the net section we will see how the give us eplicit solutions for sstems of linear equations. There are also numerous theoretical applications that go beond the scope of this book.
3.6 Determinants 5 Computing Areas and Volumes Consider the parallelogram shown on the left in Figure. Notice that its shape and size are completel ermined b the coordinates of the two points (a, b) and (c, d) once we know these points, we can draw in the rest of the parallelogram, as shown on the right. (c, d) (c, d) (0, 0) (a, b) (0, 0) (a, b) Figure It follows that the area of this parallelogram is also ermined b the four numbers a, b, c, and d, and, in fact, the area is the absolute value of the following erminant: Area of parallelogram a b ad bc c d b d Wh? Figure 2 shows the parallelogram inside a rectangle. The area of the rectangle is c Figure 2 (c, d) a a (a, b) c d b (a + c)(b + d) ab + cb + ad + cd To obtain the area of the parallelogram we subtract the combined area of the four (green) triangles and two (pink) rectangles, which is ( ) ( ) 2 2 ab + 2 2 cd + 2bc ab + cd + 2bc So, the area of the parallelogram is ab + cb + ad + cd (ab + cd + 2bc) ad bc. EXAMPLE Computing Areas Use erminants to compute the areas of the following regions: a. b. (2, 3) ( 3, 3) (0, 0) (5, ) (0, 0) (5, )
6 3.6 Determinants Solution a. We are given a parallelogram with (a, b) (5, ) and (c, d) (2, 3). (You could also reverse the choice b taking (a, b) (2, 3) and (c, d) (5, ) see Before we go on below). Therefore, Area 5 (5)(3) ()(2) 3 3 square units 2 3 b. Although the figure is not a parallelogram, it can be thought of as half a parallelogram (Figure 3). ( 3, 3) Figure 3 (0, 0) (5, ) The area of the complete parallelogram is Area of parallelogram 5 8 8 square units. 3 3 Therefore, the area of the original triangle is half of that: Area of triangle 8 9 square units. 2 Before we go on... In Eample, which point do I take as (a, b) and which point do I take as (c, d)? It makes no difference. If we reverse our choice, we get Area 2 3 (2)() (3)(5) 3 3 square units. 5 This is alwas true: Changing the order of the rows does not affect the absolute value of the erminant, onl its sign. Parallelepipeds are three-dimensional versions of parallelograms: You can form one b taking two identical parallelograms that are parallel to each other, and then joining corresponding corners (Figure 4).
3.6 Determinants 7 Figure 4 To specif a parallelepiped, one of whose corners is at the origin, we use three points as shown in Figure 5. (g, h, i) z (a, b, c) (0, 0, 0) (d, e, f) Figure 5 Notice that the three labeled points are on the ends of the three edges that contain the origin (0, 0, 0). Wh are we labeling points with 3 coordinates now? Answer We need 3 coordinates to specif a point in 3-dimensional space. Look at the point (a, b, c). We get to this point from the origin (0, 0, 0) as follows: Move a units in the -direction (towards ou if a is positive, awa from ou if a is negative), then move b units in the -direction (to the right if a is positive, to the left if a is negative), and finall move c units in the z-direction (straight up if a is positive, down if a is negative). Just as the area of a parallelogram is given b the erminant of a 2 2 matri, so the volume of a parallelepiped is given b the erminant of a 3 3 matri: a b c Volume of parallelepiped d e f g h i Does it matter in what order we write the rows? No. Changing the order of the rows in a matri effects onl the sign of its erminant, not the absolute value.
8 3.6 Determinants EXAMPLE 2 Computing Volumes Use erminants to compute the volumes of the following solids: a. z b. (,, 3) (, 0, 0) (0, 0, 0) (0, 2, 0) c a b Rectangular Solid Solution a. Since we are given the coordinates of the three points on the ends of the three edges containing the origin, we can use the formula directl: 0 0 Volume of parallelepiped 0 2 0 ()(2)(3) 6. 3 z c Notice that we arranged the three points in such a wa as to obtain an upper triangular matri, so that the erminant is just the product of the diagonal entries. As we said above, the erminant of a matri does not change in absolute value if we rearrange the rows. b. Since the figure is a rectangular solid, we know that its volume is depth width height abc (0, 0, 0) a Figure 6 b However, we were asked to compute it using erminants. To do this we first place one corner at the origin and find the coordinates of the three adjacent points. Figure 6 shows a wa of doing that. We have placed the far corner at the origin so that the point a has coordinates (a, 0, 0), the point b has coordinates (0, b, 0), and the point c has coordinates (0, 0, c). Wh? To get to the point labeled a, just move a units in the -direction, and no units in an of the other directions. Therefore, its coordinates are (a, 0, 0). The coordinates of the other points are computed in a similar wa. We now have a 0 0 Volume of parallelepiped 0 b 0 0 0 c abc, as epected.
3.6 Determinants 9 Some Shortcuts There are quicker was of calculating the erminants of matrices of certain tpes. The justifications of the following shortcuts are beond the scope of this book, but can be found in standard linear algebra tets. Shortcuts and Special Cases The erminant of a triangular matri (one in which either all the entries above the main diagonal are zero or all the entries below it are) is the product of the entries on the diagonal. Quick Eample 5 98 44 0 2 0 (5)(2)() 0 0 0 Check this b calculating minors. The erminant of a matri is the same as the erminant of its transpose: (A) (A T ) Quick Eample [ 0 30 ] [ 0 ] 0 200 4 30 200 0 30 0 0 4 Switching two rows changes the sign of the erminant, but leaves its magnitude unchanged. (The same is true if we switch two columns.) Quick Eample 99 6 0 99 3 4 3 0 0 6 0 2 3 0 0 99 99 3 4 R R 3 If a matri has a row or column of zeros, or if one row or column is a multiple of another, then its erminant is zero. Quick Eamples 99 4 0 0 0 0 3 0 3 0 0 6 0 0 2 2 0 Second column is zero R 3 is twice R 2
0 3.6 Determinants EXAMPLE 3 Shortcuts & Special Cases Compute the erminant of each of the following matrices. a. A 3 0 0 6 0 b. B 2 3 2 3 4 2 4 6 c. C 0 0 0 99 0 4 99 40 0 3 0 0 3 4 Solution The matri A 3 0 0 a. 6 0 is lower triangular (it has onl zeros above the 99 0 4 main diagonal). Therefore, its erminant is the product of the diagonal entries: (A) (3)( )(4) 2. b. Notice that in the matri B 2 3 2 4 6, Row 2 is ( 2) times Row. 99 40 Therefore its erminant is zero: (B) 0. 2 3 4 c. We notice that the second row of C 0 0 0 0 3, is almost all zero. It 0 0 3 4 would therefore be easier to compute the erminant if we first switched Rows and 2: 2 3 4 0 0 0 (C) 0 0 0 0 3 2 3 4 0 3 0 0 3 4 0 0 3 4 Rows and 2 switched [a M a 2 M 2 + a 3 M 3 a 4 M 4 ] [( ) M 3 ] 2 4 0 3 0 0 4 0 0 4 0 3 2 4 4 0 4 2 8 2 All the other terms are zero. Rows and 3 switched
3.6 Determinants P 3.6 9. EXERCISES more advanced 0 5 Let A 3 5. In Eercises 6, write the associated 2 0 minor matri and then compute the indicated minor. 2. (, 4) (2, 2) 22. ( 2, ) (3, 6). M 23 2. M 32 3. M 22 (0, 0) (0, 0) 4. M 5. M 2 6. M 2 In Eercises 7 6 compute the erminant of the given matri directl (no shortcuts). 0 2 0 7. 8. 2 0 5 0 2 0 2 2 3 0 2 9. 0 3 0. 2 0 2 0 0 3 3 0 2 3 2 3. 3 2. 2 3 4 3 4 5 3 4 5 0 0 0 2 0 2 0 0 2 0 0 3. 4. 0 2 0 0 0 3 2 0 0 0 0 3 3 4 0 0 5. 0 0 6 3 6. 0 0 0 4 2 3 0 3 In Eercises 7 22, use a erminant to compute the area of the given region. 7. 8. (, 4) In Eercises 23 28, use a erminant to compute the volume of the given solid. 23. 25. 27. (2,, 0) (3, 0, 0) z (0, 0, 2) (2,, 0) (0, 0, 3) z z (0, 0, 2) (0, 3, 0) (, 2, 0) (2, 3, 0) 24. 26. 28. (2,, 0) (, 0, 2) (3, 0, 0) (0, 0, 3) (2,, 0) z z z (0, 0, 2) (2, 3, 0) (, 3, 0) (0, 3, 0) (2, 2) (0, 0) 9. (4, ) 20. (0, 0) (2, ) In Eercises 29 44, use shortcuts to find the erminant of the given matri. 2 0 0 0 29. 0 0 2 0 0 30. 0 3 0 0 0 4 ( 2, 2) (0, 0) (2, 2) ( 2, 2) (0, 0) (2, 3) 0 0 2 3 3. 2 3 0 32. 0 0 0 0 4 0 0 4
2 3.6 Determinants 2 3 4 2 3 33. 0 3 34. 0 3 2 4 6 0 0 2 0 2 0 0 35. 0 0 36. 0 0 5 2 0 0 0 2 0 0 3 0 3 3 2 37. 4 0 0 38. 2 0 0 5 2 2 0 2 0 0 0 4 2 3 0 0 3 3 4 0 0 39. 0 0 6 3 40. 2 3 0 0 0 0 4 8 2 4 6 3 2 2 0 0 0 2 2 0 4 0 0 4. 4 4 42. 0 0 0 0 0 3 0 3 3 0 0 0 43. 0 0 2 3 4 0 0 0 4 44. 0 0 0 0 4 5 3 COMMUNICATION AND REASONING EXERCISES 45. Multiple Choice: If the n n matri B is obtained from A b switching two rows, then: (A) (B) (A) (B) (B) (A) (C) (B) /(A) (D) (B) 2(A) 46. Multiple Choice: If A is an n n matri all of whose entries are s, then: (A) (A) (B) (A) 0 (C) (A) n 2 (D) (A) n 47. Thinking of 3 3 matrices as volumes, eplain wh the erminant of a matri is zero if two rows are identical. 48. Thinking of 3 3 matrices as volumes, decide what effect doubling all the entries in one row has on the magnitude of the erminant. 3.7 Using Determinants to Solve Sstems: Cramer s Rule As we claimed, erminants can be used to write down formulas for solutions of sstems of linear equations. To see how, let us first take a look at a general sstem of two linear equations in two unknowns: A X B a + a 2 b a a 2 b Matri form a 2 + a 22 b 2 a 2 a 22 b 2 We can solve the sstem b the elimination method described in Section 2.: To eliminate, multipl the first equation b a 22 and the second b a 2 and subtract: a 22 a + a 22 a 2 a 22 b a 2 a 2 + a 2 a 22 a 2 b 2 (a 22 a a 2 a 2 ) a 22 b a 2 b 2 so a 22b a 2 b 2 a a 22 a 2 a 2 Assuming that a a 22 a 2 a 2 0 If we instead eliminate b multipling the first equation b a 2 and the second b a and subtracting, we similarl obtain a b 2 a 2 b a a 22 a 2 a 2 Again assuming that a a 22 a 2 a 2 0
3.7 Using Determinants to Solve Sstems: Cramer s Rule 3 The denominator in both cases, [ a a 22 a ] 2 a 2, ou might recognize as the erminant a a of the coefficient matri A 2. The numerators are also erminants: a 2 a 22 b a a 22 b a 2 b 2 2 a b a b 2 a b 2 a 2 b 22 a 2 b 2 Numerator of solution for Numerator of solution for In the first, we have replaced the first column of the coefficient matri A b the column B of right-hand sides, and in the second, we have replaced the second column of A b B. We can now write the solutions as follows: Cramer s Rule for Solution of a Sstem of 2 Linear Equations in 2 Unknowns The sstem of two linear equations in two unknowns A X B a + a 2 b a a 2 b Matri form a 2 + a 22 b 2 a 2 a 22 b 2 a a has a unique solution if and onl if 2 a a 2 a a 22 a 2 a 2 0, 22 in which case the solution is given b b a 2 a b b 2 a 22 a 2 b 2 Quick Eamples. The sstem + 2 3 3 + 4 5 has sstem has the unique solution b a 2 b 2 a 22 2 a b b 2 a 2 2 2. The sstem 3 2 2 5 2 ()(4) (2)(3) 2 0. Therefore the 3 4 [ 3 2 5 4 [ 3 3 5 ] ] (3)(4) (2)(5) 2 ()(5) (3)(3) 2 2 2 4 2 2
4 3.7 Using Determinants to Solve Sstems: Cramer s Rule a a has 2 ()( 2) ( )(2) 0 a 2 a 22 2 2 As the coefficient matri is singular, Cramer s rule does not appl (in fact the given sstem is inconsistent), and so we would need to analze the sstem using the methods of Chapter 2. What happens when the erminant of the coefficient matri is zero? Notice first that in this case the Cramer s rule formulas have zero in their denominators and hence make no sense. In general, the coefficient matri of a sstem of n linear equations in n unknowns has erminant zero if and onl if the sstem is inconsistent (there is no solution) or underermined (there are infinitel man solutions). In either case, we would need to analze the sstem using a method like row-reduction discussed in Section 2.2. One advantage of Cramer s rule over row reduction is that the eplicit formulas it gives allow us to write down the solution of a linear sstem even when the coefficients are parameters (algebraic variables) instead of numbers. (In such cases, attempting to solve the sstem b row-reduction might be etremel mess.) The net eample illustrates the use of Cramer s rule for solving such a sstem. EXAMPLE Using Cramer s Rule with Parameters: Regression We shall see in Chapter 5 that the equations for the slope m and intercept b of the regression line associated with a set of data points are given b solving the sstem m ( 2 ) + b m + nb for m and b. Here, n is the number of data points, is the sum of their -coordinates, is the sum of the products, and ( 2 ) is the sum of the squares of the - coordinates. What are m and b, and what condition is necessar to ensure a unique solution? Solution The erminant of the coefficient matri is [ ( 2 ] ) n ( 2 ) ( ) 2 n For a unique solution, we require that n ( 2 ) ( ) 2 0. (It can be shown that this condition holds whenever there is more than a single -coordinate.) When this condition is satisfied, the unique solution is given b b a 2 b 2 a 22 m [ ] n n ( )( ) n ( 2 ) ( ) 2
3.7 Using Determinants to Solve Sstems: Cramer s Rule 5 [ a b a 2 b 2 b ] [ ( 2 ] ) ( ) 2 ( )( ) n ( 2) ( ) 2 The method described above can be etended to sstems of n linear equations in n unknowns. To see how to etend it, is useful to look first a general sstem of three equations in three unknowns: a + a 2 + a 3 z b a 2 + a 22 + a 23 z b 2 a 3 + a 32 + a 33 z b 3 As in the case of two equations in two unknowns, it is possible to solve this sstem b elimination: First eliminate z from the first two equations b multipling the first b a 23 and the second b a 3 and subtracting. Then eliminate z from the second and third equations in a similar wa (multipl the second b a 33 and the third b a 23 and subtract). This will leave us with two equations in and : (a a 23 a 2 a 3 ) + (a 2 a 23 a 3 a 22 ) a 23 b a 3 b 2 (a 2 a 33 a 3 a 23 ) + (a 22 a 33 a 23 a 32 ) a 33 b 2 a 23 b 3 At this point we can calculate and as we did earlier: Eliminate b multipling each equation b the coefficient of in the other and subtracting to obtain, and similarl we can obtain b eliminating. To obtain z with this method, we would start all over again b first eliminating and then. If we actuall went through these remaining steps we would find that the results can again be epressed in terms of erminants: Cramer s Rule for Solution of Sstem of 3 Linear Equations in 3 Unknowns The sstem of 3 linear equations in 3 unknowns a + a 2 + a 3 z b A X B a a 2 a 3 b a 2 + a 22 + a 23 z b 2 a 3 + a 32 + a 33 z b 3 Matri form a 2 a 3 a 22 a 32 a 23 a 33 z b 2 b 3 [ a a 2 ] a 3 has a unique solution if and onl if a 2 a 22 a 23 0, in which a 3 a 32 a 33 case the solution is given b b a 2 a 3 b 2 a 22 a 23 b 3 a 32 a 33, a b a 3 a 2 b 2 a 23 a 3 b 3 a 33, z a a 2 b a 2 a 22 b 2 a 3 a 32 b 3 Notice again that the matrices in the numerators are obtained from the coefficient matri A b replacing each column in turn b the column B of right-hand sides.
6 3.7 Using Determinants to Solve Sstems: Cramer s Rule Cramer s Rule for Solution of a Sstem of n Linear Equations in n Unknowns If A is an n n matri, then the sstem of linear equations AX B has a unique solution if and onl if 0, in which case the unique solution is given b b a 2... a n b 2 a 22... a 2n......... b n a n2... a nn a b a 3... a n a 2 b 2 a 23... a 2n............ a 2 n b n a n3... a nn... a... a (n ) b a 2... a 2(n ) b 2......... a n n... a n(n ) b n EXAMPLE 2 Using Cramer s Rule: 3 Equations in 3 Unknowns Use Cramer s rule to solve the sstem 2 + z 2 + z 3 + z Solution We first compute the erminant of the coefficient matri: 2 0 2 3 (2) (0) (2)(0) (0)() + ()( ) 2 + () 3 2 3 Since the erminant is nonzero, the sstem has a unique solution. The unknowns are 0
3.7 Using Determinants to Solve Sstems: Cramer s Rule 7 () (0) ()(0) (0)(0) + ()(0) 0 0 2 2 3 (2) z () (2)(0) ()() + ()( ) (2) [ 2 0 ] 2 3 [ ] (0) (2)(0) (0)( ) + ()( ) + () 2 + () 3 Thus, the unique solution is (,, z) (0, 2, ). 2 2 2 + () 3 2 3 2 3 In the net eample we solve a sstem of four linear equations in four unknowns with the aid of a spreadsheet. From the general case for n n sstems, we can write down the solution of the 4 4 sstem as a + a 2 + a 3 z + a 4 t b a 2 + a 22 + a 23 z + a 24 t b 2 a 3 + a 32 + a 33 z + a 34 t b 3 a 4 + a 42 + a 43 z + a 44 t b 4 b a 2 a 3 a 4 a b a 3 a 4 b 2 a 22 a 23 a 24 a 2 b 2 a 23 a 24 b 3 a 32 a 33 a 34 a 3 b 3 a 33 a 34 b 4 a 42 a 43 a 44 a, 4 b 4 a 43 a 44 a a 2 b a 4 a a 2 a 3 b a 2 a 22 b 2 a 24 a 2 a 22 a 23 b 2 a 3 a 32 b 3 a 34 a 3 a 32 a 33 b 3 a z 4 a 42 b 4 a 44 a z 4 a 42 a 43 b 4
8 3.7 Using Determinants to Solve Sstems: Cramer s Rule a a 2 a 3 a 4 a provided 2 a 22 a 23 a 24 0, a 3 a 32 a 33 a 34 a 4 a 42 a 43 a 44 EXAMPLE 3 Four Equations in Four Unknowns with Ecel Use Cramer s rule to solve the sstem + z t 2 z + + z t 2 + + z + t Solution Doing this calculation b hand would be tedious. We show how to use Ecel to help. First enter the coefficients and the right-hand sides in our spreadsheet: The formulas for the solution shown above require the erminants of four more matrices, each obtained from the original coefficient matri (A:D4) b changing a single column. We therefore make four copies of (A:D4) (this takes seconds using cop-andpaste) and then paste the column (E:E4) in the appropriate place of each (again using cop-and-paste):
3.7 Using Determinants to Solve Sstems: Cramer s Rule 9 Net, we compute the erminant of the coefficient matri in cell A5 using the MDETERM function we saw on p. 4: Control+Shift+Enter Since the erminant is nonzero, the sstem has a unique solution. We net obtain the numerators of the solutions for,, z, and t b coping and pasting the formula of cell A5 into cells A0, A5, A20, and A25: Finall, the solution for is computed in cell B0 and then pasted into B5, B20, and B25. Note the use of the absolute reference to cell $A$5:
20 3.7 Using Determinants to Solve Sstems: Cramer s Rule so we conclude that (,, z, t) ( 5 6, 6 6, 2 6, 3 6) (0.8333,, 0.3333, 0.5) 3.7 EXERCISES more advanced indicates eercises that should be solved using technolog In Eercises 2 solve the given sstem of linear equations using Cramer s Rule.. + 4 3. 0. 0.2 0 0.4 + 0.2.2 5. 3 + 2 0 6. 2 + 7. + 2 + z 0 8. + 2z 0 2 z 7 9. 4 + 2z 4 0. + 2 z 3 + z 8. 0. + 0.2z 4 0.2.z 2 + z 2 2. 2 + 2 2 3 2 4. 0.5 0. 0.3 2.5 + 0.3 2.2 2 2 + 2 4 z 0 + 3 2z 5 4 + 2z 8 z 3 + z 2 0. 0.2z 6 z 6 0.. 3 In Eercises 3 8 use Cramer s Rule with technolog to solve the given sstem of linear equations in the event that is has 2. 2 3 2 6 a unique solution. If there is no unique solution, indicate wh. HINT [See Eample 3.] 3. + + 5z 4. + 2z + w + 3 + 7z + 2w 2 + + 5z + w 5. + + 5z + 2z + w + + 5z + w + 2 + 7z + 2w 2 7. + + 5z 8. + 2z + w + + 5z + w + 2 + 7z + 2w 2 + + 4w 2 2 3z + 2w 4 + 6z + w 4 2 + 4 + 9z 6 + + 4w 2 2 3z + 2w 4 + 6z + w 4 3 + 3 + 3z + 6w 4 + + 4w 2 2 3z + 2w 4 + 6z + w 4 3 + 3 + 3z + 6w 4 In Eercises 9 22, write down an equation the parameters must satisf for there to be a unique solution, and then solve for the indicated variables assuming that condition is met. HINT [See Eample.] 6.
3.7 Using Determinants to Solve Sstems: Cramer s Rule 2 9. (a + b) p + cq a b 20. a 2 p (r + s)q a 3 cp (a b)q b a (r + s)q q Solve for p and q. a 2 r Solve for p and q. 2. a + q 3 q 22. b + a 2 + q 3 2a r + 2 0 a 2 2 + q 3 2a 2 r 2 2 + a 3 a b + q 3 0 Solve for, 2, and 3. Solve for, 2, and 3. APPLICATIONS Some of the following eercises are similar or identical to eercises and eamples in Section 3.3. All should be solved using Cramer s Rule. 23. Resource Allocation You manage an ice cream factor that makes three flavors: Cream Vanilla, Continental Mocha, and Succulent Strawberr. Into each batch of Cream Vanilla go two eggs, one cup of milk, and two cups of cream. Into each batch of Continental Mocha go one egg, one cup of milk, and two cups of cream. Into each batch of Succulent Strawberr go one egg, two cups of milk, and one cup of cream. Your stocks of eggs, milk, and cream var from da to da. How man batches of each flavor should ou make in order to use up all of our ingredients if ou have the following amounts in stock? (a) 350 eggs, 350 cups of milk, and 400 cups of cream (b) 400 eggs, 500 cups of milk, and 400 cups of cream 24. Resource Allocation The Arctic Juice Compan makes three juice blends: PineOrange, using 2 quarts of pineapple juice and 2 quarts of orange juice per gallon; PineKiwi, using 3 quarts of pineapple juice and quart of kiwi juice per gallon; and OrangeKiwi, using 3 quarts of orange juice and quart of kiwi juice per gallon. The amount of each kind of juice the compan has on hand varies from da to da. How man gallons of each blend can it make on a da with the following stocks? (a) 800 quarts of pineapple juice, 650 quarts of orange juice, 350 quarts of kiwi juice. (b) 650 quarts of pineapple juice, 800 quarts of orange juice, 350 quarts of kiwi juice. Investing In Mutual Funds Eercises 25 and 26 are based on the following data on three mutual funds. 2007 Yield FHIFX (Fidelit Focused 6% High Income Fund) FFRHX (Fidelit Floating 5% Rate High Income Fund) FASIX (Fidelit Asset 7% Manager 20%) Yields are for the ear ending September, 2007 and rounded. Source: http://mone.ecite.com, October 2007 25. You invested a total of $9,000 in the three funds at the beginning of 2007, including an equal amount in FFRHX and FASIX. Your 2007 ield for the ear from the first two funds amounted to $400. How much did ou invest in each of the three funds? 26. You invested a total of $6,000 in the three funds at the beginning of 2007, including an equal amount in FHIFX and FFRHX. Your total ields for 2007 amounted to $360. How much did ou invest in each of the three funds? Investing in Stocks Eercises 27 and 28 are based on the following data on three computer-related stocks. 2 Price per Share Dividend Yield MSFT (Microsoft) $30.5% INTC (Intel) 25.8 YHOO (Yahoo) 25 0 27. You invested a total of $5,400 in Microsoft, Intel, and Yahoo shares at the above prices, and epected to earn $45 in annual dividends. If ou purchased a total of 200 shares, how man shares of each stock did ou purchase? 28. You invested a total of $5,800 in Microsoft, Intel, and Yahoo shares at the above prices, and epected to earn $54 in annual dividends. If ou purchased a total of 220 shares, how man shares of each stock did ou purchase? COMMUNICATION AND REASONING EXERCISES 29. Name one advantage and one disadvantage of Cramer s rule versus row-reduction for solving a sstem of linear equations. 30. What does it mean about a sstem of n linear equations in n unknowns when the erminant of the coefficient matri is zero? 3. Multiple Choice: If the erminant of the coefficient matri is zero for a sstem of linear equations, then: (A) Cramer s Rule ields the eact solution. (B) Cramer s Rule fails, but we can obtain the solution b row-reducing the augmented matri. (C) Cramer s Rule fails, but we can obtain the solution b using the inverse of the coefficient matri. (D) There is onl the zero solution. 32. Multiple Choice: If the erminant of the coefficient matri is nonzero for a sstem of linear equations, but the right-hand sides are zero, then: (A) There are infinitel man solutions. (B) Cramer s Rule fails, but we can obtain the solution b row-reducing the augmented matri. (C) Cramer s Rule fails, but we can obtain the solution b using the inverse of the coefficient matri. (D) There is onl the zero solution. 2 Stocks were trading at or near the given prices in September, 2007. Dividends are rounded. Source: http://mone.ecite.com, October 2007.
Answers to Odd-Numbered Eercises ANSWERS TO ODD-NUMBERED EXERCISES 3.6 [ 0.m 23 2 0 [ 3 5 5. m 2 2 ] ; M 23 0 3. m 22 5 ; M 22 2 ] ; M 2 7 7. 7 9. 8. 2 3. 5. 24 7. 6 9. 8 2. 5 23. 8 25. 2 27. 4 29. 0 3. 2 33. 0 35. 0 37. 60 39. 0 4. 0 43. 4 45. (A) 47. The erminant a 3 3 matri gives the volume of the solid parallelepiped obtained with corner points the three rows of the matri. If two are the same, then two of the three edges are on top of each other, and the solid has zero volume. 3.7. (2.5,.5) 3. (2.4,.2) 5. (6, 4) 7. (5,, 3) 9. (0, 5, 3). ( 2, 87, 4) 3. (.5, 0, 0.5, 0) 5. No unique solution as (A) 0 7. (0,, 0, 0) 9. a 2 + b 2 c 2 0; p q (b a)(a + b + c) a 2 + b 2 c 2 (a b)( a + b + c) a 2 + b 2 c 2, 2. a 2 rq( + r) 0; (, 2, 3 ) (0, 0, ) 23. (a) 00 batches of vanilla, 50 batches of mocha, 00 batches of strawberr (b) 00 batches of vanilla, no mocha, 200 batches of strawberr. 25. $5,000 in FHIFX, $2,000 in FFRHX, $2,000 in FASIX 27. 80 MSFT, 20 INTC, 00 YHOO 29. Advantage: Cramer s Rule allows us to write own the solution eplicit, and this is useful when, for instance, the coefficients are parameters. Disadvantage: Cramer s Rule applies onl to sstems in which the number of equations equals the number of unknowns and then onl when there is a unique solution. Row reduction can be used to analze an sstem of linear equations. 3. (B) 22