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1
1.1 SOLUTIONS
Notes : The key exercises are 7 (or 11 or 12), 19–22, and 25. For brevity, the symbols R1, R2,…, stand for
row 1 (or equation 1), row 2 (or equation 2), and so on. Additional notes are at the end of the section.
1 2
x x x x
Replace R2 by R2 + (2)R1 and obtain: 1 2 2
x x x
Scale R2 by 1/3: 1 2 2
x x x
Replace R1 by R1 + (–5)R2: 1 2
x x
The solution is ( x 1 , x 2 ) = (–8, 3), or simply (–8, 3).
2. 1 2 1 2
x x x x
Scale R1 by 1/2 and obtain: 1 2 1 2
x x x x
Replace R2 by R2 + (–5)R1: 1 2 2
x x x
Scale R2 by –1/3: 1 2 2
x x x
Replace R1 by R1 + (–2)R2: 1 2
x x
The solution is ( x 1 , x 2 ) = (12, –7), or simply (12, –7).
2 CHAPTER 1 • Linear Equations in Linear Algebra
3. The point of intersection satisfies the system of two linear equations: 1 2 1 2
x x x x
Replace R2 by R2 + (–1)R1 and obtain: 1 2 2
x x x
Scale R2 by –1/7: 1 2 2
x x x
Replace R1 by R1 + (–5)R2: 1 2
x x
The point of intersection is ( x 1 , x 2 ) = (4/7, 9/7). 4. The point of intersection satisfies the system of two linear equations: 1 2 1 2
x x x x
Replace R2 by R2 + (–3)R1 and obtain: 1 2 2
x x x
Scale R2 by 1/8: 1 2 2
x x x
Replace R1 by R1 + (5)R2: 1 2
x x
The point of intersection is ( x 1 , x 2 ) = (9/4, 1/4). 5. The system is already in “triangular” form. The fourth equation is x 4 = –5, and the other equations do not contain the variable x 4. The next two steps should be to use the variable x 3 in the third equation to eliminate that variable from the first two equations. In matrix notation, that means to replace R2 by its sum with 3 times R3, and then replace R1 by its sum with –5 times R3. 6. One more step will put the system in triangular form. Replace R4 by its sum with –3 times R3, which
produces
−^ −
. After that, the next step is to scale the fourth row by –1/5.
7. Ordinarily, the next step would be to interchange R3 and R4, to put a 1 in the third row and third column. But in this case, the third row of the augmented matrix corresponds to the equation 0 x 1 + 0 x 2 + 0 x 3 = 1, or simply, 0 = 1. A system containing this condition has no solution. Further row operations are unnecessary once an equation such as 0 = 1 is evident. The solution set is empty.
4 CHAPTER 1 • Linear Equations in Linear Algebra
−^ −^ −^ −
. The solution is (5, 3, –1).
−^ −^ −^ −
The solution is (2, –1, 1).
15. First, replace R4 by R4 + (–3)R1, then replace R3 by R3 + (2)R2, and finally replace R4 by R4 + (3)R3. 1 0 3 0 2 1 0 3 0 2 0 1 0 3 3 0 1 0 3 3 0 2 3 2 1 ~ 0 2 3 2 1 3 0 0 7 5 0 0 9 7 11
~ 0 1 0 3 3 ~^0 1 0 3
The resulting triangular system indicates that a solution exists. In fact, using the argument from Example 2, one can see that the solution is unique.
16. First replace R4 by R4 + (2)R1 and replace R4 by R4 + (–3/2)R2. (One could also scale R2 before adding to R4, but the arithmetic is rather easy keeping R2 unchanged.) Finally, replace R4 by R4 + R3. 1 0 0 2 3 1 0 0 2 3 (^0 2 2 0 0) ~ 0 2 2 0 0 0 0 1 3 1 0 0 1 3 1 2 3 2 1 5 0 3 2 3 1
−^ −^ −^ −
~ 0 2 2 0 0 ~^0 2 2 0
−^ −^ −^ −
The system is now in triangular form and has a solution. The next section discusses how to continue with this type of system.
1.1 • Solutions 5
17. Row reduce the augmented matrix corresponding to the given system of three equations: 1 4 1 1 4 1 1 4 1 2 1 3 ~ 0 7 5 ~ 0 7 5 1 3 4 0 7 5 0 0 0
−^ −^ −
The system is consistent, and using the argument from Example 2, there is only one solution. So the three lines have only one point in common.
18. Row reduce the augmented matrix corresponding to the given system of three equations: 1 2 1 4 1 2 1 4 1 2 1 4 0 1 1 1 ~ 0 1 1 1 ~ 0 1 1 1 1 3 0 0 0 1 1 4 0 0 0 5
The third equation, 0 = –5, shows that the system is inconsistent, so the three planes have no point in common.
19. 3 1 6^ h^ 48 ^ ~^10 6 − h 3 h −^44
Write c for 6 – 3 h. If c = 0, that is, if h = 2, then the system has no solution, because 0 cannot equal –4. Otherwise, when h ≠ 2, the system has a solution.
20. ^ −^12 4^ h^^ − 63 ^ ~ ^1 0 4 + h 2 h − 03 .
Write c for 4 + 2 h. Then the second equation cx 2 = 0 has a solution for every value of c. So the system is consistent for all h.
21. ^ −^14 3 h^ −^28 ^ ~ ^10 h +^312 − 02 .
Write c for h + 12. Then the second equation cx 2 = 0 has a solution for every value of c. So the system is consistent for all h.
22. ^ −^26 − 93^ h 5^ ^ ~ ^20 − 03 5 + h 3 h .
The system is consistent if and only if 5 + 3 h = 0, that is, if and only if h = –5/3.
23. a. True. See the remarks following the box titled Elementary Row Operations. b. False. A 5 × 6 matrix has five rows. c. False. The description given applied to a single solution. The solution set consists of all possible solutions. Only in special cases does the solution set consist of exactly one solution. Mark a statement True only if the statement is always true. d. True. See the box before Example 2.
24. a. True. See the box preceding the subsection titled Existence and Uniqueness Questions. b. False. The definition of row equivalent requires that there exist a sequence of row operations that transforms one matrix into the other. c. False. By definition, an inconsistent system has no solution. d. True. This definition of equivalent systems is in the second paragraph after equation (2).
1.1 • Solutions 7
Rearranging, 1 2 4 1 2 3 2 3 4 1 3 4
T T T
T T T
T T T
T T T
34. Begin by interchanging R1 and R4, then create zeros in the first column: 4 1 0 1 30 1 0 1 4 40 1 0 1 4 40 (^1 4 1 0 60) ~ 1 4 1 0 60 ~ 0 4 0 4 20 0 1 4 1 70 0 1 4 1 70 0 1 4 1 70 1 0 1 4 40 4 1 0 1 30 0 1 4 15 190
Scale R1 by –1 and R2 by 1/4, create zeros in the second column, and replace R4 by R4 + R3: 1 0 1 4 40 1 0 1 4 40 1 0 1 4 40 ~ 0 1 0 1 5 ~ 0 1 0 1 5 ~^0 1 0 1 0 1 4 1 70 0 0 4 2 75 0 0 4 2 75 0 1 4 15 190 0 0 4 14 195 0 0 0 12 270
Scale R4 by 1/12, use R4 to create zeros in column 4, and then scale R3 by 1/4: 1 0 1 4 40 1 0 1 0 50 1 0 1 0 50 ~ 0 1 0 1 5 ~ 0 1 0 0 27.5^ ~^0 1 0 0 27. 0 0 4 2 75 0 0 4 0 120 0 0 1 0 30 0 0 0 1 22.5 0 0 0 1 22.5 0 0 0 1 22.
The last step is to replace R1 by R1 + (–1)R3: 1 0 0 0 20. ~ 0 1 0 0 27.5. 0 0 1 0 30. 0 0 0 1 22.
The solution is (20, 27.5, 30, 22.5).
Notes : The Study Guide includes a “Mathematical Note” about statements, “If … , then … .”
This early in the course, students typically use single row operations to reduce a matrix. As a result, even the small grid for Exercise 34 leads to about 25 multiplications or additions (not counting operations with zero). This exercise should give students an appreciation for matrix programs such as MATLAB. Exercise 14 in Section 1.10 returns to this problem and states the solution in case students have not already solved the system of equations. Exercise 31 in Section 2.5 uses this same type of problem in connection with an LU factorization. For instructors who wish to use technology in the course, the Study Guide provides boxed MATLAB notes at the ends of many sections. Parallel notes for Maple, Mathematica, and the TI-83+/86/89 and HP-48G calculators appear in separate appendices at the end of the Study Guide. The MATLAB box for Section 1. describes how to access the data that is available for all numerical exercises in the text. This feature has the ability to save students time if they regularly have their matrix program at hand when studying linear algebra. The MATLAB box also explains the basic commands replace , swap , and scale. These commands are included in the text data sets, available from the text web site, www.laylinalgebra.com.
8 CHAPTER 1 • Linear Equations in Linear Algebra
1.2 SOLUTIONS
Notes : The key exercises are 1–20 and 23–28. (Students should work at least four or five from Exercises
7–14, in preparation for Section 1.5.)
1. Reduced echelon form: a and b. Echelon form: d. Not echelon: c. 2. Reduced echelon form: a. Echelon form: b and d. Not echelon: c.
−^ −
. Pivot cols 1 and 2.
. (^) 1, 2, and 4Pivot cols
5. ^0 *^ , 0 * 0 ,^00 0
^6.
7. 3 1 93 47 76 ^ ~ 0 1 30 −^45 − 157 ^ ~ ^01 03 41 73 ^ ~ 0 1 30 01 −^53
Corresponding system of equations: 1 2 3
x x x
The basic variables (corresponding to the pivot positions) are x 1 and x 3. The remaining variable x 2 is free. Solve for the basic variables in terms of the free variable. The general solution is 1 2 2 3
is free 3
x x x x
^ = −^ −
Note : Exercise 7 is paired with Exercise 10.
10 CHAPTER 1 • Linear Equations in Linear Algebra
Basic variable: x 1 ; free variables x 2 , x 3. General solution:
1 2 3 2 3
4 2 3 3 is free is free
x x x x x
−^ −^ −
Corresponding system:
1 2 4 3 4
x x x x x
Basic variables: x 1 and x 3 ; free variables: x 2 , x 4. General solution:
1 2 4 2 3 4 4
is free 3 2 is free
x x x x x x x
Corresponding system:
1 5 2 5 4 5
x x x x x x
Basic variables: x 1 , x 2 , x 4 ; free variables: x 3 , x 5. General solution:
1 5 2 5 3 4 5 5
is free 4 9 is free
x x x x x x x x
^ =^ +
Note : The Study Guide discusses the common mistake x 3 = 0.
1.2 • Solutions 11
Corresponding system:
1 3 2 3 4 5
x x x x x x
Basic variables: x 1 , x 2 , x 5 ; free variables: x 3 , x 4. General solution:
1 3 2 3 4 3 4 5
is free is free 0
x x x x x x x x
^ = −^ −
15. a. The system is consistent, with a unique solution. b. The system is inconsistent. (The rightmost column of the augmented matrix is a pivot column).
16. a. The system is consistent, with a unique solution. b. The system is consistent. There are many solutions because x 2 is a free variable.
17. 2 3 ~^2 4 6 7 0 0 7 2
h h h
(^) − The system has a solution only if 7 – 2 h^ = 0, that is, if^ h^ = 7/2.
18.^1 3 2 ~^1 3
5 h 7 0 h 15 3
−^ −^ −^ −
(^) − (^) + If^ h + 15 is zero, that is, if^ h^ = –15, then the system has no solution, because 0 cannot equal 3. Otherwise, when h ≠ −15, the system has a solution.
19.
h h k h k
a. When h = 2 and k ≠ 8, the augmented column is a pivot column, and the system is inconsistent. b. When h ≠ 2,the system is consistent and has a unique solution. There are no free variables. c. When h = 2 and k = 8, the system is consistent and has many solutions.
20. 3 1 3 h^2 k^ ^ ~^1 0 h^3 −^9 k^2 − 6 a. When h = 9 and k ≠ 6 ,the system is inconsistent, because the augmented column is a pivot column. b. When h ≠ 9, the system is consistent and has a unique solution. There are no free variables. c. When h = 9 and k = 6, the system is consistent and has many solutions.
21. a. False. See Theorem 1. b. False. See the second paragraph of the section. c. True. Basic variables are defined after equation (4). d. True. This statement is at the beginning of Parametric Descriptions of Solution Sets. e. False. The row shown corresponds to the equation 5 x 4 = 0, which does not by itself lead to a contradiction. So the system might be consistent or it might be inconsistent.
1.2 • Solutions 13
32. According to the numerical note in Section 1.2, when n = 30 the reduction to echelon form takes about 2(30)^3 /3 = 18,000 flops, while further reduction to reduced echelon form needs at most (30)^2 = 900 flops. Of the total flops, the “backward phase” is about 900/18900 = .048 or about 5%. When n = 300, the estimates are 2(300)^3 /3 = 18,000,000 phase for the reduction to echelon form and (300)^2 = 90,000 flops for the backward phase. The fraction associated with the backward phase is about (9×10 4 ) /(18×10 6 ) = .005, or about .5%.
33. For a quadratic polynomial p ( t ) = a 0 + a 1 t + a 2 t^2 to exactly fit the data (1, 12), (2, 15), and (3, 16), the coefficients a 0 , a 1 , a 2 must satisfy the systems of equations given in the text. Row reduce the augmented matrix: 1 1 1 12 1 1 1 12 1 1 1 12 1 1 1 12 1 2 4 15 ~ 0 1 3 3 ~ 0 1 3 3 ~ 0 1 3 3 1 3 9 16 0 2 8 4 0 0 2 2 0 0 1 1
The polynomial is p ( t ) = 7 + 6 t – t^2.
34. [M] The system of equations to be solved is:
0 1 2 2 3 3 4 4 5 5 0 1 2 2 3 3 4 4 5 5 0 1 2 2 3 3 4 4 5 5 0 1 2 2 3 3 4 4 5 5 0 1 2 2 3 3 4 4 5 5 0 1 2 2 3 3
a a a a a a a a a a a a a a a a a a a a a a a a a a a a a a a a a a
- ⋅ + ⋅ + ⋅ + a 4 (^) ⋅ 104 + a 5 ⋅ 10 5 = 119 The unknowns are a 0 , a 1 , …, a 5. Use technology to compute the reduced echelon of the augmented matrix:
2 3 4 5
~ 0 0 8 48 224 960 9 ~^0 0 8 48 224 960
14 CHAPTER 1 • Linear Equations in Linear Algebra
1 0 0 0 0 0 0 1 0 0 0 0 0 0 0 2 4 8 16 32 2.9 0 2 4 8 16 32 2. ~ 0 0 8 48 224 960 9 ~^0 0 8 48 224 960 0 0 0 48 576 4800 3.9 0 0 0 48 576 4800 3. 0 0 0 0 384 7680 6.9 0 0 0 0 384 7680 6. 0 0 0 0 0 3840 10 0 0 0 0 0 1.
~ 0 0 8 48 224 0 6.5000^ ~ ~^0 0 1 0 0 0 1.
Thus p ( t ) = 1.7125 t – 1.1948 t^2 + .6615 t^3 – .0701 t^4 + .0026 t^5 , and p (7.5) = 64.6 hundred lb.
Notes : In Exercise 34, if the coefficients are retained to higher accuracy than shown here, then p (7.5) = 64.8.
If a polynomial of lower degree is used, the resulting system of equations is overdetermined. The augmented matrix for such a system is the same as the one used to find p , except that at least column 6 is missing. When the augmented matrix is row reduced, the sixth row of the augmented matrix will be entirely zero except for a nonzero entry in the augmented column, indicating that no solution exists. Exercise 34 requires 25 row operations. It should give students an appreciation for higher-level commands such as gauss and bgauss , discussed in Section 1.4 of the Study Guide. The command ref (reduced echelon form) is available, but I recommend postponing that command until Chapter 2. The Study Guide includes a “Mathematical Note” about the phrase, “If and only if,” used in Theorem 2.
1.3 SOLUTIONS
Notes : The key exercises are 11–14, 17–22, 25, and 26. A discussion of Exercise 25 will help students
understand the notation [ a 1 a 2 a 3 ], { a 1 , a 2 , a 3 }, and Span{ a 1 , a 2 , a 3 }.
1. 1 3 1 ( 3)^4
+ = ^ −^ ^ + ^ −^ ^ = − + −^ ^ =^ −
u v (^) (^) − (^) + − .
Using the definitions carefully, 2 1 ( 2)^3 1 ( 2)( 3)^1 6 2 1 2 ( 2)( 1) 2 2 4
− = ^ −^ ^ + − ^ −^ ^ = ^ −^ ^ + ^ −^ −^ ^ = − +^ ^ =
u v (^) (^) − (^) − − (^) + , or, more quickly,
− = ^ −^ ^ − −^ ^ = − +^ ^ =
u v (^) (^) − (^) + . The intermediate step is often not written.
+ = ^ + ^ ^ = ^ + ^ =
u v (^) (^) − (^) + − .
Using the definitions carefully,
16 CHAPTER 1 • Linear Equations in Linear Algebra
b. To reach b from the origin, travel 2 units in the u -direction and –2 units in the v -direction. So b = 2 u – 2 v. Or, use the fact that b is 1 unit in the u -direction from a , so that b = a + u = ( u – 2 v ) + u = 2 u – 2 v c. The vector c is –1.5 units from b in the v -direction, so c = b – 1.5 v = (2 u – 2 v ) – 1.5 v = 2 u – 3.5 v d. The “map” suggests that you can reach d if you travel 3 units in the u -direction and –4 units in the v -direction. If you prefer to stay on the paths displayed on the map, you might travel from the origin to –3 v , then move 3 units in the u -direction, and finally move –1 unit in the v -direction. So d = –3 v + 3 u – v = 3 u – 4 v Another solution is d = b – 2 v + u = (2 u – 2 v ) – 2 v + u = 3 u – 4 v
w x
v
u c a
d 2 v
b
z
–2 v y
0
Figure for Exercises 7 and 8 8. See the figure above. Since the grid can be extended in every direction, the figure suggests that every vector in R^2 can be written as a linear combination of u and v. w. To reach w from the origin, travel –1 units in the u -direction (that is, 1 unit in the negative u -direction) and travel 2 units in the v -direction. Thus, w = (–1) u + 2 v , or w = 2 v – u. x. To reach x from the origin, travel 2 units in the v -direction and –2 units in the u -direction. Thus, x = –2 u + 2 v. Or, use the fact that x is –1 units in the u -direction from w , so that x = w – u = (– u + 2 v) – u = –2 u + 2 v y. The vector y is 1.5 units from x in the v -direction, so y = x + 1.5 v = (–2 u + 2 v ) + 1.5 v = –2 u + 3.5 v z. The map suggests that you can reach z if you travel 4 units in the v -direction and –3 units in the u -direction. So z = 4 v – 3 u = –3 u + 4 v. If you prefer to stay on the paths displayed on the “map,” you might travel from the origin to –2 u , then 4 units in the v -direction, and finally move –1 unit in the u -direction. So z = –2 u + 4 v – u = –3 u + 4 v
2 3 1 2 3 1 2 3
x x x x x x x x
2 3 1 2 3 1 2 3
x x x x x x x x
2 3 1 2 3 1 2 3
x x x x x x x x
x x x
Usually, the intermediate calculations are not displayed.
1.3 • Solutions 17
Note : The Study Guide says, “Check with your instructor whether you need to “show work” on a problem
such as Exercise 9.”
1 2 3 1 2 3 1 2 3
x x x x x x x x x
1 2 3 1 2 3 1 2 3
x x x x x x x x x
1 2 3 1 2 3 1 2 3
x x x x x x x x x
^
x x x
Usually, the intermediate calculations are not displayed. 11. The question Is b a linear combination of a 1 , a 2 , and a 3? is equivalent to the question Does the vector equation x 1 a 1 + x 2 a 2 + x 3 a 3 = b have a solution? The equation
1 2 3
1 2 3
x x x
a a a b
has the same solution set as the linear system whose augmented matrix is 1 0 5 2 2 1 6 1 0 2 8 6
M
= ^ − − −
Row reduce M until the pivot positions are visible: 1 0 5 2 1 0 5 2 ~ 0 1 4 3 ~ 0 1 4 3 0 2 8 6 0 0 0 0
M
The linear system corresponding to M has a solution, so the vector equation (*) has a solution, and therefore b is a linear combination of a 1 , a 2 , and a 3. 12. The equation
1 2 3
1 2 3
x x x
a a a b
has the same solution set as the linear system whose augmented matrix is
1.3 • Solutions 19
17. [ a 1 a 2 b ] =
2 7 h 0 3 h 8 0 3 h 8 0 0 h 17
−^ −^ −^ −
. The vector b is
in Span{ a 1 , a 2 } when h + 17 is zero, that is, when h = –17.
18. [ v 1 v 2 y ] =
h h h
h h
−^ −^ −
. The vector y is in
Span{ v 1 , v 2 } when 7 + 2 h is zero, that is, when h = –7/2. 19. By inspection, v 2 = (3/2) v 1. Any linear combination of v 1 and v 2 is actually just a multiple of v 1. For instance, a v 1 + b v 2 = a v 1 + b (3/2) v 2 = ( a + 3 b /2) v 1 So Span{ v 1 , v 2 } is the set of points on the line through v 1 and 0.
Note : Exercises 19 and 20 prepare the way for ideas in Sections 1.4 and 1.7.
20. Span{ v 1 , v 2 } is a plane in R^3 through the origin, because the neither vector in this problem is a multiple of the other. Every vector in the set has 0 as its second entry and so lies in the xz -plane in ordinary 3-space. So Span{ v 1 , v 2 } is the xz -plane.
21. Let y = hk^
. Then [ u v y ] = ^ −^21 21^ hk ^ ~ 0 2 22 k + hh / 2 . This augmented matrix corresponds to a consistent system for all h and k. So y is in Span{ u , v } for all h and k. 22. Construct any 3×4 matrix in echelon form that corresponds to an inconsistent system. Perform sufficient row operations on the matrix to eliminate all zero entries in the first three columns. 23. a. False. The alternative notation for a (column) vector is (–4, 3), using parentheses and commas. b. False. Plot the points to verify this. Or, see the statement preceding Example 3. If ^ − 25
were on
the line through ^ −^25
and the origin, then ^ −^52
would have to be a multiple of ^ −^25
, which is not the case. c. True. See the line displayed just before Example 4. d. True. See the box that discusses the matrix in (5). e. False. The statement is often true, but Span{ u , v } is not a plane when v is a multiple of u , or when u is the zero vector. 24. a. True. See the beginning of the subsection Vectors in R n. b. True. Use Fig. 7 to draw the parallelogram determined by u – v and v. c. False. See the first paragraph of the subsection Linear Combinations. d. True. See the statement that refers to Fig. 11. e. True. See the paragraph following the definition of Span{ v 1 , …, v p }.
20 CHAPTER 1 • Linear Equations in Linear Algebra
25. a. There are only three vectors in the set { a 1 , a 2 , a 3 }, and b is not one of them. b. There are infinitely many vectors in W = Span{ a 1 , a 2 , a 3 }. To determine if b is in W , use the method of Exercise 13.
1 2 3
−^ −^ −
a a a b The system for this augmented matrix is consistent, so b is in W. c. a 1 = 1 a 1 + 0 a 2 + 0 a 3. See the discussion in the text following the definition of Span{ v 1 , …, v p }.
26. a. [ a 1 a 2 a 3 b ] =
2 0 6 10 1 0 3 5 1 0 3 5 1 0 3 5 1 8 5 3 ~ 1 8 5 3 ~ 0 8 8 8 ~ 0 8 8 8 1 2 1 3 1 2 1 3 0 2 2 2 0 0 0 0
− − − − − − −
Yes, b is a linear combination of the columns of A , that is, b is in W. b. The third column of A is in W because a 3 = 0· a 1 + 0· a 2 + 1· a 3. 27. a. 5 v 1 is the output of 5 days’ operation of mine #1. b. The total output is x 1 v 1 + x 2 v 2 , so x 1 and x 2 should satisfy 1 1 2 2 150 2825
x + x = ^
v v.
c. [M] Reduce the augmented matrix 550 20 50030 2825150 ^ ~ 0 1 01 1.54.0
Operate mine #1 for 1.5 days and mine #2 for 4 days. (This is the exact solution.) 28. a. The amount of heat produced when the steam plant burns x 1 tons of anthracite and x 2 tons of bituminous coal is 27.6 x 1 + 30.2 x 2 million Btu. b. The total output produced by x 1 tons of anthracite and x 2 tons of bituminous coal is given by the
vector (^1 )
x x
c. [M] The appropriate values for x 1 and x 2 satisfy (^1 )
x x
To solve, row reduce the augmented matrix: 27.6 30.2 162 1.000 0 3. 3100 6400 23610 ~ 0 1.000 1. 250 360 1623 0 0 0
The steam plant burned 3.9 tons of anthracite coal and 1.8 tons of bituminous coal.
