**Unformatted text preview: **2017-6-2 UW Common Math 308 Section 3.2 YIRAN CHEN
Math 308, section G, Spring 2017
Instructor: Amos Turchet WebAssign UW Common Math 308 Section 3.2 (Homework)
Current Score : 19 / 19 Due : Thursday, April 27 2017 11:00 PM PDT The due date for this assignment is past. Your work can be viewed below, but no changes can be made. Important! Before you view the answer key, decide whether or not you plan to request an extension. Your
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Request Extension 1. 3/3 points | Previous AnswersHoltLinAlg1 3.2.001. Perform the indicated computations when possible, using the matrices given below. (If an answer
does not exist, enter DNE into any single cell.)
A = −3 1 0 6 , B = 2 −1 −2 4 4 0
, C = −1 6
3 3 (a) A + B -3 7 0 3 [3,7;0,3] (b) AB + I2 -1 -14 2 9 [1,14;2,9] (c) A + C DNE DNE DNE DNE DNE Solution or Explanation
(a) A + B = −3 (b) AB + I2 = 1 2 −1
−3 + 0 6
−2 4 1 0 6 2 −1 −2 4 = + −3 7
0 3
1 0
0 1 = −2 −14
2 8 + 1 0
0 1 = −1 −14
2 9 (c) A + C is not possible, since A and C are different sizes. 1/8 2017-6-2 UW Common Math 308 Section 3.2 2. 3/3 points | Previous AnswersHoltLinAlg1 3.2.004. Perform the indicated computations when possible, using the matrices given below. (If an answer
does not exist, enter DNE into any single cell.)
A = −3 1 2 −1 6 0 0 4 , B = , C = −2 6 1 4 −6 −1 4 , E = 3 3 −2 1 −3
0 2 6 (a) A3 -41 15 30 -11 [41,15; 30,11] (b) BCT 0 16 12 -12 26 12 [0, 16, 12; 12, 26, 12] (c) EC + I3 DNE DNE Solution or Explanation
(a) A3 = (b) BCT = −3 1 2 −1 0 4
−2 6 −3 1 2 −1
6 0
−1 4
3 3 −3 1 2 −1 = 11 −4
−8 3 −3 1 2 −1 = −41 15 30 −11 T = 0 4
−2 6 6 −1 3
0 4 3 = 0 16 12
−12 26 12 (c) EC + I3 is not possible, since EC is a 3 × 2 matrix, and I3 is a 3 × 3 matrix. 2/8 2017-6-2 UW Common Math 308 Section 3.2 3. 1/1 points | Previous AnswersHoltLinAlg1 3.2.007. Find the missing values in the given matrix equation.
2 a b −3 3 −2 −1 2 = 3 −8
5 c (a, b, c) = $$−1, 1, −13 Solution or Explanation
2 a b −3 3 −2
−1
−8 a = −1 = 2 2b − a 2a − 6
3b + 2 −13 = 3 −8
5 c c = −13, 3b + 2 = 5 b = 1, 2a − 6 = 4. 1/1 points | Previous AnswersHoltLinAlg1 3.2.010. Find the missing values in the given matrix equation.
1 a 3 c d 0 −2
11 −2 1 2 b −5 = 4 9 e −2 −4
f −5 1 (a, b, c, d, e, f) = $$4, −5, 0, 1, 4, 43 Solution or Explanation
1 a 0 −2
11 b 3 c d
−2 1 2 = 3 − 2a a + c 4 −2 −4 = 33 − 2b b + 11c 2b + 11d 3 − 2a = −5 a = 4, a + c = 4 d = 1, 4 = e e = 4, b + 11c = −5 33 − 2(−5) = f 2a + d 4 + c = 4 −5 4 e −2 −4 f −5 c = 0, 2a + d = 9 b + 11(0) = −5 9 1 2(4) + d = 9 b = −5, 33 − 2b = f f = 43 3/8 2017-6-2 UW Common Math 308 Section 3.2 5. 4/4 points | Previous AnswersHoltLinAlg1 3.2.013. Let T1 and T2 be linear transformations given by
T1
T2 x1
x2
x1
x2 3x1 + 5x2 = −2x1 + 6x2
−2x1 + 8x2 = . 5x2 Find the matrix A such that the following is true.
(a) T1(T2(x)) = Ax A = -6 49 4 14 [6,49; 4, 14] (b) T2(T1(x)) = Ax A = -22 38 -10 30 [22, 38; 10, 30] (c) T1(T1(x)) = Ax A = -1 45 -18 26 [1, 45; 18, 26] (d) T2(T2(x)) = Ax A = 4 24 0 25 [4, 24; 0, 25] Solution or Explanation
We first determine that T1(x) = A1x = 3 5
−2 6 and T2(x) = A2x = −2 8
0 5 . (a) T1(T2(x)) = T1(A2x) = A1(A2x) = (A1A2)x. So A = A1A2 = 3 5 −2 8 −2 6 0 5 = −6 49
4 14 . (b) T2(T1(x)) = T2(A1x) = A2(A1x) = (A2A1)x. So A = A2A1 = −2 8 3 5 0 5 −2 6 = −22 38
−10 30 . (c) T1(T1(x)) = T1(A1x) = A1(A1x) = (A1A1)x. So A = A1A1 = 3 5 3 5 −2 6 −2 6 = −1 45
−18 26 . 4/8 2017-6-2 UW Common Math 308 Section 3.2 (d) T2(T2(x)) = T2(A2x) = A2(A2x) = (A2A2)x. So A = A2A2 = −2 8 −2 8 0 5 0 5 = 4 24
0 25 . 6. 1/1 points | Previous AnswersHoltLinAlg1 3.2.016. Expand the given matrix expression and combine as many terms as possible. Assume that all
matrices are n × n.
(A + I)(A2 + A)
$$A3+2A2+A Solution or Explanation
(A + I)(A2 + A) = A(A2 + A) + I(A2 + A) = A(A2) + A(A) + (A2 + A) = A3 + A2 + A2 + A = A3 +
2A2 + A 7. 1/1 points | Previous AnswersHoltLinAlg1 3.2.019. The given matrix equation is not true in general. Explain why. Assume that all matrices are n × n.
(A + B)2 = A2 + 2AB + B2
(A + B)2 = A2 + AB − AB + B2 = A2 + B2. This is only equal to A2 + 2AB + B2 when A =
0nn or B = 0nn, which in general is not true. (A + B)2 = A2 + AB + BA + B2. This is only equal to A2 + 2AB + B2 when AB = BA, which
in general is not true. (A + B)2 = A2 − BA + BA + B2 = A2 + B2. This is only equal to A2 + 2AB + B2 when A =
0nn or B = 0nn, which in general is not true.
(A + B)2 = A2 + AB + BA + B2. This is only equal to A2 + 2AB + B2 when AB = −BA, which
in general is not true. Solution or Explanation
(A + B)2 = (A + B)(A + B) = A(A + B) + B(A + B) = A2 + AB + BA + B2. This is only equal to A2 + 2AB + B2 when AB + BA = 2AB AB = BA, which in general is not true. 5/8 2017-6-2 UW Common Math 308 Section 3.2 8. 1/1 points | Previous AnswersHoltLinAlg1 3.2.031. Find an example that meets the given specifications.
3 × 3 matrices A and B such that AB ≠ BA
0 0 0
A = 0 0 0 1 0 0 B = 1 2 3 4 5 6 7 8 9 [0, 0, 0; 0, 0, 0; 0, 0, 1] Solution or Explanation
For example, A = 0 0 0 0 0 0 0 0 0 , B = 0 0 0 . Then AB = 0 0 0 , and BA = 0 0 0 0 0 0 . 1 0 0 0 0 1 1 0 0 0 0 0 0 0 0 9. 1/1 points | Previous AnswersHoltLinAlg1 3.2.034. Find an example that meets the given specifications.
3 × 3 nonzero matrices A and B such that AB = 033
0 0 1
A = 0 0 0 0 0 0 B = 1 0 0 0 0 0 0 0 0 [1, 0, 0; 0, 0, 0; 0, 0, 0] Solution or Explanation
For example, A = 0 0 1 1 0 0 0 0 0 , B = 0 0 0 . Then AB = 0 0 0 . 0 0 0 0 0 0 0 0 0
0 0 0 6/8
2017-6-2 UW Common Math 308 Section 3.2 10.1/1 points | Previous AnswersHoltLinAlg1 3.2.037. Find an example that meets the given specifications.
2 × 2 matrices A, B, and C that are nonzero, where A ≠ B but AC = BC.
A = 1 9
9 1 , B = 9 1 1 9 C = 1 2 1 2 [1,1;1,1] Solution or Explanation
For example, A = 1 9
9 1 , B = 9 1
1 9 , C = 1 1
1 1 . Then AC = BC = 10 10
10 10 . 11.1/1 points | Previous AnswersHoltLinAlg1 3.2.040. Determine if the statement is true or false, and justify your answer. You may assume that A and B
are n × n matrices.
If A and B are diagonal matrices, then so is A − B.
True. If i = j, then Aij = Bij = 0, so (A − B)ij = 0, and A − B is diagonal. True. If i ≠ j, then Aij = Bij = 0, so (A − B)ij = 0, and A − B is diagonal. False. If i = j, then (A − B)ij = Aij − Bij may not be zero, so A − B is not necessarily
diagonal.
False. If i ≠ j, then Aij = −Bij, so (A − B)ij = 2Aij, and A − B is not diagonal.
False. If i ≠ j, then (A − B)ij = Aij − Bij may not be zero, so A − B is not necessarily
diagonal. Solution or Explanation
True. If i ≠ j, then Aij = Bij = 0, so (A − B)ij = 0, and A − B is diagonal. 7/8 2017-6-2 UW Common Math 308 Section 3.2 12.1/1 points | Previous AnswersHoltLinAlg1 3.2.046. Determine if the statement is true or false, and justify your answer. You may assume that A and B
are n × n matrices.
If AB = BA, then either A = In or B = In.
True. AB = BA can only be true if either A = In or B =
In. False. For example, A = B = 0nn. 01
−1 0
False. For example, A =
, B =
.
10
01
False. AB = BA for any matrices A ≠ B.
01
10
False. For example, A =
, B =
.
10
01 Solution or Explanation
False. For example, A = B = 0nn. 8/8 ...

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