MATH307 - Homework 3 Solutions PDF

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textbook: Linear Algebra and Differential Equations - Peterson & Sochacki...

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HW 3 Solns MATH 307 October 2, 2020 1.4 10. Proof.  2 4 2 −4

  −10 −6 − 2 5

15 −20 25 5



=



12 −11 −3 −29

14 −3



11. Note that by theorem 1.13, (AB)T = B T AT .    1    1 (12 + 6 − 2) (−2 − 6 − 4) −2 3 −4  2 −2  = 1 5 1 (1 + 10 − 3) (1 − 10 − 1) −3 1   16 −12 . 8 −8

(1) (2)

12. As (AB)T = B T AT , we just use the solution from exercise 11. 14. 

1  2 −3

 1  1 −2 1 1

2 −3 −2 1





1+1 = 2−2 −3 + 1  2 0 8 =0 −2 −8

 2 − 2 −3 + 1 4 + 4 −6 − 2  −6 − 2 9 + 1  −2 −8  10

18. Note that A is symmetric. So, AT = A. By theorem 1.13, (AT )−1 = (A−1 )T . Thus, (A−1 )T = (AT )−1 = (A)−1 = A−1 . This shows that A−1 is symmetric. 20. Using theorem 1.13 (B T B )T = B T (B T )T = B T B. So, (B T B) is symmetric. 1.5 1. After expanding by row 1, we get the determinant of A is       1 −2  4 −2  4     2 + + 3  2 1  −3 1  −3 This gives that

 1 2

det(A) = 2(1 + 4) + (4 − 6) + 3(8 + 3) = 41.

(3)

(4)

2. After expanding by row 2, we get the determinant of A is        −1 3  2 3    +  + 2  2 −1  −4   −3 2  2 1  −3 1 This gives that

det(A) = −4(−1 − 6) + (2 + 9) + 2(4 − 3) = 41. 5. After expanding by column 2, we get the determinant of A is        4 −2   2 3    +   − 2 2 3  −3 1  −3 1  4 −2  This gives that

det(A) = (4 − 6) + (2 + 9) − 2(−4 − 12) = 41. 6. After expanding by column 3, we get the determinant of A is            4 1  + 2  2 −1  + 2 −1  3  −3 2  4 1  −3 2 This gives that

det(A) = 3(8 + 3) + 2(4 − 3) + (2 + 4) = 41. 8. We expand by the third column and get    3 4 0    3    2  1 5 2 = 2 −2  1  1 −3 0

 4  −3 

= −4(−9 − 4) = −4(−13) = 52

10. We expand by the last column and get that      0 3 2 5  0 3 2 5        0  0 −2 0 0  0 −2 0 0         3 7 3 −2 + 2 4 0 1  = 1 −2  3 4      4 5 0 1  3 7 3 −2

2 3 0

   0 5   2    −2 −2 + 1  3 1

2 0 3

 5  1   −2

= −2 [(0 − 16 + 0 − 0 − 6 − 60) + (0 + 6 + 30 + 8)] = −2[(−82) + 44] = −2[38] = 76

12.   2   −1   1   3

   1 −1 −1 3 1   −1 2 2 −1 4 = −  −1 3 1  2 −1 3 2 −1 5 2   1 −1  0 1 = −  0 1 0 5   1 −1  0 1 = −  0 0 0 0   1 −1  0 1 = −  0 0 0 0

 3 1  −1 4 3 1  −1 5 

 3 1  2 5  3 −1 −10 2   3 1  2 5  −5 −6 −20 23   3 1  2 5  −5 −6 0 1

= −1(1 · 1 · −5 · 1) =5

1.6 2.

 5  3

So the matrix in question is invertible.

 −1  = 20 + 3 = 23 = 6 0. 4 

4.  2  1  6

−1 1 0

 −3  3  = (−1)(3)(6) − (−3)(1)(6) 0 

(5)

= −18 + 18

(6)

= 0.

(7)

Thus the matrix in question is not invertible...