Evaluate the following determinants and negative of determinants of order two: (a) (1) 3 4 5 (ii) 3 6 © ® -|³ 2, (1) M-|3 −2. M-| | (ii) (iii) 4 Use c|ª b|= ad — bc and −|ª $|= be — ad. Thus: (a). (i) = 7 (ii) = -10, (iii) = 7. (a). (i) = 7 (ii) = 10, (iii) = 7. (a). (i) = -7 (ii) = 10, (iii) = 7. O(a). (i)=-7 (1) = 10, (iii) = -7. Find (b) * Evaluate the following determinants and negative of determinants of order two: 3 4 (a) (1) 5 36 (b) (i) · -| 2 ® -|3 ¯½ (ii) ® -| | ·|a b|= ad − be and −|ª b|= be- ad. Thus: Use (b). (i)= 18 (ii) = -29, (iii) = 4. (b). (i)= 18 (ii) = -29, (iii) = -4. (b). (i)=-18 (ii) = -29, (iii) = -4. (b). (i)=8 (ii) = 29, (iii) = 4.

Elementary Linear Algebra (MindTap Course List)
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Chapter3: Determinants
Section3.1: The Determinants Of A Matrix
Problem 70E: The determinant of a 22 matrix involves two products. The determinant of a 33 matrix involves six...
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Question
Evaluate the following determinants and negative of determinants of order two:
(a) (1)
3 4
5
(ii)
3
6
© ® -|³ 2,
(1)
M-|3 −2. M-| |
(ii)
4
Use
c|ª
b|= ad — bc and −|ª $|= be— ad.
Thus:
(a). (i) = 7 (ii) = -10, (iii) = 7.
(a). (i) = 7 (ii) = 10, (iii) = 7.
(a). (i) = -7 (ii) = 10, (iii) = 7.
O(a). (i)=-7 (1) = 10, (iii) = -7.
Find (b) *
Evaluate the following determinants and negative of determinants of order two:
3 4
(a) (1)
5
(b) (i)
2| ® -|3 ¯½
(ii)
® -| |
· -|
|«
Use
|= ad − be and −|ª
b|= be- ad. Thus:
(b). (i)= 18 (ii) = -29, (iii) = 4.
(b). (i)= 18 (ii) = -29, (iii) = -4.
(b). (i)=-18 (ii) = -29, (iii) = -4.
36
O (b). (1) = 8 (11) = 29, (iii) = 4.
Transcribed Image Text:Evaluate the following determinants and negative of determinants of order two: (a) (1) 3 4 5 (ii) 3 6 © ® -|³ 2, (1) M-|3 −2. M-| | (ii) 4 Use c|ª b|= ad — bc and −|ª $|= be— ad. Thus: (a). (i) = 7 (ii) = -10, (iii) = 7. (a). (i) = 7 (ii) = 10, (iii) = 7. (a). (i) = -7 (ii) = 10, (iii) = 7. O(a). (i)=-7 (1) = 10, (iii) = -7. Find (b) * Evaluate the following determinants and negative of determinants of order two: 3 4 (a) (1) 5 (b) (i) 2| ® -|3 ¯½ (ii) ® -| | · -| |« Use |= ad − be and −|ª b|= be- ad. Thus: (b). (i)= 18 (ii) = -29, (iii) = 4. (b). (i)= 18 (ii) = -29, (iii) = -4. (b). (i)=-18 (ii) = -29, (iii) = -4. 36 O (b). (1) = 8 (11) = 29, (iii) = 4.
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