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Unformatted text preview: Math 330 Homework 3.1 & 3.2 (Pages 190,191,199)
(3.1# 10) Compute the given determinant. Be as efficient as possible with the cofactor method 1 2 5 2 0 0 3 0 2 6 7 5 5 0 4 4 SOLUTION: We expand along the 2nd row. The only nonzero entry is at position (2, 3) and (1)2+3 = 1 So 1 2 2 = 3 2 6 5 5 0 4 We then expand along the bottom row to get just two 2 2 determinants: = 3 5 2 2 1 2 +4 6 5 2 6 = 3 [5((2)(5)  (6)(2)) + 4((1)(6)  (2)(2))] = 3 [5(10 + 12) + 4(6 + 4)] = 3[10  8] = 6 As an alternative method, to compute the 3 3 determinant I would recommend row reduction:
R2R22R1 & R3R35R1 NoChange 1 2 2 1 2 2 R3R3+5R2 3 0 2 1 NoChange 3 0 2 1 0 10 6 0 0 1 Which equals, as before, 3(1)(2)(1) = 6. (3.1# 22) Explore the effect of an elementary row operation on the determinant of the matrix. State the row operation and how it effects the determinant. a b c d SOLUTION: goes to a + kc b + kd c d The determinant of the first matrix is just ad  bc whereas the determinant of the second matrix is 1 (a + kc)d  c(b + kd) = ad + kcd  bc  kcd = ad  bc This confirms that when i = j, then the row operation Ri Ri + Rj does not change the determinant. (3.2 # 8) Find the determinant by row reduction to Echelon form. SOLUTION: 1 3 3 4 0 1 2 5 2 5 4 3 3 7 5 2 1 3 3 4 R4R4+2R3 0 1 2 5 No Change 0 1 2 5 0 2 4 10 1 3 3 4 0 1 2 5 0 1 2 5 0 0 0 0 R3R32R1 & R4R4+3R1 No Change From which we can see that the matrix is noninvertible and therefore has a determinant of zero. (3.2 # 12) Combine the methods of row reduction and cofactor expansion to compute the determinant. 1 3 5 4 SOLUTION: 2 4 4 2 3 3 6 4 0 0 6 3 R2R2+3R1 & R3R3+5R1 & R4R4+4R1 No Change 1 2 3 0 0 10 12 0 = 0 14 21 6 0 10 16 3 Now we can expand along the first column to get 10 12 0 10 12 10 12 +3 = 1 14 21 6 = 1 6 14 21 10 16 10 16 3 = 1 [6((10)(16)  (12)(10)) + 3((10)(21)  (14)(12))] = 1 [6(40) + 3(42)] = 114 2 (3.2 # 22) Use the determinant to find out if the matrix is invertible. 5 0 1 1 3 2 0 5 3 SOLUTION: The determinant of the matrix is the same as the determinant of 0 15 9 15 9 R1R15R2 = = ((15)(3)  9(5)) = 0 1 3 2 = 1 5 3 0 5 3 Because the determinant is zero we conclude that the matrix is not invertible. 3 ...
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 Spring '08
 Johnson,J
 Math, Linear Algebra, Algebra, Determinant

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