MATH 330 H3132

# MATH 330 H3132 - Math 330 Homework 3.1& 3.2(Pages...

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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 co-factor 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: R2R2-2R1 & R3R3-5R1 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 R3R3-2R1 & R4R4+3R1 No Change From which we can see that the matrix is non-invertible 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 R1R1-5R2 = = -((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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