MATH 1104 Midterm: MATH 1104 Term Test 4 Winter
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Math1104f - test 4 ( march 18, wednesday, 11:35 am - 12:25 pm) Closed book, no graphing calculator: (2 points) let a be 3 3 matrix, det(a) = 3. Solution. det(2a) = 23det(a) = 8(3) = 24: (2 points) let a, b and c be 3 3 invertible matrices, det(a) = 2, det(b) = 4, det(c) = 3. = 6: (2 points) let a = sponding to . 5 3 5 11: (3 points) calculate det a, where a = Since b is an upper triangular matrix. det b = 5(1)(2)(4) : thus det(a) = 40, (3 points) let (cid:126)a1 = Calculate the area of the parallelogram formed by (cid:126)a1 and (cid:126)a2. , det(a) = (cid:175)(cid:175)(cid:175)(cid:175)(cid:175) = 3(2) 2(5) = 4. (cid:34) (cid:35) (cid:34) (cid:34) (cid:35) (cid:35) 2 2 (cid:175)(cid:175)(cid:175)(cid:175)(cid:175) 3 5 area = |det(a)| = | 4| = 4.