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Math 152 Test #1 version A B2: For each of the systems described below as augmented matrices that have been put into reduced row echelon form, determine whether the systems have • no solutions • a single solution (in this case, find the solution) • an infinite number of solutions (in this case, find a parametric description of all solutions). Each part is worth 1 mark. 1 2 3 0 (a) 0 0 0 1 2 3 1 0 21 3 5 (b) 4 0 1 0 0 0 1 2 0 9 3 (c) 0 0 1 8 2 1 0 5 (d) 0 1 9 2 3 1 0 3 2 (e) 4 0 1 0 5 5 0 0 0 0 6 Test #1, Version A, MATH 2 3 0 0 0 1 solutions: the last 2 3 line means 0x + 0y + 0z = 1, which is not true for any real values of x, y, and z. 1 0 21 3 5 (b) 40 1 0 0 0 21 Precisely one solution: 3 1 2 0 9 3 (c) 0 0 1 8 2 Infinitely many solutions. The easiest way to parameterize this is to let x2 = s and x4 = t. Then the first line says x1 = 3 2s 9t, and the second line says x3 = 2 8t. So: 2 3 2 3 2 3 2 3 2 3 x1 3 2s 9t 3 2 9 6x2 7 6 7 607 617 607 s 6 7=6 7 6 7 6 7 6 7 4x3 5 4 2 8t 5 = 425 + s 4 0 5 + t 4 85 x4 t 0 0 1 (a) 1 0 No Solution will look di↵erent if di↵erent variables are chosen for parameters. 1 0 5 (d) 0 1 9 5 Precisely one solution: 9 2 3 1 0 3 2 55 (e) 40 1 0 0 0 0 0 Infinitely many solutions. The easiest way to parameterize this is to let x3 = s. Then: 2 3 2 3 2 3 2 3 x1 2 3s 2 3 4 x 2 5 = 4 5 5 = 4 55 + s 4 0 5 x3 s 0 1 Solutions might look di↵erent, but they will all have a direction vector that is a scalar multiple of [ 3, 0, 1]. 2 MWF Version 2: One corner of a parallelepiped in R3 is at the point A = (1, 1, 1). The three corners adjacent to that corner lie on the points B = (1, 1, 1), C = ( 1, 2, 4), and D = (2, 0, 3). (a) [2] What is the volume of the parallelepiped? (b) [1] What is the area of the triangle ABC? (c) [2] What are the other four corners of the parallelepiped? Solution: Let ! b = AB = (1, 1, 1) ! c = AC = ( 1, 2, 4) ! d = AC = (2, 0, 3) (1, 1, 1) = (0, 2, 2), (1, 1, 1) = ( 2, 1, 3), (1, 1, 1) = (1, 1, 2). (a) The volume is the absolute value of 0 0 @ 2 det 1 2 1 1 1 2 3 A. 2 Expanding this determinant in the first row, we obtain 0 · (2 + 3) ( 2) · ( 4 3) + ( 2) · (2 1) = 16 . Answer: The volume in of the parallelepiped is 16. (b) The area of ABC is 1 ||b ⇥ c|| = 2 0 i 1 ||b ⇥ c|| = det @0 2 1 1 j k 1 2 2A || = ||( 2 2)i 2 1 2 1 = || 4i + 4j 4k|| = 2|| 2 (0 i+j 4)j(0 4)k|| p k|| = 2 3. (c) Denote the parallelepiped in the problem by P . Let us translate this parallelepiped ! by the vector OA, so that the vertex A gets translated to the origin, O = (0, 0, 0), and the vetrices B, C and D, adjacent to A get translated to (0, 2, 2), ( 2, 1, 3), (1, 1, 2), respectively. I will denote this new (translated) parallelepiped by P0 and its vertices O = (0, 0, 0), (0, 2, 2), ( 2, 1, 3), (1, 1, 2), by B0 , C0 and D0 , respectively. Note that P0 is the parallelepiped spanned by b, c and d. Its vertices A0 , B0 , C0 , D0 , E0 , F0 , G0 , H0 are given as follows: ! OA0 = 0 = (0, 0, 0), ! OB0 = b = (0, 2, 2), ! OC0 = c = ( 2, 1, 3), ! OD0 = d = (1, 1, 2), ! OE0 = b + c = ( 2, 1, 1), ! OF0 = b + d = (1, 3, 0), ! OG0 = c + d = ( 1, 0, 5). ! OH0 = b + c + d = ( 1, 2, 3). ! The vertices of P are obtained by adding OA = (1, 1, 1) back to each of these. The first four, ! ! OA = OA0 + (1, 1, 1) = (1, 1, 1), ! ! OB = OB0 + (1, 1, 1) = (1, 1, 1), ! ! OC = OC0 + (1, 1, 1) = ( 1, 2, 4), ! ! OD = OD0 + (1, 1, 1) = (2, 0, 3), yield back the four corners A, B, C, D in the problem. The other four corners are given by ! ! OE = OE0 + (1, 1, 1) = ( 1, 0, 2), ! ! OF = OF0 + (1, 1, 1) = (2, 2, 1), ! ! OG = OG0 + (1, 1, 1) = (0, 1, 6), ! ! OH = OH0 + (1, 1, 1) = (0, 1, 4).