# solutions_to_midterm_Fall2019.pdf - MATH 204 Instructor...

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Unformatted text preview: MATH 204 Instructor: Dominic Fall 2019: Midterm Solutions (1) Solve by Cramer's Rule only: Solution −1 A= 2 3 2 0 −4 Andoh −x + 2y − 3z = 1 2x + z=0 3x − 4y + 4z = 2 −3 1 1 , A1 = 0 4 2 2 0 −4 −3 −1 1 , A2 = 2 4 3 1 −3 −1 0 1 , A3 = 2 2 4 3 2 0 −4 1 0 2 −1 2 −3 −1 2 |A| = 2 0 1 2 0 = [−1 · 0 · −1 + 2 · 1 · 3 + (−3) · 2 · −4] − [3 · 0 · −3 + (−4) · 1 · −1 + 4 · 2 · 2] 3 −4 4 3 −4 = 10 1 2 −3 1 2 |A1 | = 0 0 1 0 0 = [1 · 0 · −4 + 2 · 1 · 2 + (−3) · 0 · −4] − [−3 · 0 · 2 + 1 · 1 · (−4) + 2 · 0 · 4] 2 −4 4 2 −4 =8 −1 1 −3 −1 1 |A2 | = 2 0 1 2 0 = [−1 · 0 · −4 + 1 · 1 · 3 + (−3) · 2 · 2] − [−3 · 0 · 3 + (−1) · 1 · 2 + 1 · 2 · 4] 3 2 4 3 2 = −15 −1 2 1 −1 2 |A3 | = 2 0 1 0 0 = [−1 · 0 · (−4) + 2 · 1 · 3 + 1 · 0 · (−4)] − [1 · 0 · 3 + (−1) · 1 · (−4) + 2 · 0 · 2] 3 −4 2 3 −4 = −16 x y = z (2) Compute the determinant of A |A1 | |A| |A2 | |A| |A3 | |A| 1 4 |A| = −2 3 (3) Compute the inverse of A −2 −6 4 −6 3 3 −9 9 8/10 −15/10 = −16/10 −2 −6 4 −6 3 3 −9 9 1 2 −3 2 1 1 2 R4 =r4 −3r1 0 −−−−−−−→ −3 R2 =r2 −4r1 0 2 R3 =r3 −2r1 0 −2 2 0 0 3 −9 −3 0 1 4 A= −2 3 Solution 1 A= 3 −2 1 −2 = (1)(2)(−3)(−1) = 6 −1 −1 1 2 1 0 0 3 1 of 3 MATH 204 Instructor: Dominic Fall 2019: Midterm Solutions Andoh Solution 1 1 2 1 3 1 0 0 −2 0 3 0 1 0 1 1 0 1 3 3/2 2 0 2 7 1 R =r1 +r3 −−1−−− −−→ 0 R2 =r2 −3r3 0 A−1 −3/2 = 9/2 −1 (4) If (2A − 5I)−1 =  5 2 1 0 R =r2 −3r1 0 −−2−−− −−→ 0 R3 =r3 +2r1 1 0 0 0 R =r1 −r2 −1/2 0 −−1−−− −−→ R3 =r3 −r2 0 1 0 1 0 −3/2 −9/2 −1 0 0 1 0 0 1 3/2 −7/2 1 0 −2 2 1 −6 7 1 0 0 1 0 0 0 0 r2 R3 = −2 1 0 −−−−−→ 0 1 −1/2 1/2 0 3/2 −1/2 0 −1 1 1 1 −3 2 −1 3 1 1 −3 1 1 −3 1 3/2 −7/2 1  −4 , determine A 1 Solution −1 (2A − 5I)  = 5 2 −4 1     −1 1 5 −4 1 4 2A − 5I = = 2 1 13 −2 5       1 4 1 1 4 +5 5 0 13 2A = + = 13 2 5 0 5 − 13 13 −2 5 13 + 5  66    4 1 1 33 2 2 2 = A= 13 − 22 70 13 −1 35 2 (5) Determine the values of a and b, such that the given sytem of equations have: (a) exactly one solution. (b) no solutions. (c) innite solutions. x + 2y = 3 ax + by = −9 Solution  1 2 a b 3 −9  R =r −ar 2 1 −−2−−− −−→  1 0 2 3 (b − 2a) (−9 − 3a)  innite solution: a = −3; b = 2a = −6 one solution: b 6= 2a; for all a no solution: b = 2a; a 6= −3 (6) Solve the system of Equations: 2x1 + x2 + x3 + 2x4 5x − 2x + x − 3x 1 2 3 4 −x + 3x + 2x + 2x 1 2 3 4 3x1 + 2x2 + 3x3 − 5x4 = −1 =0 =1 = 12 2 of 3 MATH 204 Instructor: Dominic Fall 2019: Midterm Solutions Andoh Solution 2 1 1 5 −2 1 −1 3 2 3 2 3 2 x1 x2 5 ⇒ x3 = −1 3 x4 9 −13 1 13 −49/4 = 17 −16 59/4 1 2 −1 0 = 1 12 −1 2 −1 −3 0 2 1 −5 12 −1 11/4 31/4 0 1 −29/4 12 −3 1 17 68 = 4 −3 −51 −2 −34 x1 2 x2 −3 2 x3 x4 −5 1 −2 3 2 = 1 1 2 3 −7 −12 20 3 1 17 3 of 3 ...
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