Question

1. Graph the system of linear equations. Solve the system and interpret your answer 3y 2 -+2y 3 2. Solve the system of linear equations for and y (Cos ) x(sin 0) y = 1 (sin 0) x (cos 0) y = 1 3. Use back substitution to solve the system. 6r23r =-3 r22r3 1 3-2 4. Slove the given system by Gaussian elimination.. 4x1-2+x3-1 +2x2-3r3 = 2 2x 3= 1 5. Identify the element ary row operation (s) being performed on the matrix A to obtain the matrix B [ 3 -1 -4 3-1-4 (а) А%3D -4 B = 5 3 7 0 -5 -2 -2 -2 -1 3 -1 -2 3 (b) А- 2 -5 1 -7 B = 0 -9 7 -11 5 4 -7 6 0 -6 8 -4 6. Slove the given system by Gaussian elimination 4x1 2+x3-1 12x2 3r3 2 2x1 23 1 7. Slove the given system by Gauss Jordan Method 4r 223 (a) 12r2+3 = 2 3x 3r2 = 5 =-1 2x123+24=-6 3a14r2+4= 1 5r22x3 +6a4-3 5x12r2 T3-43 (b) 22r3-4-1 2x1223-2r4-2 (c) -1+2x2-4r3 +4= 1 3r 3r4-3

Answer 1

7-x+2y = 3 a +39 52 = 1 + 1 = 1 ((x+y = 3 = 2 + 12 = 1 sy = 5 * 4 = 1 a = 2-3y = 2-3 = -1 x+3y=2 The intersection point of the lines is the solution. So the solution is, d=-1, y = 2 Coso 2 Coso x + Sino y = 1 - sind x + coso y = 1 x Sind + Meloso ay sin 20 y of Cosby= sind + Coso. y y = sind+ Coso x=1- Sindy - sin 20-sindlose coso - Cos2o - Sindloso - Corso-sino coso ay a= Coso-sind, y = Sindt loso 3. + 6 x y 24 + 6x 5+ 3x(-2) = -37% = -3+6-30 =-27 3 xen-3 x2 + 2x3 =1 ay X₂ + 2x (-2) = 1 ay = 1+ = 5 3=- 2 => 24 =-27, 2=5, 3 = -2

d 14. 4.06 - 2 + z = – 1 2 14 -1 11-17 x + 2x₂ - 3X3 = 2 = 1 2 - 32 zxi - de +13= 1 ) [2 - 1 RkRi 11 2 -3 127 11 2 - 3 → R₂ R₂-4R, 4 -1 1 -1 o 9 13 R3 R3 -22, Lo -5 7 1-3 R2 11 2 -3 12 R, TR, -2R2 1 0 - 5107 R2 RatsRo 1 -13 1 toglu. R₂5 Lo -5 7 1-3 37 3112 To 0-2 3 1 Rovan) F1 0-50TR h2+3 R3 TI 001-17 10 10-12 loo 19. HON Ro Rzx ( - ) 101-18 1 1 R1 R1 + 2 R3 looll-g) So the solution is Mi-1, M2=-12, ₂= -9