resoluçao cap 3 steinbruch, Exercises of Linear Algebra

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Resolvendo para xi. x+1=—-4>x=-5 Resolvendo para y: y= -1 > y Resolvendo para z:2+3=—-l3z=—44 3.16 Problemas Propostos: 1. Dados os vetores i’ = (1,a,-2a-1), 2 = (a,a—1, 1) ew = (a,-1,1), determine a, de modo iad = (+ a). Solugio: (1, a, -2a —1).(@,a— 1,1) = [(1,a,—-2a — 1) + (a,a —1,1)].(@,-1, 1) (a+ ala—1)—-2a—-1) = [fa+ 1),a+a—-1,2a—-141]-{a,-1,1) Qa—1= [a+ 1,20,-24].(a,-1, 1) a? —2a—-1=ala+ 1)— (2a-1}-2a a? —a? -2a-a+2at+M=14+1 ata— AB=B-A=(-4+1,1-0,1—-2) = (-3,1,-]) ae BC =C-B=(0+4,1-1,3-1) =(4,0,2) AC=C-A =(0+1,1-0,3-2)=(1,1,1) =a BC.AB = 4(-3)+0.1+2(-1) =-12-2=-14 = (BC.AB)AC = (14.1, -14.1, -14.1) = (-14,-14, -14). Portanto, av- # = (-14, -14,-14) + (-3,1,-1) > 15)] 3. Determinar o vetor @, sabendo que (3,7,1) + 2a = (6, 10,4) - a Solugio: (3, 7,1) + 2(x, y,z) = (6, 10,4) - (x,y,z) (3, 7,1) + (2x, 2y,22z) = (6-x,10—- y,4—2) (3+ 2x,7+2y,1+ 22) = (6—-2,10- y,4—- 2) Parax, temos: 34+ 24 =6-x>3x=35x=1 Para y, temos: 7 +2y = 10-y= y= Para z, temos: 1+ 22 =4-z =z=1 # = (1,1,1)