Matrices De Orden 3

Resolver los sistemas:
10) {█(3x+y+z=1@x+2y-z=1@x+y+2z=-17)┤ D=[█(■(3&1 1@1 &2 -1)@1 1 2@3 1 1@1 2 -1)]= [12+1+ (-1)] – [2+(-3)+2]= 12–1
DX[█(■(1&1 1@1& 2-1)@-17 1 2)]= (1) |■(2&-1@1&2)| – (1)|■(1&-1@-17&2)| + (1)|■(1&2@-17&1)|
= (1) (4+1) – (1) (2- 17) + (1) (1+34)
= (1) (5) – (1) (-15) + (1) (35)
= 5 +15+35
DY[█(3 11@█(■(1 1&-1@1 -17&2)))]= (3)|■(1&-1@-17&2)|- (1)|■(1&-1@1&2)| + (1)|■(1&1@1&-17)|
= (3) (2-17) – (1) (2+1) + (1) (-17-1)
= (3) (-15) – (1) (3) + (1) (-18)
= - 45 – 3 – 18
DZ[█(3 11@■(1 2& 1@1 1&-17))]= (3)|■(2&1@1&-17)|- (1)|■(1&1@1&-17)|+ (1)|■(1&2@1&1)|
= (3) (-34-1) – (1) (-17-1) + (1) (1-2)
= (3) (-35) – (1) (-18) + (1) (-1)
= -105 + 18 – 1
X=DX/D=55/11= 5
Y=DY/D=(-66)/11= -6 C.S{█(X=5@Y=-6@Z=-8 )┤
Z=DZ/D=(-88)/11= -8

21){█(5x-3z=2@2z-y=-5@x+2y-4z=8)┤ D[█(■(5& 0 -3@0&-1 2)@1 2 -4@5 0 -3 @0-1 2 )]= [20+0+0] - [3+20+0]= 20-23
DX[█(2 0 -3@■(-5 -1& 2@8 2&-4))] = (2)|■(-1&2@2&-4)|-(0)|■(-5&2@8&-4)|+(-3)|■(-5&-1@8&2)|
= (2) (4-4) – (0) (20-16) + (-3) (-10+8)
= (2)(0) – (0) (4) + (-3) (-2)
= 0 – 0 + 6
DY[█(5 2 -3@■(0&-5 2@1&8 -4))]= (5)|■(-5&2@8&-4)|-(2) |■(0&2@1&-4)|+ (-3)|■(0&-5@1&8)|
= (5) (20-16) – (2) (0-2) + (-3) (0+5)
= (5) (4) – (2)(-2) + (-3) (5)
= 20 + 4 -15
DZ[█(5 0 2@■(0&-1 -5@1&2 8))]= (5)|■(-1&-5@2&8)|-(0) |■(0&-5@1&8)|+ (2)|■(0&-1@1&2)|
= (5) (-8+10) – (0) (0+5) + (2) (0+1)
= (5) (2) – (0) (5) + (2) (1)
= 10– 0 + 2
X=DX/D=6/(-3)= -2
Y=DY/D=9/(-3)= -3 C.S{█(X=-2@Y=-3@Y=-4)┤
Z=DZ/D=12/(-3)= -4


3){█(2X+3Y+4Z=3@2X+6Y+8Z=5@4X+9Y-4Z=4)┤ D[█(2 3 4@2 6 8@ 49 -4@■(2&3 4@2&6 8))]=[-48+72+96]-[96+144-24]=120-216
DX[█(3 3 4@■(5&6 8@4&9 -4))]= (3)|■(6&8@9&-4)|- (3)|■(5&8@4&-4)|+(4)|■(5&6@4&9)|
= (3) (-24-72) – (3) (-20-32) +...