dan Elimination. 1. x1 + x2 - x3 = -2 2x1 - x2 + x3 = -5 -x1 + 2x2 -3x3 = -

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Chapter1: Systems Of Linear Equations
Section1.1: Introduction To Systems Of Linear Equations
Problem 90E: Consider the system of linear equations in x and y. ax+by=ecx+dy=f Under what conditions will the...
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Solve the given system of linear equations using the Gauss-Jordan Elimination.

1.  x1 + x2 - x3 = -2

    2x1 - x2 + x= -5

    -x1 + 2x-3x3 = -4

These is the example or guide  of gauss jordan elimination that might help...

 

Solve the given system of linear equations using the Gauss - Jor dan
Elimination.
1) 3x - 2y t82=9
-2x +2y + z - 3
x + 2y-32 - 8
%3D
%3D
SOUTION:
- 2R2 +R, D R,
8 R2 t Rg =D Rg
5/3
-2
8.
9.
O -4/3
-2
2
3
19/6
2
-3
8.
O 31/3
3/3
R3 (swap)
- 3
R,>
R3 D R,
-2
2
3
- 4/3 5/3
3 -2
8.
-5% 1%
19/6
2R, +R2 D R2
- 3 R, + R3 > R3
* Rg tR, D R,
2 -3
8
* R, + R2 D R2
-5
19
O -8
LI
-15
4
R2
+ R2
X + Oy + o2 :3 x = 3
Ox + y + 02 = 4 D Y: 4
Ox + Oy + 2 *
2
- 3
8.
-5/6 19/6
LI
-15
2.
Transcribed Image Text:Solve the given system of linear equations using the Gauss - Jor dan Elimination. 1) 3x - 2y t82=9 -2x +2y + z - 3 x + 2y-32 - 8 %3D %3D SOUTION: - 2R2 +R, D R, 8 R2 t Rg =D Rg 5/3 -2 8. 9. O -4/3 -2 2 3 19/6 2 -3 8. O 31/3 3/3 R3 (swap) - 3 R,> R3 D R, -2 2 3 - 4/3 5/3 3 -2 8. -5% 1% 19/6 2R, +R2 D R2 - 3 R, + R3 > R3 * Rg tR, D R, 2 -3 8 * R, + R2 D R2 -5 19 O -8 LI -15 4 R2 + R2 X + Oy + o2 :3 x = 3 Ox + y + 02 = 4 D Y: 4 Ox + Oy + 2 * 2 - 3 8. -5/6 19/6 LI -15 2.
2.)2×, - 2x2 tメ, : 3
%3D
3 x, + x2 -X, =
X,-3x, +2x, こ0
%3D
SOUTION:
R」+R2 ヤ R」
R, + 2R2D R,
17/5
5/8
メ」
X2 X 3
2
-2
3
3
2.
リ4
-/4.
-3
Ri
A R,
支 Ry
2
17/6
ソ/2
-3
2
R,+
R3 R,
- 3R, +R2 Rz
-R, +R3 D R3
/2
Rz
Rz D R2
2
3/2
S/2
-|
-5/2
3/2
4
-2
X」+0×at0メ 2 ×=2
0×,+ メ2 t0X,= 0 |×2-0
0x,+ Ox2t X,= -1 |xっ-|
/2
> R2
Y2
%3D
/2
5/8
-/2
- 1
0x,t Oxzt メ,
-2
3/2
ー|oleO
の
Transcribed Image Text:2.)2×, - 2x2 tメ, : 3 %3D 3 x, + x2 -X, = X,-3x, +2x, こ0 %3D SOUTION: R」+R2 ヤ R」 R, + 2R2D R, 17/5 5/8 メ」 X2 X 3 2 -2 3 3 2. リ4 -/4. -3 Ri A R, 支 Ry 2 17/6 ソ/2 -3 2 R,+ R3 R, - 3R, +R2 Rz -R, +R3 D R3 /2 Rz Rz D R2 2 3/2 S/2 -| -5/2 3/2 4 -2 X」+0×at0メ 2 ×=2 0×,+ メ2 t0X,= 0 |×2-0 0x,+ Ox2t X,= -1 |xっ-| /2 > R2 Y2 %3D /2 5/8 -/2 - 1 0x,t Oxzt メ, -2 3/2 ー|oleO の
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