2x1 3x1 - X11 - + 3x3 4x2+8x3 + 3x4 - 4x5 = - - x22x3 + x4 - == x5= -2x15x29x3 - 3x4 - 5x5 5 сл 2 -8

College Algebra
7th Edition
ISBN:9781305115545
Author:James Stewart, Lothar Redlin, Saleem Watson
Publisher:James Stewart, Lothar Redlin, Saleem Watson
Chapter6: Matrices And Determinants
Section: Chapter Questions
Problem 6T: Use Gaussian elimination to find the complete solution of the system, or show that no solution...
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Section 3.4: Number 2(j) only!

 

 

2. Use Gaussian elimination to solve the following systems of linear equa-
tions.
-
x1 + 2x2 x3 =
(a) 2x12x2 + x3 =
3x1 +5x2
x1 + 2x2
x1 - 2x2
2x13x2
1
(b)
X3=1
x3 = 6
3x15x2
= 7
2x3
=-1
X1
+5x3 = 9
+2x4=
= 6
x36x4
= 17
-
X2
-
(d)
3x1 +5x2
-
(c)
2x14x2 x3 + 2x4 = 12
2x1
X1
2x1
-
7x311x4= 7
-
x2+6x3 +
6x4
= -2
2x3 + 3x4 =
-63
-2x1 + x2
-
4x3
3x4 == 0
3x12x29x3 +
10x4 = -5
-
x1 - 4x2 x3 + x4 =
3
-
x12x2 x3 + 3x4 =
(e) 2x18x2 + x3 - 4x4 =
-x14x22x3 +5x4 =
-
2x12x2 x3 + 6x42x5 = 1
(g)
-
X1 x2
x3 2x4 -
x5 = 2
4x14x2+5x3+7x4x5 = 6
9
(f) 2x14x2
6
X2
2
x3 + 6x4 = 5
+ 2x4 = 3
=
Transcribed Image Text:2. Use Gaussian elimination to solve the following systems of linear equa- tions. - x1 + 2x2 x3 = (a) 2x12x2 + x3 = 3x1 +5x2 x1 + 2x2 x1 - 2x2 2x13x2 1 (b) X3=1 x3 = 6 3x15x2 = 7 2x3 =-1 X1 +5x3 = 9 +2x4= = 6 x36x4 = 17 - X2 - (d) 3x1 +5x2 - (c) 2x14x2 x3 + 2x4 = 12 2x1 X1 2x1 - 7x311x4= 7 - x2+6x3 + 6x4 = -2 2x3 + 3x4 = -63 -2x1 + x2 - 4x3 3x4 == 0 3x12x29x3 + 10x4 = -5 - x1 - 4x2 x3 + x4 = 3 - x12x2 x3 + 3x4 = (e) 2x18x2 + x3 - 4x4 = -x14x22x3 +5x4 = - 2x12x2 x3 + 6x42x5 = 1 (g) - X1 x2 x3 2x4 - x5 = 2 4x14x2+5x3+7x4x5 = 6 9 (f) 2x14x2 6 X2 2 x3 + 6x4 = 5 + 2x4 = 3 =
(h)
-
3x1
X1
-
-
X2
X2
5x12x2
2x1-
-
X3 x4 + 2x5
5
-
X3
-
2x4-
x5 =
2
x3
-
X2
-
3x43x510
2x4x5=
5
2x1
(j)
X1
-
-2x15x29x3
3x1
-
(i)
X1 x2 + 2x3 + 3×4 +
2x13x26x3 + 9x4 +
7x12x2+4x3 + 8x4 + x5 =
+ 3x3
3x14x2+8x3 + 3x4
x22x3x4
-
3x45x5=-8
x22x3 + 4x4 +
x5
2
×5
-1
4x5 = -5
6
-
4x5
5
8
-
x5 =
2
Transcribed Image Text:(h) - 3x1 X1 - - X2 X2 5x12x2 2x1- - X3 x4 + 2x5 5 - X3 - 2x4- x5 = 2 x3 - X2 - 3x43x510 2x4x5= 5 2x1 (j) X1 - -2x15x29x3 3x1 - (i) X1 x2 + 2x3 + 3×4 + 2x13x26x3 + 9x4 + 7x12x2+4x3 + 8x4 + x5 = + 3x3 3x14x2+8x3 + 3x4 x22x3x4 - 3x45x5=-8 x22x3 + 4x4 + x5 2 ×5 -1 4x5 = -5 6 - 4x5 5 8 - x5 = 2
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