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Unformatted text preview: − 2 7
14
−7
−4 1
7
−6
0
6 −42
3 −18
1
0
0
0 1
7
6
0
0 −42
0
0 
2
 −5  −2  −7 
2

5
.
 −7  − 13
2 The rank of A is 3 while the rank of the augmented matrix [Ab] is 4, so the system
has no solution. (OR: From the last row, we can see that there is no solution.)
(c) 2 2 2 −8  1
2 2 2 −8 
1
0  4 → 0 2 2 16 
2.
[Ab] = 4 6 6
6 6 10 −4  2
0 0 4 20  −1
This is upper triangular form. We can ﬁnd the solutions by back substitution. Variable
x4 = t is free and 4x3 +20x4 = −1, so x3 = − 1 −5x4 = − 1 −5t. Then x2 = 1−x3 −8x4 =
4
4 5
4 − 3t and x1 = 1
2 − x2 − x3 + 4x4 = − 1 + 12t. The solution
2 1 1
−2
12
− 2 + 12t −3 5 − 3t 5 4 x = 41 − − 5t = − 1 + t −5
4
4
t
0
1 is . (d) 2 −3 −1
1
[Ab] = 0
2
3
0 1 −1 0 −1
→
0
2
0
3 0 −1
0
1
→
0
0
0
0 1 −1
0
1
→
0
0
0
0 1 −1
0
1
→
0
0
0
0 
5
1 −1
0 −1  1  −1 4 −1  5 → 2 −3 0

1
2 −1 −...
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This note was uploaded on 02/13/2014 for the course MATH 2050 taught by Professor Dryuan during the Fall '12 term at Memorial University.
 Fall '12
 DrYuan
 Math, Linear Algebra, Algebra

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