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Unformatted text preview: 3 + 4 = 1 and x1 − x2 − x4 = 1, so x1 = 1 + x2 + x4 = 1 + 1 = 2. 2
x1 x2 1 The solution is x3 = 1 .
0
x4
(e) [Ab] = [1 − 124]. This is row echelon form. The free variables are y = t and z = s,
so x = 4 + y − 2z = 4 + t − 2s. In vector form the solution is x
4 + t − 2s
4
1
−2 y = = 0 + t 1 + s 0 .
t
z
s
0
0
1 4. The question asks if there are scalars a and b such that 2
0
−1
0 −1 −11 = a −1 + b 4 ? This is −1
4
−3
5
9
5
9 a
b 2
= −11 .
−3 Gaussian elimination gives 0 −1 
2
1 −4 
11
1 −4  11 −1
4  −11 → 0 −1 
2 → 0
1  −2 5
9  −3
0 29  −58
0
0
0
There is a unique solution: b = −2, a = 3. Then the given vector is indeed a linear
combination of the other two...
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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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