Unformatted text preview: 1, 1) = 0 and (a, b, c) · (1, 1, −1) = 0. That is all simultaneous solutions to a + b + c = 0
and a + b − c = 0, a line. A basis for this line is the vector (−1, 1, 0).
22) P is the nullspace of the matrix [1 1 1 1]. We can read oﬀ a basis for P ⊥ , it is the normal 1
1
vector, namely .
1
1 26) Pick two perpendicular vectors and then take their cross product to get the third. So 1
3
1
4 works. Now
A = 1 −2
1 −1 −5 1
1
1
1
3
1
300
4 = 0 14 0 .
AT A = 3 −2 −1 1 −2
1
4 −5
1...
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This note was uploaded on 01/30/2014 for the course MATH 2940 taught by Professor Hui during the Fall '05 term at Cornell.
 Fall '05
 HUI
 Math, Linear Algebra, Algebra

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