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Unformatted text preview: Math 136 Assignment 9 Solutions 1. Calculate the area of the parallelogram determined by the following vectors.
a) 2
−5
,
3
2 Solution: A = det 2 −5
32 = 2(2) − (−5)(3) = 19 1
−3
b) 0 , 1 −5
1
Solution: From our work in deriving the formula for the volume of a parallelepiped, we
found that the area of the parallelogram induced by u, v ∈ R3 is
A= u×v
Hence, we get 5
−3
1
√
A = 0 × 1 = 14 = 222
1
1
−5 2. Find the volume of the parallelepiped determined by the following vectors 5
1
1
−1, 4 , 2 a)
1
−1
−6 0 6 31
5
1
1
2 = det 0 3 −4 = −117
Solution: V = det −1 4
1 −1 6
1 −1 −6 1
1,
b)
4 1
3,
4 −2
1
−5 1 1 −2
1 1 −2
Solution: V = det 1 3 1 = det 0 2 3 = 6
4 4 −5
00 3 1 2 3. Use Cramer’s Rule to solve the following systems.
a)
x1 − 5x2 − 2x3 = −2
2x1
+ 3 x3 = 3
4x1 + x2 − x3 = 1 1 −5 −2
3 , so det A = −77 and
Solution: The coeﬃcient matrix is A = 2 0
4 1 −1
x1 = x2 = 1 −2 −2
2
1
23
3=
−77 4 1 −1
11 x2 = x1
30/77
Hence, x2 = 2/11 .
57/77
x3 −2 −5 −2
1
30
3
0
3=
−77 1
77
1 −1 1 −5 −2
57
1
20
3=
−77 4 1
77
1 3 b)
x1
+ 2 x3
3x1 − x2 + 3x3
−2x1 + x2 1
0
Solution: The coeﬃcient matrix is A = 3 −1
−2 1 = −2
=5
=0 2
3, so det A = −1 and
0 −2 0 2
1
5 −1 3 = −16
x1 =
−1 0
10
1 −2 2
1
3
5 3 = −32
x2 =
−1 −2 0 0
x2 = 1
0 −2
1
3 −1 5 = 7
−1 −2 1
0 x1
−16
Therefore, x2 = −32.
7
x3 2 −1 0 1
0 −1 3 2
−1
−1 4. Let A = 0 1 0 0. Use the cofactor method to calculate (A )23 and (A )42 .
0 2 03
−1 3 2
32
= −18. So
We have det A = 2 1 0 0 = −2
03
2 03
−1 201
1
18
2+3
0 3 2=
=
(−1)
=1
−18
18
003 −1 2 −1 0
1
2+4
0 1 0 =0
=
(−1)
−18
020 (A )23 (A )42 ...
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This note was uploaded on 07/13/2011 for the course MATH 136 taught by Professor All during the Winter '08 term at Waterloo.
 Winter '08
 All
 Math, Vectors

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