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Unformatted text preview: 1. (a) The volume of the parallelepiped spanned by u, v and w is the absolute value of
determinant of the matrix spanned by u, v and w . Calculate
123
149
1 8 27 = 123
026
0 6 24 =1· 26
6 24 123
149
1 8 27 = 12. = 12. (b) Note that w′ = w − v . Therefore,
123
149
0 4 18 = Therefore, the volume of the parallelepiped spanned by u, v and w′ is also 12.
2. We denote log as ln.
Let f (x, y ) = ex log(1 + y ). Then
∂f
= ex log(1 + y ),
∂x ∂f
ex
=
.
∂y
1+y At the base point (x0 , y0 ) = (0, 0), we have
∂f
(0, 0) = e0 log(1 + 0) = 0,
∂x ∂f
e0
(0, 0) =
= 1.
∂y
1+0 Therefore, by linear approximation, we have
· e0.02 log 1.01 = f (0.02, 0.01) = f (0, 0) + 1 ∂f
∂f
(0, 0) · 0.02 +
(0, 0) · 0.01 = 0.01.
∂x
∂y ...
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 '08
 OBLOMKOV
 Determinant

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