**Unformatted text preview: **ith
a
a
b
·
aa
√
. The shortest distance is (−9)2 + (−1)2 + (−1)2 + 112 = 204.
a
6) 12) (a) Here, as in problem 6, T is just the dot product of with itself. This is m.
aa
a
1
Similarly T = m bi so x = m m bi .
ab
ˆ
i=1
i=1
− x and ||e||2 = m (bi − x)2 and the standard deviation is the square root of
(b) = b ˆb
e
ˆ
i=1
this last quantity.
(c) = − x = (1, 2, 6) − 3(1, 1, 1) = ( 2, −1, 3). This dots to 0 with (3, 3, 3). The
e
b ˆa
− 111
T
aa
1
projection matrix P = T = 1 1 1 .
aa
3
111
22) We need to consider
E (C, D) = (C − 2D − 4)2 + (C − D − 2)2 + (C − −1)2 + (C + D − 0)2 + (C + 2D)2
so
∂...

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