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(c) The basis vectors [ 1, 0 ] and [ 0, 1 ] are transformed to [ 2, 0 ] and [ 0, 2 ], respectively.
Thus, M performs a uniform scale, scaling both the x and y dimensions by 2.
(d) The basis vectors [ 1, 0 ] and [ 0, 1 ] are transformed to [ 4, 0 ] and [ 0, 7 ], respectively.
Thus, M performs a nonuniform scale, scaling the x dimension by 4 and the y di-
mension by 7.
(e) The basis vectors [ 1, 0 ] and [ 0, 1 ] are transformed to [−1, 0 ] and [ 0, 1 ], respectively.
Thus, M performs a reflection across the y axis, negating x values and leaving y values
untouched.
(f) The basis vectors [ 1, 0 ] and [ 0, 1 ] are transformed to [ 0, −2 ] and [ 2, 0 ], respectively.
Thus, M is performing a combination of transformations: it is rotating clockwise by
90
◦
and scaling both dimensions uniformly by 2. This can be confirmed by multiplying
the appropriate matrices from parts (a) and (c), which perform these transformations
individually:
2
0 −1
1
0
0 −2
2
=
.
0
0
2
0
2
4
0
3
5
−b
z
b
y
This matrix is skew symmetric, as desired, since M
T
= −M.
8. M =
b
z
0
−b
x
−b
y
b
x
0
9. (a) 3
(b) 1
(c) 4
(d) 2
10. The result vector element w
i
is the product of the ith row of M multiplied by the column
vector v. To have w
i
= v
i
−v
i−1
, the ith row of M needs to capture the ith element of v,
as well as the negative of the (i−1)th element, but exclude all others. This means that
<
:
1
if j = i,
m
ij
=
−1
if j = i−1,
0
otherwise.
Thus,
2
4
3
5
1
0
0
0
0
0
0
0
0
0
−1
1
0
0
0
0
0
0
0
0
0
−1
1
0
0
0
0
0
0
0
0
0
−1
1
0
0
0
0
0
0
0
0
0
−1
1
0
0
0
0
0
M =
.
0
0
0
0
−1
1
0
0
0
0
0
0
0
0
0
−1
1
0
0
0
0
0
0
0
0
0
−1
1
0
0
0
0
0
0
0
0
0
−1
1
0
0
0
0
0
0
0
0
0
−1
1
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