Math 2270, Fall 2015
Instructor: Thomas Goller
6 October 2015
Quiz #4 Solutions
1 2 3
(1) Compute 4 5 6 . (3 points)
7 8 9
Possible solution:
1 2 3
1
4 5 6 = 0
7 8 9
0
2
3
6
3
1
6 = 0
12
0
2
3
0
3
6 =0
0
Possible solution:
1 2 3
5 6
4 5 6 =
8 9
7 8 9
2
4 6
4 5
+3
7 9
7 8
= (45 48) 2(36
= 3 + 12 9
=0
42) + 3(32
35)
Possible solution: The columns of the matrix are dependent since
2 3 2 3 2 3
2
1
3
2 455 = 445 + 465 ,
8
7
9
so the matrix is not invertible and hence its determinant is 0.
(2) Let A and B be 5 ⇥ 5 matrices, with det A = 2 and det B = 3. Use properties of
determinants to compute each of the following: (2 points)
Solution:
(a) det BA = (det B)(det A) = 3( 2) =
(b) det 2A = 25 det A =
26 =
64
6
Math 2270, Fall 2015
Instructor: Thomas Goller
6 October 2015

0
(3) Compute the area of the parallelogram whose vertices are
,
0
points)

4
,
3

1
,
2

3
. (2
5




0
1
0
1
Solution: Let S be the unit square with vertices
,
,
,
and let T : R2 ! R2
0
0
1 1
4
1
be the linear transformation defined by T (~x) = A~x, where A =
. Then T (S) is
3 2
the parallelogram we are interested in and its area is given by
area of T (S) = (area of S) | det A| = 1 ·
4
3
1
2
= |8
( 3)| = 11.
(4) Bonus problem: Let a and b be positive numbers. Compute the area of the region in R2
bounded by the ellipse whose equation is
x21 x22
+ 2 = 1.
a2
b
(1 bonus point)
Solution: Let S be the region in R2 bounded by the unit circle, which has equation
u21 + u22 = 1. Let T : R2 ! R2 be the linear transformation defined by T (~u) = A~u, where
a 0
A=
. Then T (S) is the ellipse we are interested in. (To see this, note that the
0 b

✓ ◆ 
u1
u1
au1
image of a vector
on the unit circle is the vector T
=
, which satisfies
u2
u2
bu2
the equation for the ellipse:
(au1 )2 (bu2 )2
+
= u21 + u22 = 1.
2
2
a
b
Thus T maps the unit circle to the ellipse. Since T is linear, T must map the interior of
the unit circle to the interior of the ellipse.) Now by the area formula,
area of T (S) = (area of S) | det A| = ⇡ · 12 ·
a 0
0 b
= ⇡ab.