Answer Key for Exam #1
1. The vectors
3
0
4
1
1
3
and
1
1
1
4
4
1
each have have length 6:
p32+ 02+ 42+ 12+ +12+ 32= 6 = p12+ 12+ 12+ 42+ 42+ 12.
Their dot product is 3 + 0 + 4 + 4 + 4 + 3 = 18, so the angle θbetween them satisfies cosθ=18
6·6=1
2.
Therefore θ=π
3, or 60◦.
2. We use elimination on an augmented matrix. First two steps: subtract twice the first row from the second,
and three times the first row from the third. Next two steps: multiply the second row by −1, and then add
five times the second row to the third. This gives
1 2 3 10
2 3 1 7
3 1 2 19
−→
1 2 3 10
0−1−5−13
0−5−7−11
−→
1 2 3 10
0 1 5 13
0 0 18 54
.
The third equation gives x3= 3. Then the second equation becomes x2+ 15 = 13, so x2=−2. Finally the
first equation becomes x1−4 + 9 = 10, so x1= 5.
3. The inverse of ³2 3
3 4 ´is
1
2·4−3·3µ4−3
−3 2 ¶=µ−4 3
3−2¶,
and the inverse of ³3 4
4 5 ´is
1
3·5−4·4µ5−4
−4 3 ¶=µ−5 4
4−3¶,
so you might guess that A−1=
−4 0 0 3
0−5 4 0
0 4 −3 0
3 0 0 −2
, and once guessed this is easy to check. Alternatively
we can use row operations to reduce [A I] to £I A−1¤. I’ll try to avoid fractions: subtract the first row
from the fourth, and the second row from the third to get
2003 1000
0340 0100
0450 0010
3004 0001
−→
2 0 0 3 1 0 0 0
0 3 4 0 0 1 0 0
0 1 1 0 0 −110
1 0 0 1 −1 0 0 1
.
Next, subtract twice the fourth row from the first, and three times the third row from the second:
2 0 0 3 1 0 0 0
0 3 4 0 0 1 0 0
0 1 1 0 0 −110
1 0 0 1 −1 0 0 1
−→
0 0 0 1 3 0 0 −2
0 0 1 0 0 4 −3 0
0 1 1 0 0 −1 1 0
1 0 0 1 −1 0 0 1
.
Now subtract the second row from the third and the first row from the fourth:
0 0 0 1 3 0 0 −2
0 0 1 0 0 4 −3 0
0110 0−1 1 0
1001 −1 0 0 1
−→
0 0 0 1 3 0 0 −2
0 0 1 0 0 4 −3 0
0 1 0 0 0 −5 4 0
1 0 0 0 −4 0 0 3
.
