Answer Key for Exam #1
1. The vectors
1
1
1
1
and
1
1
0
1
have lengths p12+ 12+ 12+ 12= 2 and p12+ 12+ 02+ 12=√3,
respectively, and their dot pro duct is 1 + 1 + 0 + 1 = 3. So the angle θbetween them satisfies cos θ=3
2√3=
√3
2. Therefore θ=π
6, or 30◦.
2. We use elimination on an augmented matrix. Subtracting the first row from the second and third, and
then adding the second row to the third, we have
1 1 2 3
1 1 3 7
1 9 1 15
−→
1 1 2 3
0 0 1 4
0 8 −1 12
−→
1 1 2 3
0 0 1 4
0 8 0 16
.
The last step wasn’t necessary since the system was already triangular, but it doesn’t hurt. The second
equation gives x3= 4 and the third one says x2= 2. Then the first equation becomes x1+ 2 + 8 = 3, so
x1=−7.
3. Here we have to reduce [A I] to £I A−1¤by row operations. Using the first pivot to eliminate the other
entries in the first column of Awe have
1 2 3 1 0 0
2 3 1 0 1 0
3 1 2 0 0 1
−→
1 2 3 1 0 0
0−1−5−2 1 0
0−5−7−3 0 1
.
Multiplying the second row by −1 to make the second pivot equal 1, and then using it to clear out the other
entries in the second column, we get
1 2 3 1 0 0
0−1−5−2 1 0
0−5−7−3 0 1
−→
1 0 −7−3 2 0
0 1 5 2 −1 0
0 0 18 7 −5 1
.
To put off the fractions as long as possible we can multiply the first and second rows by 18:
1 0 −7−3 2 0
0 1 5 2 −1 0
0 0 18 7 −5 1
−→
18 0 −126 −54 36 0
0 18 90 36 −18 0
0 0 18 7 −5 1
.
If we now add 7 times the third row to the first, and subtract 5 times the third row from the second, we get
18 0 −126 −54 36 0
0 18 90 36 −18 0
0 0 18 7 −5 1
−→
18 0 0 −5 1 7
0 18 0 1 7 −5
0 0 18 7 −5 1
.
Therefore A−1=1
18
−5 1 7
1 7 −5
7−5 1
.
4(a) Step 1: subtract twice the first row of Afrom the second, and thrice the first row from the third. Step
2: subtract 4 times the second row from the third:
A=
1 4 3
2 9 8
3 16 18
−→
1 4 3
0 1 2
0 4 9
−→
1 4 3
0 1 2
0 0 1
=U.
