Answer Key for Exam #1
1. The vectors
1
3
−5
1
and
2
−2
2
2
have lengths 6 and 4, respectively:
p12+ 32+ (−5)2+ 12=√36 = 6 and p22+ (−2)2+ 22+ 22=√16 = 4.
Their dot product is 1·2+3(−2)+(−5)2+1·2 = −12, so the angle θbetween them satisfies cos θ=−12
6·4=−1
2.
Therefore θ= 120◦, or θ=2π
3.
2. We use elimination on an augmented matrix. First two steps: subtract three times the first row from the
second, and four times the first row from the third. Next two steps: multiply the second row by −1, and
then add twice the new second row to the third. This gives
1 5 6 22
3 4 1 3
4−2 5 37
−→
1 5 6 22
0−11 −17 −63
0−22 −19 −51
−→
1 5 6 22
0 11 17 63
0 0 15 75
.
The third equation gives x3= 5. Then the second equation becomes 11x2+ 85 = 63, so x2=−2. Finally
the first equation becomes x1−10 + 30 = 22, so x1= 2.
3(a) Step 1: subtract three times the first row of Afrom the second row, and subtract the first row from the
third. Step 2: subtract twice the second row from the third.
A=
1 2 1
3 7 6
1 4 8
−→
1 2 1
0 1 3
0 2 7
−→
1 2 1
0 1 3
0 0 1
=U.
Step 1 puts 3 in the second row, first column of L; and 1 in the third row, first column. Then step 2 puts a
2 in the second row, third column of L, so L=
100
310
121
.
3(b) We may now solve A~x =
5
8
−12
in two steps. First we solve L~y =~
b, which in this case is
1 0 0
3 1 0
1 2 1
y1
y2
y3
=
5
8
−12
and we get y1= 5; 15 + y2= 8, so y2=−7; and 5 −14 + y3=−12, so y3=−3. Finally we solve U ~x =~y,
which in this case is
1 2 1
0 1 3
0 0 1
x1
x2
x3
=
5
−7
−3
.
Here we have x3=−3; x2−9 = −7, so x2= 2; x1+ 4 −3 = 5, so x1= 4.
3(c) Here we have to reduce [L I] to £I L−1¤by row operations, which is not too hard:
100 100
310 010
121 001
−→
100 100
010 −3 1 0
021 −1 0 1
−→
1 0 0 1 0 0
0 1 0 −3 1 0
0 0 1 5 −2 1
.
