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**Unformatted text preview: **Math 115 - Lab 3 - Fall 2008
Topics:
• Gaussian elimination
• Homogeneous equations
• Matrix addition, scalar multiplication and transposition
• Matrix multiplication
• Matrix inversion
You are to provide full solutions to the following problems. You are allowed, and encouraged to
collaborate with your classmates, use your notes and textbook and ask the TA for guidance. Direct
copying of solutions is not encouraged, nor is it allowed or ethical. Last name:
Student number: First name: Math 115 - Lab 3 - Fall 2008 Student number: 1. Find all solutions to the following homogeneous systems. If there is more than one
solution, then please express the solution set in vector equation form (i.e. a linear
combination of vectors).
a)
5x
+z=0
x + 2y
=0
y +z=0
Solution: 501
120 1 2 0 ∼ 0 1 1 , swapping rows
011
501 1
20
∼ 0
1 1 R 3 = R 3 − 5R 1
0 −10 1 12 0
∼ 0 1 1 R3 = R3 + 10R2
0 0 11 120
1
R3
∼ 0 1 1 R3 =
11
001 and from this it is clear that this system has only the trivial solution. 1 b)
x + 2y + 3z = 0
x
+ 2z = 0
2x + 6y + 7z = 0
Solution: 123
102
1 0 2 ∼ 1 2 3
267
267 102
0 2 1
∼
063 102
0 1 1
∼
2
063 102
0 1 1
∼
2
000 , swapping rows R 2 = R 2 − R 1 ; R 3 = R 3 − 2R 1 R2 = 1 R2
2 R 3 = R 3 − 6R 2 So letting z = t, we obtain the following.
1
R2 reads y + t = 0
2
1
=⇒ y = − t
2
R1 reads x + 2t = 0
=⇒ x = −2t
Therefore the solution set is −2
−2t
0
−4 − 1 t = 0 + t − 1 = t −1 , for t ∈ R
2
2
t
0
1
2 2 2. Compute the following sums, or indicate ‘cannot add’.
a)
107
3
−2 1 2 −4 4
0
1
3 −3
2 + Solution:
107
−3 1 2
2 + −4 4
0
1
3 −3
2 = −3 4
7
−1 4 −1 b)
sin 1 e−3 + Solution: sin 1 + e−3 = sin 1 + e−3 c)
107
3
−2 1 2 1
−4
2
+ 4
3
0 −3 Solution: Cannot add, since the matrices have diﬀerent sizes. 3 3. Compute the following if possible, or indicate ‘cannot compute’.
a)
π
2
π
3 1
π 2
3 108 Solution:
π
2
π
3 = 2
3 108 1
(π)π
2
1π
(π) 3 = 1
π 1
2
1
3 1
(π)2
3
1
( π )108 2
3
108
π b)
T
−5 −1
7
0
8
1 Solution:
T
−5 −1
7
0
8
1 4 = −5 7 8
−1 0 1 c)
T 11
−1
1 2 1 1 −1 11
0
1 Solution:
T 11
−1
1 2 1 1 −1 =
11
0
1 = 121
111
10
01 −1
1 1 −1 0
1 d)
T
11
2 1
11 −1
10
1 −1 1 Solution: Cannot compute, since the matrices are the wrong size to multiply
together. 5 12
1 −1 0
, C = 3 4 .
0
11
56 211
,B=
130 e) (A + B )C , where A =
Solution 1:
(A + B )C = 211
130 = 301
141 = 8 12
18 24 + 1
3
5 1 −1 0
0
11 2
4
6 12
3 4
56 Solution 2:
(A + B )C = AC + BC
= 10 14
10 14 = 12 3 4 +
56 211
130 = 8 12
18 24 + 6 −2 −2
8 10 1 −1 0
0
11 12
3 4
56 f)
−1 1 −1
0
2
Solution:
1 −1
0
2 −1 1
1 · 2 − 0 · (−1)
1 21
2 01 =
= 1
0 = 21
01 1
2
1
2 Check:
1 −1
0
2 1
0 1
2
1
2 = 10
01 as required. g)
1 −1
−1
1 −1 Solution: The determinant of this matrix is
1 · 1 − (−1) · (−1) = 0
and therefore this matrix has no inverse. 7 4. Compute the inverses for the following matrices, or prove that no inverse exists.
a) 3
12 1 −1 3 1
24
Solution: 3
1
1 1
∼1
3 1
∼0
0 1
∼0
0 1
∼0
0 1
0
∼
0 1
0
∼
0 1
0
∼
0 12
−1 3
24 100
010
001 −1 3
24
12 010
001
100 , swapping rows −1
3
3
1
4 −7 0
10
0 −1 1
1 −3 0 −1
3
1
1
3
4 −7 0
1
0 −1
3
1 −3 −1
3
1
1
3
0 − 25
3
−1
1
0 3
1
3 1 −1 0
10
01
00
10
01 0 R 2 = R 2 − R 1 ; R 3 = R 3 − 3R 1 R2 = 1 R2
3
0 0
1
0
1 R 3 = R 3 − 4R 2
0 −1
3
3
5
4
1 −3 −3 0
10
1 R 3 = (− 3 )R 3
0 −1
3
3
25
3
5
4
− 25
25
25 9
10
12
25
25 − 25
1
10
7 R 1 = R 1 − 3R 3 ; R 2 = R 2 − 1 R 3
25 − 25
25
3
3
5
4
− 25
25
25 10
5
0 − 25
25
10
7
1 R1 = R1 + R2
25 − 25
25 3
− 25 5
25 1
3 4
25 Check: 3
12
10
0 −5
25 0 0
100
1
1
1 −1 3 1 −10
7 =
0 25 0 = 0 1 0 25
25
1
24
−3
5
4
0 0 25
001 8 b) −1 1 2 4 0 7
1 3 13
Solution: ∼ ∼ ∼ ∼ −1 1 2 40 7
1 3 13 1 −1 −2
4
0
7
1
3 13 1 −1 −2
0
4 15
0
4 15 1 −1 −2
0
1 15
4
0
4 15 1 −1 −2
0
1 15
4
0
0
0 100
010
001 −1 0 0
010
001 −1 0 0
410
101 −1
1
1 00
1
40
01 −1
00
1
1
40
−3 −1 1 R1 = (−1)R1 R 2 = R 2 − 4R 1 ; R 3 = R 3 − R 1 R2 = 1 R2
4 R 3 = R 3 − 4R 2 So we can now see that no inverse exists for the given matrix. 9 Bonus Question: Compute the values of a, b, c, given the following matrices. 1
0 −1
0
b2
A= 3
2
0 , A−1 = 0 −1 c −1 −1
a
−1
12
Solution: Since AA−1 1
0
0 = In for any invertible n × n matrix A, we have 00
1
0 −1
0
b2
1 0 = 3
2
0 0 −1 c 01
−1 −1
a
−1
12 1
b−1
0
= 0
3b − 2
2c + 6 −a a − b + 1 2a − c − 2 • From row 3, column 1, we obtain that 0 = −a ⇐⇒ a = 0.
• From row 1, column 2, we obtain that 0 = b − 1 ⇐⇒ b = 1.
• From row 2, column 3, we obtain that 0 = 2c + 6 ⇐⇒ c = −3.
These substitutions satisfy the remaining identities, so they are correct and we are done. 10 ...

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