Matrices Exercises—Solutions 1. Perform the indicated operations

Matrices Exercises—Solutions 1. Perform the indicated operations
Matrices Exercises—Solutions
1. Perform the indicated operations
⎡
⎤ ⎡
⎤
0 4
−2 4
(a) ⎣ 1 0 ⎦ + ⎣ 1 5 ⎦
7 2
4 6
Solution:
⎡
⎤ ⎡
0 4
−2
⎣ 1 0 ⎦+⎣ 1
7 2
4
∙
¸
∙
¸
2 −5 1
6 1 0
(b) 2
−3
1 −3 4
−7 2 9
Solution:
∙
¸
∙
2 −5 1
6
2
−3
1 −3 4
−7
⎤ ⎡
⎤ ⎡
⎤
4
0−2 4+4
−2 8
5 ⎦=⎣ 1+1 0+5 ⎦=⎣ 2 5 ⎦
6
7+4 2+6
11 8
1 0
2 9
¸
=
=
=
=
=
∙
¸ ∙
3 5
2
(c)
·
1 4
−1
Solution:
∙
¸ ∙
3 5
2
·
1 4
−1
⎡
∙
¸
1 0 2
(d)
·⎣
6 4 1
8
4
¸
8
4
¸
=
2
0
1
5
11 −5
∙
∙
∙
∙
∙
∙
4 −10 2
2 −6 8
¸
∙
18 3 0
−
−21 6 27
¸
4 − 18
−10 − 3 2 − 0
2 − (−21) −6 − 6 8 − 27
¸
−14
−13
2
2 − (−21) −6 − 6 8 − 27
¸
−14 −13
2
2 + 21 −12 −19
¸
−14 −13
2
23 −12 −19
3 · 2 + 5 · (−1) 3 · 8 + 5 · 4
1 · 2 + 4 · (−1) 1 · 8 + 4 · 4
⎤
4 1
7 0 ⎦
2 8
¸
=
∙
6 − 5 24 + 20
2 − 4 8 + 16
¸
=
∙
¸
1 44
−2 24
Solution:
⎡
⎤
∙
¸
2
0 4 1
1 0 2
5 7 0 ⎦
·⎣ 1
6 4 1
11 −5 2 8
∙
¸
1 · 2 + 0 · 1 + 2 · 11 1 · 0 + 0 · 5 + 2 · (−5) 1 · 4 + 0 · 7 + 2 · 2 1 · 1 + 0 · 0 + 2 · 8
=
6 · 2 + 4 · 1 + 1 · 11 6 · 0 + 4 · 5 + 1 · (−5) 6 · 4 + 4 · 7 + 1 · 2 6 · 1 + 4 · 0 + 1 · 8
∙
¸
2 + 0 + 22 0 + 0 − 10 4 + 0 + 4 1 + 0 + 16
=
12 + 4 + 11 0 + 20 − 5 24 + 28 + 2 6 + 0 + 8
∙
¸
24 −10 8 17
=
27
15 54 14
⎡
⎤
∙ ¸
2 0
5
⎣
⎦
−4 3 ·
(e)
6
8 1
Solution:
⎡
⎤
⎡
⎤ ⎡
⎤ ⎡
⎤
∙ ¸
2 0
2·5+0·6
10 + 0
10
5
⎣ −4 3 ⎦ ·
= ⎣ −4 · 5 + 3 · 6 ⎦ = ⎣ −20 + 18 ⎦ = ⎣ −2 ⎦
6
8·5+1·6
40 + 6
46
8 1
1
¸
.
(f)
£
⎤
0 1
5 −6 9 · ⎣ −7 0 ⎦
2 5
⎡
⎤
0 1
£
¤
£
¤
5 −6 9 · ⎣ −7 0 ⎦ =
5 · 0 − 6 · (−7) + 9 · 2 5 · 1 − 6 · 0 + 9 · 5
2 5
£
¤
0 + 42 + 18 5 − 0 + 45
=
£
¤
60 50 .
=
¤
⎡
⎤
2
£
¤
(g) ⎣ 0 ⎦ · 1 4 −5
3
⎡
⎡
⎤
2
Solution: Note that ⎣ 0 ⎦ is a 3 × 1 matrix and
3
£
¤
1 4 −5 is a 3 × 3 matrix.
⎡ ⎤
⎡
2
2·1 2·4
£
¤
⎣ 0 ⎦ · 1 4 −5 = ⎣ 0 · 1 0 · 4
3
3·1 3·4
£
1 4 −5
¤
⎤
2
is a 1 × 3 matrix. So ⎣ 0 ⎦ ·
3
⎤ ⎡
⎤
2 · (−5)
2 8 −10
0 · (−5) ⎦ = ⎣ 0 0
0 ⎦.
3 · (−5)
3 12 −15
2. Use Gauss-Jordan elimination method to solve the following system of equations.
x + 2y − 3z = 9
2x − 3y + z = −3
−2x − y + 2z = −7
Solution: The augmented of this system of equations is:
⎤
⎡
9
1
2 −3
⎣ 2 −3
1 −3 ⎦
−2 −1
2 −7
Next we find the reduced row-echelon form of the augmented matrix. Now
⎤
⎡
9
1
2 −3
⎣ 2 −3
1 −3 ⎦
−2 −1
2 −7
⎡
⎤
1
2 −3
9
(−2)R1 + R2 → R2
⎣ 0 −7
7 −21 ⎦
2R1 + R3 → R3
0
3 −4
11
⎡
⎤
1 2 −3
9
3 ⎦
(− 17 )R2 → R2 ⎣ 0 1 −1
0 3 −4 11
⎡
⎤
1 0 −1 3
(−2)R2 + R1 → R1
⎣ 0 1 −1 3 ⎦
(−3)R2 + R3 → R3
0 0 −1 2
⎡
⎤
1 0 −1
3
3 ⎦
(−1)R3 → R3 ⎣ 0 1 −1
0 0
1 −2
⎡
⎤
1 0 0
1
R3 + R1 → R1
⎣ 0 1 0
1 ⎦
R3 + R2 → R2
0 0 1 −2
This implies that x = 1, y = 1, and z = −2.
2
⎡
3. Use Gauss-Jordan elimination method to solve the following system of equations.
2x + 4y − 2z = 2
−3x + 2y − z = 4
−x + 6y − 3z = 6
Solution: The augmented of this system of equations is:
⎡
⎤
2 4 −2 2
⎣ −3 2 −1 4 ⎦
−1 6 −3 6
Next we find the reduced row-echelon form of the augmented
⎡
2 4
⎣ −3 2
−1 6
⎡
1 2
( 12 )R1 → R1 ⎣ −3 2
−1 6
⎡
1 2
3R1 + R2 → R2
⎣ 0 8
R1 + R3 → R3
0 8
⎡
1 2
1
⎣ 0 1
R
→
R
2
2
8
0 8
⎡
1 0
(−2)R2 + R1 → R1
⎣ 0 1
(−8)R2 + R3 → R3
0 0
matrix. Now
⎤
−2 2
−1 4 ⎦
−3 6
⎤
−1 1
−1 4 ⎦
−3 6
⎤
−1 1
−4 7 ⎦
−4 7
⎤
−1 1
− 12 78 ⎦
−4 7
⎤
0 − 34
7 ⎦
− 12
8
0
0
The system of equations corresponding to this system of equations is:
x = − 34
y − 12 z =
7
8
We have two equations and three variables. The system of equations has infinite solutions. The first
equation implies that x = − 34 . From the second equation, y − 12 z = 78 . This implies that y = 78 + 12 z.
Hence solutions are of the form:
3 7 1
(− , + z, z),
4 8 2
where z is any real number.
4. Find the inverse of the following matrix:
⎡
⎤
−4 −5
3
3 −2 ⎦ .
A=⎣ 3
−1 −1
1
Solution:
⎡
−4 −5
⎣ 3
3
−1 −1
3
−2
1
1
0
0
0
1
0
⎤
0
0 ⎦
1
R1 ←→ R3
3
⎡
−1
⎣ 3
−4
−1
3
−5
1
−2
3
0
0
1
0
1
0
⎤
1
0 ⎦
0
0
0
1
0
1
0
⎤
−1
0 ⎦
0
(−1)R1 −→ R1
⎡
1
⎣ 3
−4
1
3
−5
−1
−2
3
(−3)R1 + R2 −→ R2
4R1 + R3 −→ R3
⎡
1
⎣ 0
0
1
0
−1
−1
1
−1
0
0
1
0
1
0
⎤
−1
3 ⎦
−4
−1
−1
1
0
1
0
0
0
1
⎤
−1
−4 ⎦
3
0
−1
0
0
0
1
⎤
−1
4 ⎦
3
0
0
1
⎤
−5
4 ⎦
3
R2 ←→ R3
⎡
1
⎣ 0
0
1
−1
0
(−1)R2 −→ R2
⎡
1
⎣ 0
0
1
1
0
−1
1
1
(−1)R2 + R1 −→ R1
⎡
1
⎣ 0
0
0
1
0
−2
1
1
1
−1
0
2R3 + R1 −→ R1
(−1)R3 + R2 −→ R2
⎡
1
⎣ 0
0
0
1
0
0
0
1
1
−1
0
2
−1
1
⎤
1
1 ⎦
3
A−1
⎡
1
= ⎣ −1
0
4
2
−1
1
⎤
1
1 ⎦
3
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