Download MAT1503 Linear Algebra Assignment 2 and more Exams Linear Algebra in PDF only on Docsity!
MAT150 3 Linear Algebra
Assignment 2
Question 1
a b c
d e f = 8
g h i
(g+2a) (h+2b) (i+2c)
3a 3b 3c.
2d 2e 2f
See page 189 of Anton’s elementary linear algebra
= 3
= 3 x 2
(g+2a) (h+2b) (i+2c)
a b c
2d 2e 2f
(g+2a) (h+2b) (i+2c)
a b c
d e f
Factor out 3 from the 2
nd row. Theorem 2.2.3a
Factor out 2 from the 3
rd row. Theorem 2.2.3a
= (-) 3x
a b c
(g+2a) (h+2b) (1+2c)
d e f
Interchange the 1
st and 2
nd rows. Theorem 2.2.3b
= (-)(-) 3x
a b c
d e f
(g+2a) (h+2b) (1+2c)
Interchange the 3
rd and the 4
th row. Theorem 2.2.3b
a b c
= (-)(-)3x2 d e f
g h i
= (-) (-) 3 x 2 x 8
st row wad added to the 3
rd row. Theorem 2.2.3c
= 5
3 × 125
= 15 625
Question 4
Det [A
4 ] = det [A] x det [A] x det [A] x det [A]
= 5 x 5 x 5 x 5
= 625
See page 202 of Anton’s elementary linear algebra.
Question 5
The system can be solved using Cramer’s rule because the determinant of the coefficient matrix
is not zero.
3 - 4 - 1 See page 208 of Anton’s elementary linear algebra
A= 3 2 - 1
A
1
A
2
3
A 3 =
Using co-factor expansion to calculate the determinants (See page 177 of Anton’s elementary
linear algebra)
det (A) =3 (2-(-1)) - (-4)(3-(-1)) + (-1)(3-2)
det (A 1
det (A 2
det (A 3
Therefore:
x 1 =
det 𝐴 1
=
=
det 𝐴 24 8
x 2 =
det 𝐴 2
=
=
det 𝐴 24 6
x =
det 𝐴 3
=
det 𝐴 24
Question 6
Cos 90
⁰ = 0
⇒ u ۰ v = 0 when u and v are orthogonal
x = , y = , z =
2
2
× (u 1 , u 2
1 2
Part (b)
u = (1, 2, 3)
To find the norm of u:
‖u‖ = √ 1
2
2
2
See page 253 of Anton’s linear elementary algebra
To find unit vectors, multiply u with the reciprocal of the norm (See page 255 of Anton’s
linear elementary algebra)
u= (1, 2, 3)
𝟏
√𝟏𝟒
𝟐
√𝟏𝟒
𝟑
√𝟏𝟒
Question 8
𝑈 × 𝑉 = |𝑢 1 𝑢 2 𝑢 3 | = 𝑖(𝑢
2
3
3
2
1
3
3
1
1
2
2
1
1
2
3
𝑈 × 𝑉 = (𝑢
2
3
3
2
1
3
3
1
1
2
2
1
(𝑈 × 𝑉) × 𝑊 = |
2
3
3
2
1
3
3
1
1
2
2
1
Question 9
(x, y, z) = (2, 3, - 1) + 𝜆(1, 0,1)
x ⇒ 2+ λ (1) = 4
y ⇒ 3 + λ (0) = 3
z ⇒ - 1 + λ (1) = a
Solve for λ ⇒ 2+ λ (1) = 4
2 + λ= 4
λ= 2
Solve for a ⇒ - 1 + λ (1) = a
1=a
a=
Question 10
According to theorem 3.3.4, page 286 of Anton’s elementary linear algebra
Where x o , y o , z o is a point from the plane. Taking the point as the origin, (0, 0, 0), the distance
between the origin and the plane is
| 𝑎 ×0+𝑏×0+𝑐×0+𝑑| 𝑑
D=
2 +𝑏
2 +𝑐
2
=
2
2
2
In depth proof:
From the given plane ax + by +cz +d =
The normal vector 𝑛⃑` = (a, b, c)
A point on the plane p when x=0, y=0 is P (0, 0,
The origin is P 0
A vector between the origin P o and a point on the plane P is 𝑃
⃑`
o
Representing the above diagrammatically
