Linear Algebra and Analytic Geometry

Replenishment date: 18.07.2023
Content: Linear Algebra and Analytic Geometry.pdf (104.52 KB)
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Description
Lesson 1
Lesson 2. Complex numbers. Part 2
Lesson 3. Complex numbers. Practice
Lesson 4
Lesson 5 Practice
Lesson 6 Completion
Lesson 7. Analytic properties of vectors lecture Part 1
Lesson 8. Analytic properties of vectors lecture Part 2
Lesson 9
Lesson 10
Lesson 11
Lesson 12
Lesson 13
Lesson 14
Lesson 15
Lesson 16
Lesson 17
Lesson 18
Lesson 19
Lesson 20
Lesson 21
Lesson 22
Lesson 23
Lesson 24
Lesson 25
Lesson 26
Lesson 27
Lesson 28
Lesson 29
Lesson 30
Lesson 31
Lesson 32
Lesson 33
final examination
Additional Information
Vectors are called collinear if:
Answer type: Single choice • with the choice of one correct answer from several proposed options

they lie either on the same line or on parallel lines and are co-directed
they lie on the same line
they lie on parallel lines
they lie on the same line and on parallel lines
Vectors are called coplanar if
Answer type: Single choice • with the choice of one correct answer from several proposed options

they are parallel to the same plane or lie in the same plane and co-directed.
they are perpendicular to each other.
they are pairwise parallel.
they are parallel to the same plane or lie in the same plane.
Choose the correct matrix determinants:
Answer type: Multiple choice • with the choice of several correct answers from the proposed options

The determinant of the transposed matrix is ​​equal to the determinant of the original matrix
The determinant exists for any matrix
Multiplying all elements of a row or column of a determinant by some number λ is equivalent to multiplying the determinant by this number
If the matrix contains a zero row (column), then the determinant of this matrix is ​​zero:
Choose the correct properties of the vector product:
Answer type: Multiple choice • with the choice of several correct answers from the proposed options

a × b = b × a
The module of the cross product of two vectors a and b is equal to the area of ​​the parallelogram built on these vectors.
If the vectors are collinear, then the cross product is zero.
Calculate the area of ​​a triangle with vertices at points A(4;5), B(6;2) and C(2;3):
Response type: Text response

The double cross product is calculated by the formula:
Answer type: Single choice • with the choice of one correct answer from several proposed options

[a, [b, c]] = b[a ⋅ c] + c[a ⋅ b]
[a, [b, c]] = b(a ⋅ c) − c(a ⋅ b)
[a, [b, c]] = b(a ⋅ c) + c(a ⋅ b)
[a, [b, c]] = b[a ⋅ c] − c[a ⋅ b]
The division of complex numbers a + bi and c + di in algebraic form is performed according to the formula:
Answer type: Single choice • with the choice of one correct answer from several proposed options

(ac − bd + i(ad − bc)) / (c² + d²)
(ac + bd + i(ad − bc)) / (c² − d²)
(ac + bd + i(ad − bc)) / (c² + d²)
(ac + bd + i(ad + bc)) / (c² + d²)
The division of complex numbers r₁(cos cos α₁ + i sin sin α₁) and r₂(cos cos α₂ + i sin sin α₂) in trigonometric form is performed by the formula:
Answer type: Single choice • with the choice of one correct answer from several proposed options

r₁/r₂ × (cos cos (φ₁ − φ₂) − i sin sin (φ₁ − φ₂))
r₁/r₂ × (sin sin (φ₁ − φ₂) + i cos cos (φ₁ − φ₂))
r₁/r₂ × (cos cos (φ₁ + φ₂) + i sin sin (φ₁ + φ₂))
r₁/r₂ × (cos cos (φ₁ − φ₂) + i sin sin (φ₁ − φ₂))
What pair of numbers can have the sum 10 + 4i:
Answer type: Multiple choice • with the choice of several correct answers from the proposed options

6 + 2i and 4 + 2i
8 + i and 2 + 3i
12 − i and −2 + 3i
−8 + 6i and −2 − 10i
What is the complex number obtained by multiplying 3 ⋅ (cos cos 5° + i sin sin 5°) and 4(cos cos 25° + i sin sin 25°):
Answer type: Multiple choice • with the choice of several correct answers from the proposed options

12 × (cos cos 30° + i sin sin 30°)
7 × (cos cos 30° + i sin sin 30°)
6√3+6i
3√3+3i
The canonical equation of a two-sheeted hyperboloid has the form:
Answer type: Single choice • with the choice of one correct answer from several proposed options

x²/a² + y²/b² - z²/c² = -1
x²/a² + y²/b² − z²/c² = 1
x²/a² + y²/b² − 2z² = 0
x²/a² + y²/b² + z² = 1
x²/a² − y²/b² = 2z
The canonical equation of the cone is:
Answer type: Single choice • with the choice of one correct answer from several proposed options

x²/a² + y²/b² - z²/c² = -1
x²/a² + y²/b² = 2z
x²/a² + y²/b² − 2z² = 0
x²/a² + y²/b² − z²/c² = 1
x²/a² − y²/b² = 2z
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