W3. Matrices, Determinants, Vector Cross Product, and Triple Products

Author

Salman Ahmadi-Asl

Published

September 18, 2025

Quiz | Flashcards

1. Summary

1.1 Introduction to Matrices

A matrix is a rectangular grid of numbers or expressions arranged in rows and columns. It serves as a powerful tool for organizing data and representing systems of linear equations or transformations. A matrix is denoted by a capital letter (e.g., \(A\)), and its individual entries are called elements.

The dimensions of a matrix are expressed as m x n, where m is the number of rows and n is the number of columns. An element at the intersection of the i-th row and j-th column is denoted by \(a_{ij}\).

For example, a 2x3 matrix A is written as: \[ A = \begin{pmatrix} a_{11} & a_{12} & a_{13} \\ a_{21} & a_{22} & a_{23} \end{pmatrix} \]

1.2 Types of Matrices

Matrices with specific structures and properties are given special names:

  • Square Matrix: A matrix where the number of rows equals the number of columns (m = n). These are the only matrices that have determinants and inverses.
  • Identity Matrix (\(I\)): A square matrix with ones on the main diagonal (top-left to bottom-right) and zeros everywhere else. It is the matrix equivalent of the number 1; multiplying a matrix by the identity matrix leaves it unchanged (\(AI = IA = A\)).
  • Zero Matrix: A matrix where every element is zero.
  • Diagonal Matrix: A square matrix where all elements not on the main diagonal are zero.
  • Symmetric Matrix: A square matrix that is unchanged by transposition (\(A = A^T\)). This means that for any i and j, the element \(a_{ij}\) is equal to \(a_{ji}\).
  • Triangular Matrix: A square matrix where all elements are zero either above or below the main diagonal. If the elements below the diagonal are zero, it is an upper triangular matrix. If the elements above are zero, it is a lower triangular matrix.
  • Hermitian Matrix: An extension of symmetric matrices for complex numbers. A matrix is Hermitian if it is equal to its own conjugate transpose (taking the transpose and then the complex conjugate of each element).
1.3 Basic Matrix Operations
  • Addition and Subtraction: These operations are only defined for matrices of the same dimensions. The operation is performed element-wise, meaning the corresponding elements of the matrices are added or subtracted. For \(C = A \pm B\), the elements are \(c_{ij} = a_{ij} \pm b_{ij}\).
  • Scalar Multiplication: To multiply a matrix by a scalar (a single number), you multiply every element in the matrix by that scalar. For \(B = \lambda A\), the elements are \(b_{ij} = \lambda a_{ij}\).
  • Transpose (\(A^T\)): The transpose of a matrix is found by interchanging its rows and columns. The first row becomes the first column, the second row becomes the second column, and so on. If A is an m x n matrix, \(A^T\) is an n x m matrix. A key property is that the transpose of a product reverses the order: \((AB)^T = B^T A^T\).
  • Trace (\(tr(A)\)): The trace is defined only for a square matrix and is the sum of the elements on its main diagonal. A crucial property is its cyclic invariance under multiplication: \(tr(ABC) = tr(BCA) = tr(CAB)\).
1.4 Matrix Multiplication

Matrix multiplication is a more complex operation. For the product \(AB\) to be defined, the number of columns in matrix A must equal the number of rows in matrix B. If A is an m x n matrix and B is an n x p matrix, the resulting product C will be an m x p matrix.

The element \(c_{ij}\) in the resulting matrix is the dot product of the i-th row of A and the j-th column of B. There are two key ways to conceptualize matrix-vector multiplication (\(Ax\)):

  1. Dot Product View: Each element of the resulting vector is the dot product of a corresponding row of matrix A with the vector x.
  2. Linear Combination View: The resulting vector is a weighted sum (a linear combination) of the columns of matrix A, where the weights are the elements of vector x.

A fundamental property of matrix multiplication is that it is not commutative: in general, \(AB \neq BA\). The order matters greatly. Think of it like real-world actions: putting on socks then shoes is very different from putting on shoes then socks. However, matrix multiplication is associative (\((AB)C = A(BC)\)) and distributive (\(A(B+C) = AB + AC\)).

1.5 The Determinant of a Matrix

The determinant, denoted \(det(A)\) or \(|A|\), is a unique scalar value that can only be calculated from a square matrix. It provides crucial information:

  • Invertibility: A matrix has an inverse if and only if its determinant is non-zero. A matrix with a zero determinant is called a singular matrix.
  • Geometric Meaning: The determinant represents the scaling factor of the linear transformation described by the matrix. For a 2x2 matrix, its absolute value tells you how much area is scaled. For a 3x3 matrix, it’s the scaling factor for volume. A negative determinant indicates that the transformation also involves a reflection (an “inversion” of orientation).

Key properties of determinants include:

  • \(det(AB) = det(A)det(B)\)
  • \(det(A^T) = det(A)\)
  • \(det(A^{-1}) = 1 / det(A)\)
  • For an n x n matrix, \(det(kA) = k^n det(A)\)
1.6 The Matrix Inverse

The inverse of a square matrix A, written as \(A^{-1}\), is the matrix that “undoes” the operation of A. When multiplied together, they yield the identity matrix: \(AA^{-1} = A^{-1}A = I\). This is analogous to how a number multiplied by its reciprocal equals 1 (e.g., \(5 \times \frac{1}{5} = 1\)). An inverse exists only if the matrix is non-singular (\(det(A) ≠ 0\)).

The adjugate method is a common way to find the inverse of a 3x3 matrix:

  1. Matrix of Minors (\(M\)): For each element \(a_{ij}\), its minor \(M_{ij}\) is the determinant of the 2x2 sub-matrix that remains after deleting the i-th row and j-th column.
  2. Matrix of Cofactors (\(C\)): The cofactor \(C_{ij}\) is the minor with a sign applied, determined by a “checkerboard” pattern. The formula is \(C_{ij} = (-1)^{i+j}M_{ij}\).
  3. Adjugate Matrix (\(adj(A)\)): This is the transpose of the cofactor matrix, \(adj(A)\) = \(C^T\).
  4. Calculate Inverse: The inverse is found by multiplying the adjugate matrix by the reciprocal of the determinant: \(A^{-1} = \frac{1}{\det(A)}\text{adj}(A)\).
1.7 The Vector Cross Product

The cross product is an operation between two vectors in 3D space, written as a x b. The result is a new vector that is geometrically perpendicular (orthogonal) to both of the original vectors.

The direction of the resulting vector is given by the right-hand rule: point the fingers of your right hand in the direction of vector a, then curl them toward vector b. Your thumb will point in the direction of a x b. Key properties of the cross product:

  • Anti-commutative: Reversing the order flips the direction of the resulting vector: a x b = - (b x a).
  • Parallel Vectors: The cross product of two parallel vectors is the zero vector (0).
  • Magnitude: The magnitude, \(|| \mathbf{a} \times \mathbf{b} ||\), equals the area of the parallelogram formed by vectors a and b. The area of the triangle with these sides is half this value.
1.8 Triple Products

Triple products combine dot and cross products for three vectors.

  • Scalar Triple Product: This product, \(\mathbf{a} \cdot (\mathbf{b} \times \mathbf{c})\), results in a scalar. Its absolute value represents the volume of the parallelepiped with sides a, b, and c. If the result is 0, the vectors are coplanar (they all lie in the same plane), meaning the parallelepiped is flat and has no volume. The operation is cyclically symmetric: \(\mathbf{a} \cdot (\mathbf{b} \times \mathbf{c}) = \mathbf{b} \cdot (\mathbf{c} \times \mathbf{a}) = \mathbf{c} \cdot (\mathbf{a} \times \mathbf{b})\).
  • Vector Triple Product: This product, \(\mathbf{a} \times (\mathbf{b} \times \mathbf{c})\), results in a vector. The resulting vector lies in the same plane as b and c. It can be simplified with the “BAC-CAB” identity: \(\mathbf{a} \times (\mathbf{b} \times \mathbf{c}) = (\mathbf{a} \cdot \mathbf{c})\mathbf{b} - (\mathbf{a} \cdot \mathbf{b})\mathbf{c}\).

Another key identity is the Jacobi identity, which shows a cyclic relationship between triple products: \[ \mathbf{a} \times (\mathbf{b} \times \mathbf{c}) + \mathbf{b} \times (\mathbf{c} \times \mathbf{a}) + \mathbf{c} \times (\mathbf{a} \times \mathbf{b}) = \mathbf{0} \]

2. Definitions

  • Matrix: A rectangular array of numbers arranged in rows and columns.
  • Scalar: A single numerical quantity that scales a vector or matrix.
  • Square Matrix: A matrix with an equal number of rows and columns.
  • Identity Matrix (\(I\)): A square matrix with ones on the main diagonal and zeros elsewhere, which acts as the multiplicative identity.
  • Determinant (\(det(A)\)): A scalar value computed from a square matrix that determines its invertibility and geometric scaling properties.
  • Singular Matrix: A square matrix whose determinant is zero, meaning it has no inverse.
  • Matrix Inverse (\(A^{-1}\)): For a non-singular square matrix A, it is the unique matrix that yields the identity matrix when multiplied with A.
  • Transpose (\(A^T\)): A matrix formed by interchanging the rows and columns of the original matrix.
  • Trace (\(tr(A)\)): The sum of the elements on the main diagonal of a square matrix.
  • Adjugate Matrix (\(adj(A)\)): The transpose of the matrix of cofactors.
  • Coplanar Vectors: A set of vectors that all lie within the same two-dimensional plane.
  • Vector Cross Product (\(\mathbf{a} \times \mathbf{b}\)): A binary operation on two vectors in 3D space, resulting in a vector perpendicular to both.
  • Scalar Triple Product (\(\mathbf{a} \cdot (\mathbf{b} \times \mathbf{c})\)): An operation whose absolute value is the volume of the parallelepiped defined by three vectors.
  • Vector Triple Product (\(\mathbf{a} \times (\mathbf{b} \times \mathbf{c})\)): An operation involving three vectors that results in a new vector coplanar with the second and third vectors.

3. Formulas

  • Matrix Addition: \(C = A + B \implies c_{ij} = a_{ij} + b_{ij}\)
  • Scalar Multiplication: \(B = \lambda A \implies b_{ij} = \lambda a_{ij}\)
  • Matrix Multiplication Transpose: \((AB)^T = B^T A^T\)
  • Determinant of a Product: \(det(AB) = det(A)det(B)\)
  • Determinant of an Inverse: \(det(A^{-1}) = 1 / det(A)\)
  • Determinant of a Scalar Multiple: For an \(n \times n\) matrix \(A\), \(det(kA) = k^n det(A)\)
  • Trace of a Matrix: \(tr(A) = \sum_{i=1}^n a_{ii}\)
  • 2x2 Determinant: For \(A = \begin{pmatrix} a & b \\ c & d \end{pmatrix}\), \(\det(A) = ad - bc\)
  • 2x2 Matrix Inverse: \[ A^{-1} = \frac{1}{ad-bc} \begin{pmatrix} d & -b \\ -c & a \end{pmatrix} \]
  • General Matrix Inverse: \[ A^{-1} = \frac{1}{\det(A)} \text{adj}(A) \]
  • Cofactor: \(C_{ij} = (-1)^{i+j}M_{ij}\)
  • Adjugate Matrix: \(adj(A) = C^T\)
  • Vector Between Points: \(\vec{AB} = B - A\)
  • Cross Product (Determinant Form): \[ \mathbf{a} \times \mathbf{b} = \det \begin{pmatrix} \mathbf{i} & \mathbf{j} & \mathbf{k} \\ a_1 & a_2 & a_3 \\ b_1 & b_2 & b_3 \end{pmatrix} \]
  • Magnitude of Cross Product: \(|| \mathbf{a} \times \mathbf{b} || = ||\mathbf{a}|| \cdot ||\mathbf{b}|| \cdot \sin(\theta)\)
  • Dot Product: \(\mathbf{a} \cdot \mathbf{b} = ||\mathbf{a}|| \cdot ||\mathbf{b}|| \cdot \cos(\theta)\)
  • Pythagorean Identity: \(\sin^2(\theta) + \cos^2(\theta) = 1\)
  • Cross Product Property (Self): \(\mathbf{a} \times \mathbf{a} = \mathbf{0}\)
  • Cross Product Property (Anti-Commutativity): \(\mathbf{a} \times \mathbf{b} = -(\mathbf{b} \times \mathbf{a})\)
  • Area of a Triangle: \(Area = \frac{1}{2} || \mathbf{a} \times \mathbf{b} ||\)
  • Scalar Triple Product (Determinant Form): \[ \mathbf{a} \cdot (\mathbf{b} \times \mathbf{c}) = \det \begin{pmatrix} a_1 & a_2 & a_3 \\ b_1 & b_2 & b_3 \\ c_1 & c_2 & c_3 \end{pmatrix} \]
  • Volume of a Parallelepiped: \(Volume = |\mathbf{a} \cdot (\mathbf{b} \times \mathbf{c})|\)
  • Vector Triple Product (BAC-CAB Rule): \(\mathbf{a} \times (\mathbf{b} \times \mathbf{c}) = (\mathbf{a} \cdot \mathbf{c})\mathbf{b} - (\mathbf{a} \cdot \mathbf{b})\mathbf{c}\)
  • Jacobi Identity: \(\mathbf{a} \times (\mathbf{b} \times \mathbf{c}) + \mathbf{b} \times (\mathbf{c} \times \mathbf{a}) + \mathbf{c} \times (\mathbf{a} \times \mathbf{b}) = \mathbf{0}\)

4. Examples

4.1. Matrix Expression Definitions (Lab 3, Task 1a)

Given the matrix sizes P(4x5), Q(4x5), R(5x2), S(4x2), and T(5x4), determine if the expression \(PR + S\) is defined, and if so, what is the size of the resulting matrix.

Click to see the solution
  1. Analyze the product \(PR\):
    • For matrix multiplication, the number of columns in the first matrix must equal the number of rows in the second.
    • P has size (4x5). R has size (5x2). The inner dimensions match (5=5).
    • The product \(PR\) is defined and its size is (4x2).
  2. Analyze the sum \(PR + S\):
    • For matrix addition, both matrices must have the same size.
    • \(PR\) has size (4x2). S has size (4x2). The dimensions match.
    • The sum is defined.
  3. Determine the final size:
    • The resulting matrix will have the same size as the operands, which is (4x2).
Answer: The expression is defined, and the resulting matrix has a size of (4x2).
4.2. Matrix Expression Definitions (Lab 3, Task 1b)

Given the matrix sizes P(4x5), Q(4x5), R(5x2), S(4x2), and T(5x4), determine if the expression \(P - 3T^T\) is defined, and if so, what is the size of the resulting matrix.

Click to see the solution
  1. Analyze the transpose \(T^T\):
    • The transpose operation swaps the number of rows and columns.
    • T has size (5x4).
    • \(T^T\) has size (4x5).
  2. Analyze the subtraction \(P - 3T^T\):
    • Scalar multiplication (\(3T^T\)) does not change the size of the matrix. So, \(3T^T\) has size (4x5).
    • For subtraction, both matrices must have the same size.
    • P has size (4x5). \(3T^T\) has size (4x5). The dimensions match.
    • The expression is defined.
  3. Determine the final size:
    • The resulting matrix will have the same size as the operands, which is (4x5).
Answer: The expression is defined, and the resulting matrix has a size of (4x5).
4.3. Matrix Expression Definitions (Lab 3, Task 1c)

Given the matrix sizes P(4x5), Q(4x5), R(5x2), S(4x2), and T(5x4), determine if the expression \(S^T(QT)\) is defined, and if so, what is the size of the resulting matrix.

Click to see the solution
  1. Analyze the product \(QT\):
    • Q has size (4x5). T has size (5x4). The inner dimensions match (5=5).
    • The product \(QT\) is defined and its size is (4x4).
  2. Analyze the transpose \(S^T\):
    • S has size (4x2).
    • \(S^T\) has size (2x4).
  3. Analyze the product \(S^T(QT)\):
    • \(S^T\) has size (2x4). \(QT\) has size (4x4). The inner dimensions match (4=4).
    • The product is defined.
  4. Determine the final size:
    • The size of the final matrix is (2x4).
Answer: The expression is defined, and the resulting matrix has a size of (2x4).
4.4. Matrix Expression Definitions (Lab 3, Task 1d)

Given the matrix sizes P(4x5), Q(4x5), R(5x2), S(4x2), and T(5x4), determine if the expression \(QP^T + S\) is defined, and if so, what is the size of the resulting matrix.

Click to see the solution
  1. Analyze the transpose \(P^T\):
    • P has size (4x5).
    • \(P^T\) has size (5x4).
  2. Analyze the product \(QP^T\):
    • Q has size (4x5). \(P^T\) has size (5x4). The inner dimensions match (5=5).
    • The product \(QP^T\) is defined and its size is (4x4).
  3. Analyze the sum \(QP^T + S\):
    • For addition, the matrices must have the same size.
    • \(QP^T\) has size (4x4). S has size (4x2). The dimensions do not match.
    • The sum is not defined.
Answer: The expression is not defined.
4.5. Matrix Expression Definitions (Lab 3, Task 1e)

Given the matrix sizes P(4x5), Q(4x5), R(5x2), S(4x2), and T(5x4), determine if the expression \(T(9Q + P)\) is defined, and if so, what is the size of the resulting matrix.

Click to see the solution
  1. Analyze the sum \(9Q + P\):
    • Scalar multiplication (\(9Q\)) does not change the size. \(9Q\) is (4x5).
    • For addition, matrices must have the same size. Q is (4x5) and P is (4x5). The dimensions match.
    • The sum is defined, and the resulting matrix, let’s call it \(M\), has size (4x5).
  2. Analyze the product \(TM\):
    • T has size (5x4). \(M\) (from the sum) has size (4x5). The inner dimensions match (4=4).
    • The product is defined.
  3. Determine the final size:
    • The size of the final matrix is (5x5).
Answer: The expression is defined, and the resulting matrix has a size of (5x5).
4.6. Matrix Expression Computations (Lab 3, Task 2a)

Given matrices \(D = \begin{bmatrix} 1 & 5 & 2 \\ -1 & 0 & 1 \\ 3 & 2 & 4 \end{bmatrix}\) and \(E = \begin{bmatrix} 6 & 1 & 3 \\ -1 & 1 & 2 \\ 4 & 1 & 3 \end{bmatrix}\), compute \(tr(D - 3E)\).

Click to see the solution
  1. Compute \(3E\):
    • \(3E = 3 \begin{bmatrix} 6 & 1 & 3 \\ -1 & 1 & 2 \\ 4 & 1 & 3 \end{bmatrix} = \begin{bmatrix} 18 & 3 & 9 \\ -3 & 3 & 6 \\ 12 & 3 & 9 \end{bmatrix}\)
  2. Compute \(D - 3E\):
    • \(D - 3E = \begin{bmatrix} 1 & 5 & 2 \\ -1 & 0 & 1 \\ 3 & 2 & 4 \end{bmatrix} - \begin{bmatrix} 18 & 3 & 9 \\ -3 & 3 & 6 \\ 12 & 3 & 9 \end{bmatrix} = \begin{bmatrix} 1-18 & 5-3 & 2-9 \\ -1-(-3) & 0-3 & 1-6 \\ 3-12 & 2-3 & 4-9 \end{bmatrix} = \begin{bmatrix} -17 & 2 & -7 \\ 2 & -3 & -5 \\ -9 & -1 & -5 \end{bmatrix}\)
  3. Compute the trace: The trace (\(tr\)) is the sum of the diagonal elements.
    • \(tr(D - 3E) = -17 + (-3) + (-5) = -25\)
Answer: \(tr(D - 3E) = -25\).
4.7. Matrix Expression Computations (Lab 3, Task 2b)

Given matrices \(A = \begin{bmatrix} 3 & 0 \\ -1 & 2 \\ 1 & 1 \end{bmatrix}\) and \(B = \begin{bmatrix} 4 & -1 \\ 0 & 2 \end{bmatrix}\), compute \(AB\).

Click to see the solution
  1. Check dimensions: A is (3x2) and B is (2x2). The inner dimensions match, so the product is defined. The result will be a (3x2) matrix.
  2. Perform multiplication:
    • \(AB = \begin{bmatrix} (3)(4)+(0)(0) & (3)(-1)+(0)(2) \\ (-1)(4)+(2)(0) & (-1)(-1)+(2)(2) \\ (1)(4)+(1)(0) & (1)(-1)+(1)(2) \end{bmatrix}\)
  3. Calculate the new matrix:
    • \(= \begin{bmatrix} 12 & -3 \\ -4 & 1+4 \\ 4 & -1+2 \end{bmatrix} = \begin{bmatrix} 12 & -3 \\ -4 & 5 \\ 4 & 1 \end{bmatrix}\)
Answer: \(AB = \begin{bmatrix} 12 & -3 \\ -4 & 5 \\ 4 & 1 \end{bmatrix}\).
4.8. Matrix Expression Computations (Lab 3, Task 2c)

Given matrices \(A = \begin{bmatrix} 3 & 0 \\ -1 & 2 \\ 1 & 1 \end{bmatrix}\), \(B = \begin{bmatrix} 4 & -1 \\ 0 & 2 \end{bmatrix}\), and \(C = \begin{bmatrix} 1 & 4 & 2 \\ 3 & 1 & 5 \end{bmatrix}\), compute \((AB)C\).

Click to see the solution
  1. Compute \(AB\): From the previous task, we know \(AB = \begin{bmatrix} 12 & -3 \\ -4 & 5 \\ 4 & 1 \end{bmatrix}\).
  2. Check dimensions for \((AB)C\): \(AB\) is (3x2) and \(C\) is (2x3). The inner dimensions match. The result will be a (3x3) matrix.
  3. Perform multiplication:
    • \((AB)C = \begin{bmatrix} (12)(1)+(-3)(3) & (12)(4)+(-3)(1) & (12)(2)+(-3)(5) \\ (-4)(1)+(5)(3) & (-4)(4)+(5)(1) & (-4)(2)+(5)(5) \\ (4)(1)+(1)(3) & (4)(4)+(1)(1) & (4)(2)+(1)(5) \end{bmatrix}\)
  4. Calculate the new matrix:
    • \(= \begin{bmatrix} 12-9 & 48-3 & 24-15 \\ -4+15 & -16+5 & -8+25 \\ 4+3 & 16+1 & 8+5 \end{bmatrix} = \begin{bmatrix} 3 & 45 & 9 \\ 11 & -11 & 17 \\ 7 & 17 & 13 \end{bmatrix}\)
Answer: \((AB)C = \begin{bmatrix} 3 & 45 & 9 \\ 11 & -11 & 17 \\ 7 & 17 & 13 \end{bmatrix}\).
4.9. Matrix Expression Computations (Lab 3, Task 2d)

Given matrix \(C = \begin{bmatrix} 1 & 4 & 2 \\ 3 & 1 & 5 \end{bmatrix}\), compute \(CC^T\).

Click to see the solution
  1. Find the transpose \(C^T\):
    • \(C^T = \begin{bmatrix} 1 & 3 \\ 4 & 1 \\ 2 & 5 \end{bmatrix}\)
  2. Check dimensions for \(CC^T\): \(C\) is (2x3) and \(C^T\) is (3x2). The inner dimensions match. The result will be a (2x2) matrix.
  3. Perform multiplication:
    • \(CC^T = \begin{bmatrix} (1)(1)+(4)(4)+(2)(2) & (1)(3)+(4)(1)+(2)(5) \\ (3)(1)+(1)(4)+(5)(2) & (3)(3)+(1)(1)+(5)(5) \end{bmatrix}\)
  4. Calculate the new matrix:
    • \(= \begin{bmatrix} 1+16+4 & 3+4+10 \\ 3+4+10 & 9+1+25 \end{bmatrix} = \begin{bmatrix} 21 & 17 \\ 17 & 35 \end{bmatrix}\)
Answer: \(CC^T = \begin{bmatrix} 21 & 17 \\ 17 & 35 \end{bmatrix}\).
4.10. Matrix Expression Computations (Lab 3, Task 2e)

Given matrix \(D = \begin{bmatrix} 1 & 5 & 2 \\ -1 & 0 & 1 \\ 3 & 2 & 4 \end{bmatrix}\), compute \(\det(2D)\).

Click to see the solution
  1. Use the determinant property: For an \(n \times n\) matrix \(D\) and a scalar \(k\), \(\det(kD) = k^n \det(D)\).
  2. Apply the property: Here, \(n=3\) and \(k=2\). So, \(\det(2D) = 2^3 \det(D) = 8 \det(D)\).
  3. Calculate \(\det(D)\):
    • \(\det(D) = 1(0\cdot4 - 1\cdot2) - 5(-1\cdot4 - 1\cdot3) + 2(-1\cdot2 - 0\cdot3)\)
    • \(= 1(-2) - 5(-4-3) + 2(-2)\)
    • \(= -2 - 5(-7) - 4 = -2 + 35 - 4 = 29\)
  4. Calculate \(\det(2D)\):
    • \(\det(2D) = 8 \cdot \det(D) = 8 \cdot 29 = 232\).
Answer: \(\det(2D) = 232\).
4.11. Matrix Expression Computations (Lab 3, Task 2f)

Given matrices \(D = \begin{bmatrix} 1 & 5 & 2 \\ -1 & 0 & 1 \\ 3 & 2 & 4 \end{bmatrix}\) and \(E = \begin{bmatrix} 6 & 1 & 3 \\ -1 & 1 & 2 \\ 4 & 1 & 3 \end{bmatrix}\), compute \(\det(DE)\).

Click to see the solution
  1. Use the determinant property: The determinant of a product of matrices is the product of their determinants: \(\det(DE) = \det(D) \det(E)\).
  2. Calculate \(\det(D)\): From the previous task, we know \(\det(D) = 29\).
  3. Calculate \(\det(E)\):
    • \(\det(E) = 6(1\cdot3 - 2\cdot1) - 1(-1\cdot3 - 2\cdot4) + 3(-1\cdot1 - 1\cdot4)\)
    • \(= 6(3-2) - 1(-3-8) + 3(-1-4)\)
    • \(= 6(1) - 1(-11) + 3(-5) = 6 + 11 - 15 = 2\)
  4. Calculate \(\det(DE)\):
    • \(\det(DE) = \det(D) \det(E) = 29 \cdot 2 = 58\).
Answer: \(\det(DE) = 58\).
4.12. Find Matrix A (Lab 3, Task 3a)

Find matrix A given \((7A)^{-1} = \begin{bmatrix} -3 & 7 \\ 1 & -2 \end{bmatrix}\).

Click to see the solution
  1. Isolate \(A\): Let \(M = \begin{bmatrix} -3 & 7 \\ 1 & -2 \end{bmatrix}\). We are given \((7A)^{-1} = M\).
    • Taking the inverse of both sides: \(7A = M^{-1}\).
    • \(A = \frac{1}{7} M^{-1}\).
  2. Find the inverse of \(M\): For a \(2 \times 2\) matrix \(\begin{pmatrix} a & b \\ c & d \end{pmatrix}\), the inverse is \(\frac{1}{ad-bc} \begin{pmatrix} d & -b \\ -c & a \end{pmatrix}\).
    • \(\det(M) = (-3)(-2) - (7)(1) = 6 - 7 = -1\).
    • \(M^{-1} = \frac{1}{-1} \begin{bmatrix} -2 & -7 \\ -1 & -3 \end{bmatrix} = \begin{bmatrix} 2 & 7 \\ 1 & 3 \end{bmatrix}\).
  3. Calculate \(A\):
    • \(A = \frac{1}{7} \begin{bmatrix} 2 & 7 \\ 1 & 3 \end{bmatrix} = \begin{bmatrix} 2/7 & 1 \\ 1/7 & 3/7 \end{bmatrix}\).
Answer: \(A = \begin{bmatrix} 2/7 & 1 \\ 1/7 & 3/7 \end{bmatrix}\).
4.13. Find Matrix A (Lab 3, Task 3b)

Find matrix A given \((5A^T)^{-1} = \begin{bmatrix} -3 & -1 \\ 5 & 2 \end{bmatrix}\).

Click to see the solution
  1. Isolate \(A\): Let \(M = \begin{bmatrix} -3 & -1 \\ 5 & 2 \end{bmatrix}\). We are given \((5A^T)^{-1} = M\).
    • Take the inverse of both sides: \(5A^T = M^{-1}\).
    • \(A^T = \frac{1}{5} M^{-1}\).
    • Take the transpose of both sides: \((A^T)^T = (\frac{1}{5} M^{-1})^T\).
    • \(A = \frac{1}{5} (M^{-1})^T\).
  2. Find the inverse of \(M\):
    • \(\det(M) = (-3)(2) - (-1)(5) = -6 + 5 = -1\).
    • \(M^{-1} = \frac{1}{-1} \begin{bmatrix} 2 & 1 \\ -5 & -3 \end{bmatrix} = \begin{bmatrix} -2 & -1 \\ 5 & 3 \end{bmatrix}\).
  3. Find the transpose of \(M^{-1}\):
    • \((M^{-1})^T = \begin{bmatrix} -2 & 5 \\ -1 & 3 \end{bmatrix}\).
  4. Calculate \(A\):
    • \(A = \frac{1}{5} \begin{bmatrix} -2 & 5 \\ -1 & 3 \end{bmatrix} = \begin{bmatrix} -2/5 & 1 \\ -1/5 & 3/5 \end{bmatrix}\).
Answer: \(A = \begin{bmatrix} -2/5 & 1 \\ -1/5 & 3/5 \end{bmatrix}\).
4.14. Find Matrix A (Lab 3, Task 3c)

Find matrix A given \((I + 2A)^{-1} = \begin{bmatrix} -1 & 2 \\ 4 & 5 \end{bmatrix}\).

Click to see the solution
  1. Isolate \(A\): Let \(M = \begin{bmatrix} -1 & 2 \\ 4 & 5 \end{bmatrix}\). We are given \((I + 2A)^{-1} = M\).
    • Take the inverse of both sides: \(I + 2A = M^{-1}\).
    • \(2A = M^{-1} - I\).
    • \(A = \frac{1}{2}(M^{-1} - I)\).
  2. Find the inverse of \(M\):
    • \(\det(M) = (-1)(5) - (2)(4) = -5 - 8 = -13\).
    • \(M^{-1} = \frac{1}{-13} \begin{bmatrix} 5 & -2 \\ -4 & -1 \end{bmatrix} = \begin{bmatrix} -5/13 & 2/13 \\ 4/13 & 1/13 \end{bmatrix}\).
  3. Subtract the identity matrix \(I\):
    • \(M^{-1} - I = \begin{bmatrix} -5/13 & 2/13 \\ 4/13 & 1/13 \end{bmatrix} - \begin{bmatrix} 1 & 0 \\ 0 & 1 \end{bmatrix} = \begin{bmatrix} -18/13 & 2/13 \\ 4/13 & -12/13 \end{bmatrix}\).
  4. Calculate \(A\):
    • \(A = \frac{1}{2} \begin{bmatrix} -18/13 & 2/13 \\ 4/13 & -12/13 \end{bmatrix} = \begin{bmatrix} -9/13 & 1/13 \\ 2/13 & -6/13 \end{bmatrix}\).
Answer: \(A = \begin{bmatrix} -9/13 & 1/13 \\ 2/13 & -6/13 \end{bmatrix}\).
4.15. Find the Inverse of a Matrix (Lab 3, Task 4a)

Find the inverse of the matrix \(X = \begin{bmatrix} 3 & 4 & -1 \\ 2 & 0 & 7 \\ 1 & -3 & -2 \end{bmatrix}\).

Click to see the solution
  1. Calculate the determinant:
    • \(\det(X) = 3(0(-2) - 7(-3)) - 4(2(-2) - 7(1)) + (-1)(2(-3) - 0(1))\)
    • \(= 3(21) - 4(-4-7) - 1(-6) = 63 - 4(-11) + 6 = 63 + 44 + 6 = 113\).
  2. Find the matrix of cofactors:
    • \(C = \begin{bmatrix} +(21) & -(-11) & +(-6) \\ -(-11) & +(-5) & -(-13) \\ +(28) & -(23) & +(-8) \end{bmatrix} = \begin{bmatrix} 21 & 11 & -6 \\ 11 & -5 & 13 \\ 28 & -23 & -8 \end{bmatrix}\).
  3. Find the adjugate (transpose of cofactors):
    • \(\text{adj}(X) = C^T = \begin{bmatrix} 21 & 11 & 28 \\ 11 & -5 & -23 \\ -6 & 13 & -8 \end{bmatrix}\).
  4. Calculate the inverse: \(X^{-1} = \frac{1}{\det(X)}\text{adj}(X)\).
    • \(X^{-1} = \frac{1}{113} \begin{bmatrix} 21 & 11 & 28 \\ 11 & -5 & -23 \\ -6 & 13 & -8 \end{bmatrix}\).
Answer: \(X^{-1} = \frac{1}{113} \begin{bmatrix} 21 & 11 & 28 \\ 11 & -5 & -23 \\ -6 & 13 & -8 \end{bmatrix}\).
4.16. Find the Inverse of a Matrix (Lab 3, Task 4b)

Find the inverse of the matrix \(Y = \begin{bmatrix} 1 & 5 & -2 \\ 3 & -1 & 4 \\ -3 & 6 & -7 \end{bmatrix}\).

Click to see the solution
  1. Calculate the determinant:
    • \(\det(Y) = 1((-1)(-7) - 4(6)) - 5(3(-7) - 4(-3)) + (-2)(3(6) - (-1)(-3))\)
    • \(= 1(7 - 24) - 5(-21 + 12) - 2(18 - 3)\)
    • \(= 1(-17) - 5(-9) - 2(15) = -17 + 45 - 30 = -2\).
  2. Find the matrix of cofactors:
    • \(C = \begin{bmatrix} +(-17) & -(-9) & +(15) \\ -(-23) & +(-13) & -(21) \\ +(18) & -(10) & +(-16) \end{bmatrix} = \begin{bmatrix} -17 & 9 & 15 \\ 23 & -13 & -21 \\ 18 & -10 & -16 \end{bmatrix}\).
  3. Find the adjugate (transpose of cofactors):
    • \(\text{adj}(Y) = C^T = \begin{bmatrix} -17 & 23 & 18 \\ 9 & -13 & -10 \\ 15 & -21 & -16 \end{bmatrix}\).
  4. Calculate the inverse: \(Y^{-1} = \frac{1}{\det(Y)}\text{adj}(Y)\).
    • \(Y^{-1} = \frac{1}{-2} \begin{bmatrix} -17 & 23 & 18 \\ 9 & -13 & -10 \\ 15 & -21 & -16 \end{bmatrix} = \begin{bmatrix} 17/2 & -23/2 & -9 \\ -9/2 & 13/2 & 5 \\ -15/2 & 21/2 & 8 \end{bmatrix}\).
Answer: \(Y^{-1} = \begin{bmatrix} 8.5 & -11.5 & -9 \\ -4.5 & 6.5 & 5 \\ -7.5 & 10.5 & 8 \end{bmatrix}\).
4.17. Scalar Multiplication of a Matrix (Lecture 3, Example 1)

Multiply the matrix \(A = \begin{bmatrix} 3 & -1 & 2 \\ 0 & 4 & -3 \\ 1 & 2 & 0 \end{bmatrix}\) by the scalar value \(-2\).

Click to see the solution
  1. Understand the operation: To multiply a matrix by a scalar, every element inside the matrix must be multiplied by that scalar.
  2. Perform the multiplication:
    • \(-2A = -2 \cdot \begin{bmatrix} 3 & -1 & 2 \\ 0 & 4 & -3 \\ 1 & 2 & 0 \end{bmatrix}\)
    • \(= \begin{bmatrix} -2 \cdot 3 & -2 \cdot (-1) & -2 \cdot 2 \\ -2 \cdot 0 & -2 \cdot 4 & -2 \cdot (-3) \\ -2 \cdot 1 & -2 \cdot 2 & -2 \cdot 0 \end{bmatrix}\)
  3. Calculate the new elements:
    • \(= \begin{bmatrix} -6 & 2 & -4 \\ 0 & -8 & 6 \\ -2 & -4 & 0 \end{bmatrix}\)
Answer: The resulting matrix is \(-2A = \begin{bmatrix} -6 & 2 & -4 \\ 0 & -8 & 6 \\ -2 & -4 & 0 \end{bmatrix}\).
4.18. Matrix Addition (Lecture 3, Example 2)

Add the following matrices: \(A = \begin{bmatrix} 1 & 3 & -1 \\ 2 & 0 & 4 \\ -3 & 1 & 2 \end{bmatrix}\) and \(B = \begin{bmatrix} 2 & -1 & 3 \\ 1 & 4 & -2 \\ 0 & 1 & 1 \end{bmatrix}\).

Click to see the solution
  1. Check dimensions: To add two matrices, they must have the same dimensions. Both A and B are \(3 \times 3\) matrices, so they can be added.
  2. Add corresponding elements: The sum is found by adding the elements in the same position from each matrix.
    • \(A + B = \begin{bmatrix} 1+2 & 3+(-1) & -1+3 \\ 2+1 & 0+4 & 4+(-2) \\ -3+0 & 1+1 & 2+1 \end{bmatrix}\)
  3. Calculate the new matrix:
    • \(= \begin{bmatrix} 3 & 2 & 2 \\ 3 & 4 & 2 \\ -3 & 2 & 3 \end{bmatrix}\)
Answer: The sum is \(A+B = \begin{bmatrix} 3 & 2 & 2 \\ 3 & 4 & 2 \\ -3 & 2 & 3 \end{bmatrix}\).
4.19. Determinant of a 3x3 Matrix (Lecture 3, Example 3)

Calculate the determinant of the matrix \(A = \begin{bmatrix} 1 & 2 & 3 \\ 0 & 1 & 4 \\ 5 & 6 & 0 \end{bmatrix}\).

Click to see the solution
  1. Use the cofactor expansion method: The determinant of a \(3 \times 3\) matrix can be calculated by expanding along any row or column. We’ll use the first row.
  2. Apply the formula:
    • \(\det(A) = 1 \cdot \det\begin{pmatrix} 1 & 4 \\ 6 & 0 \end{pmatrix} - 2 \cdot \det\begin{pmatrix} 0 & 4 \\ 5 & 0 \end{pmatrix} + 3 \cdot \det\begin{pmatrix} 0 & 1 \\ 5 & 6 \end{pmatrix}\)
  3. Calculate the determinants of the \(2 \times 2\) matrices:
    • \(= 1 \cdot (1 \cdot 0 - 4 \cdot 6) - 2 \cdot (0 \cdot 0 - 4 \cdot 5) + 3 \cdot (0 \cdot 6 - 1 \cdot 5)\)
  4. Simplify the expression:
    • \(= 1(0 - 24) - 2(0 - 20) + 3(0 - 5)\)
    • \(= 1(-24) - 2(-20) + 3(-5)\)
    • \(= -24 + 40 - 15 = 1\)
Answer: The determinant of matrix A is \(1\).
4.20. Transpose of a Matrix (Lecture 3, Example 4)

Find the transpose of the matrix \(A = \begin{bmatrix} 1 & 2 & 3 \\ 4 & 5 & 6 \end{bmatrix}\).

Click to see the solution
  1. Understand transposition: The transpose of a matrix, denoted \(A^T\), is obtained by swapping the rows and columns. If \(A\) is an \(m \times n\) matrix, then \(A^T\) is an \(n \times m\) matrix.
  2. Swap rows and columns:
    • The first row of \(A\) becomes the first column of \(A^T\).
    • The second row of \(A\) becomes the second column of \(A^T\).
  3. Construct the new matrix:
    • \(A^T = \begin{bmatrix} 1 & 4 \\ 2 & 5 \\ 3 & 6 \end{bmatrix}\)
Answer: The transpose is \(A^T = \begin{bmatrix} 1 & 4 \\ 2 & 5 \\ 3 & 6 \end{bmatrix}\).
4.21. Compute the Area of a Triangle (Tutorial 3, Task 1)

Compute the area of a triangle with vertices at \(A(1, 2, 0)\), \(B(3, 0, -3)\), and \(C(5, 2, 6)\).

Click to see the solution
  1. Form two vectors from the vertices: Let’s form vectors \(\vec{AB}\) and \(\vec{AC}\).
    • \(\vec{AB} = B - A = (3-1, 0-2, -3-0) = (2, -2, -3)\)
    • \(\vec{AC} = C - A = (5-1, 2-2, 6-0) = (4, 0, 6)\)
  2. Calculate the cross product: The area of the parallelogram formed by these two vectors is the magnitude of their cross product. The triangle’s area is half of that.
    • \(\vec{AB} \times \vec{AC} = \det \begin{pmatrix} \mathbf{i} & \mathbf{j} & \mathbf{k} \\ 2 & -2 & -3 \\ 4 & 0 & 6 \end{pmatrix}\)
    • \(= \mathbf{i}((-2)(6) - (-3)(0)) - \mathbf{j}((2)(6) - (-3)(4)) + \mathbf{k}((2)(0) - (-2)(4))\)
    • \(= \mathbf{i}(-12 - 0) - \mathbf{j}(12 + 12) + \mathbf{k}(0 + 8)\)
    • \(= -12\mathbf{i} - 24\mathbf{j} + 8\mathbf{k} = (-12, -24, 8)\)
  3. Find the magnitude of the cross product:
    • \(||\vec{AB} \times \vec{AC}|| = \sqrt{(-12)^2 + (-24)^2 + 8^2}\)
    • \(= \sqrt{144 + 576 + 64} = \sqrt{784} = 28\)
  4. Calculate the area of the triangle: The area is half the magnitude.
    • Area \(= \frac{1}{2} ||\vec{AB} \times \vec{AC}|| = \frac{1}{2}(28) = 14\)
Answer: The area of the triangle is \(14\) square units.
4.22. Compute the Volume of a Parallelepiped (Tutorial 3, Task 2)

Compute the volume of the parallelepiped formed by the vectors: \(\vec{a} = (1, 1, 0)\), \(\vec{b} = (0, 1, 1)\), \(\vec{c} = (1, 0, 1)\).

Click to see the solution
  1. Recall the volume formula: The volume of a parallelepiped formed by three vectors is the absolute value of their scalar triple product, which can be calculated as the absolute value of the determinant of the matrix formed by these vectors.
  2. Set up the determinant:
    • Volume \(= |\vec{a} \cdot (\vec{b} \times \vec{c})| = \left| \det \begin{pmatrix} 1 & 1 & 0 \\ 0 & 1 & 1 \\ 1 & 0 & 1 \end{pmatrix} \right|\)
  3. Calculate the determinant:
    • \(= |1(1 \cdot 1 - 1 \cdot 0) - 1(0 \cdot 1 - 1 \cdot 1) + 0(0 \cdot 0 - 1 \cdot 1)|\)
    • \(= |1(1) - 1(-1) + 0|\)
    • \(= |1 + 1| = 2\)
Answer: The volume of the parallelepiped is \(2\) cubic units.
4.23. Calculate the Dot Product (Tutorial 3, Task 3)

Given: \(||\vec{a}|| = 3\), \(||\vec{b}|| = 26\), and \(||\vec{a} \times \vec{b}|| = 72\). Calculate the dot product \(\vec{a} \cdot \vec{b}\).

Click to see the solution
  1. Use the magnitude of the cross product: We know that \(||\vec{a} \times \vec{b}|| = ||\vec{a}|| \cdot ||\vec{b}|| \cdot \sin(\theta)\), where \(\theta\) is the angle between the vectors.
    • \(72 = 3 \cdot 26 \cdot \sin(\theta)\)
    • \(72 = 78 \cdot \sin(\theta)\)
    • \(\sin(\theta) = 72/78 = 12/13\)
  2. Find \(\cos(\theta)\): Using the trigonometric identity \(\sin^2(\theta) + \cos^2(\theta) = 1\).
    • \(\cos^2(\theta) = 1 - \sin^2(\theta) = 1 - (12/13)^2 = 1 - 144/169 = 25/169\)
    • \(\cos(\theta) = \pm\sqrt{25/169} = \pm 5/13\)
  3. Calculate the dot product: The dot product is defined as \(\vec{a} \cdot \vec{b} = ||\vec{a}|| \cdot ||\vec{b}|| \cdot \cos(\theta)\).
    • \(\vec{a} \cdot \vec{b} = 3 \cdot 26 \cdot (\pm 5/13)\)
    • \(\vec{a} \cdot \vec{b} = 78 \cdot (\pm 5/13) = \pm 30\)
Answer: The dot product \(\vec{a} \cdot \vec{b}\) can be either \(30\) or \(-30\).
4.24. Prove a Cross Product Identity (Tutorial 3, Task 4)

Assume that \(\vec{a}\), \(\vec{b}\), and \(\vec{c}\) are three vectors such that \(\vec{a} + \vec{b} + \vec{c} = \vec{0}\). Prove that \(\vec{a} \times \vec{b} = \vec{b} \times \vec{c} = \vec{c} \times \vec{a}\).

Click to see the solution
  1. Start with the given equation: \(\vec{a} + \vec{b} + \vec{c} = \vec{0}\).
  2. Take the cross product with \(\vec{a}\):
    • \(\vec{a} \times (\vec{a} + \vec{b} + \vec{c}) = \vec{a} \times \vec{0}\)
    • \(\vec{a} \times \vec{a} + \vec{a} \times \vec{b} + \vec{a} \times \vec{c} = \vec{0}\)
    • Since \(\vec{a} \times \vec{a} = \vec{0}\), we have \(\vec{a} \times \vec{b} + \vec{a} \times \vec{c} = \vec{0}\).
    • This implies \(\vec{a} \times \vec{b} = -(\vec{a} \times \vec{c})\). Using the anti-commutative property of the cross product, this becomes \(\vec{a} \times \vec{b} = \vec{c} \times \vec{a}\). (Equality 1)
  3. Take the cross product with \(\vec{b}\):
    • \(\vec{b} \times (\vec{a} + \vec{b} + \vec{c}) = \vec{b} \times \vec{0}\)
    • \(\vec{b} \times \vec{a} + \vec{b} \times \vec{b} + \vec{b} \times \vec{c} = \vec{0}\)
    • Since \(\vec{b} \times \vec{b} = \vec{0}\), we have \(\vec{b} \times \vec{a} + \vec{b} \times \vec{c} = \vec{0}\).
    • This implies \(\vec{b} \times \vec{c} = -(\vec{b} \times \vec{a}) = \vec{a} \times \vec{b}\). (Equality 2)
  4. Combine the results: From Equality 1 and Equality 2, we have shown that \(\vec{a} \times \vec{b} = \vec{b} \times \vec{c} = \vec{c} \times \vec{a}\).
Answer: The proof is complete.
4.25. Prove Vectors are Parallel (Tutorial 3, Task 5)

Suppose \(\vec{a}\), \(\vec{b}\), \(\vec{c}\), and \(\vec{d}\) are vectors for which: 1. \(\vec{a} \times \vec{c} = \vec{b} \times \vec{d}\) 2. \(\vec{a} \times \vec{b} = \vec{c} \times \vec{d}\)

Prove that if \(\vec{a} - \vec{d}\) and \(\vec{b} - \vec{c}\) are nonzero, then they are parallel.

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  1. Condition for parallelism: Two nonzero vectors are parallel if their cross product is the zero vector. We need to prove that \((\vec{a} - \vec{d}) \times (\vec{b} - \vec{c}) = \vec{0}\).
  2. Expand the cross product:
    • \((\vec{a} - \vec{d}) \times (\vec{b} - \vec{c}) = \vec{a} \times \vec{b} - \vec{a} \times \vec{c} - \vec{d} \times \vec{b} + \vec{d} \times \vec{c}\)
  3. Rearrange using anti-commutative property: \(\vec{u} \times \vec{v} = -(\vec{v} \times \vec{u})\).
    • \(= (\vec{a} \times \vec{b}) - (\vec{a} \times \vec{c}) + (\vec{b} \times \vec{d}) - (\vec{c} \times \vec{d})\)
  4. Group the terms:
    • \(= [(\vec{a} \times \vec{b}) - (\vec{c} \times \vec{d})] - [(\vec{a} \times \vec{c}) - (\vec{b} \times \vec{d})]\)
  5. Substitute the given conditions:
    • From (2), \(\vec{a} \times \vec{b} = \vec{c} \times \vec{d}\), so the first bracket is \(\vec{0}\).
    • From (1), \(\vec{a} \times \vec{c} = \vec{b} \times \vec{d}\), so the second bracket is also \(\vec{0}\).
    • \(= \vec{0} - \vec{0} = \vec{0}\)
Answer: Since the cross product of \((\vec{a} - \vec{d})\) and \((\vec{b} - \vec{c})\) is the zero vector, and they are both nonzero, they must be parallel.
4.26. Prove Vectors are Coplanar (Tutorial 3, Task 6)

Suppose \(\vec{a}\), \(\vec{b}\), and \(\vec{c}\) are three vectors with the property that: \((\vec{a} \times \vec{b}) + (\vec{b} \times \vec{c}) + (\vec{c} \times \vec{a}) = \vec{0}\). Prove that \(\vec{a}\), \(\vec{b}\), and \(\vec{c}\) are coplanar.

Click to see the solution
  1. Condition for coplanarity: Three vectors are coplanar if their scalar triple product is zero. We need to prove that \(\vec{a} \cdot (\vec{b} \times \vec{c}) = 0\) (or any cyclic permutation).
  2. Take the dot product of the given equation with \(\vec{c}\):
    • \(\vec{c} \cdot [(\vec{a} \times \vec{b}) + (\vec{b} \times \vec{c}) + (\vec{c} \times \vec{a})] = \vec{c} \cdot \vec{0} = 0\)
  3. Distribute the dot product:
    • \(\vec{c} \cdot (\vec{a} \times \vec{b}) + \vec{c} \cdot (\vec{b} \times \vec{c}) + \vec{c} \cdot (\vec{c} \times \vec{a}) = 0\)
  4. Analyze the scalar triple products:
    • The term \(\vec{c} \cdot (\vec{b} \times \vec{c})\) is the volume of a parallelepiped formed by three vectors where two are identical (\(\vec{b}\) and two \(\vec{c}\)’s). This volume is zero. Geometrically, the vector \((\vec{b} \times \vec{c})\) is orthogonal to \(\vec{c}\), so their dot product is zero.
    • Similarly, \(\vec{c} \cdot (\vec{c} \times \vec{a}) = 0\).
  5. Simplify the equation:
    • The equation reduces to \(\vec{c} \cdot (\vec{a} \times \vec{b}) + 0 + 0 = 0\).
    • \(\vec{c} \cdot (\vec{a} \times \vec{b}) = 0\).
  6. Conclusion: By the cyclic property of the scalar triple product, \(\vec{c} \cdot (\vec{a} \times \vec{b}) = \vec{a} \cdot (\vec{b} \times \vec{c})\). Since this is zero, the vectors \(\vec{a}\), \(\vec{b}\), and \(\vec{c}\) are coplanar.
Answer: The proof is complete.
4.27. Calculate the Area of a Triangle (Tutorial 3, Task 7)

Assume that \(\vec{a}\) and \(\vec{b}\) are vectors of length 5 that have an angle of \(45°\) between them. Calculate the area of a triangle formed by the vectors: \(\vec{u} = \vec{a} - 2\vec{b}\) and \(\vec{v} = 3\vec{a} + 2\vec{b}\).

Click to see the solution
  1. Area formula: The area of a triangle formed by vectors \(\vec{u}\) and \(\vec{v}\) is \(\frac{1}{2}||\vec{u} \times \vec{v}||\).
  2. Calculate the cross product \(\vec{u} \times \vec{v}\):
    • \(\vec{u} \times \vec{v} = (\vec{a} - 2\vec{b}) \times (3\vec{a} + 2\vec{b})\)
    • \(= (\vec{a} \times 3\vec{a}) + (\vec{a} \times 2\vec{b}) - (2\vec{b} \times 3\vec{a}) - (2\vec{b} \times 2\vec{b})\)
    • \(= 3(\vec{a} \times \vec{a}) + 2(\vec{a} \times \vec{b}) - 6(\vec{b} \times \vec{a}) - 4(\vec{b} \times \vec{b})\)
  3. Simplify the cross product: Since \(\vec{a} \times \vec{a} = \vec{0}\), \(\vec{b} \times \vec{b} = \vec{0}\), and \(\vec{b} \times \vec{a} = -(\vec{a} \times \vec{b})\):
    • \(= \vec{0} + 2(\vec{a} \times \vec{b}) - 6(-(\vec{a} \times \vec{b})) - \vec{0}\)
    • \(= 2(\vec{a} \times \vec{b}) + 6(\vec{a} \times \vec{b}) = 8(\vec{a} \times \vec{b})\)
  4. Calculate the magnitude:
    • \(||\vec{u} \times \vec{v}|| = ||8(\vec{a} \times \vec{b})|| = 8 ||\vec{a} \times \vec{b}||\)
  5. Calculate \(||\vec{a} \times \vec{b}||\):
    • \(||\vec{a} \times \vec{b}|| = ||\vec{a}|| \cdot ||\vec{b}|| \cdot \sin(\theta)\)
    • Given \(||\vec{a}|| = 5\), \(||\vec{b}|| = 5\), and \(\theta = 45°\). \(\sin(45°) = \frac{\sqrt{2}}{2}\).
    • \(||\vec{a} \times \vec{b}|| = 5 \cdot 5 \cdot \frac{\sqrt{2}}{2} = \frac{25\sqrt{2}}{2}\).
  6. Calculate the final area:
    • Area \(= \frac{1}{2} ||\vec{u} \times \vec{v}|| = \frac{1}{2} \cdot 8 ||\vec{a} \times \vec{b}|| = 4 ||\vec{a} \times \vec{b}||\).
    • Area \(= 4 \cdot \frac{25\sqrt{2}}{2} = 50\sqrt{2}\).
Answer: The area of the triangle is \(50\sqrt{2}\) square units.
4.28. Prove the Vector Triple Product Identity (Homework 3, Task 1)

Assume that \(\vec{a}\), \(\vec{b}\), and \(\vec{c}\) are three vectors. Prove the vector triple product identity: \(\vec{a} \times (\vec{b} \times \vec{c}) = (\vec{a} \cdot \vec{c})\vec{b} - (\vec{a} \cdot \vec{b})\vec{c}\).

Click to see the solution
  1. State the identity: This is a standard identity in vector algebra known as the vector triple product expansion or Lagrange’s formula.
  2. Use the “BAC-CAB” rule: The identity can be remembered by the mnemonic “BAC-CAB”, which corresponds to the structure of the right-hand side.
    • \(\vec{a} \times (\vec{b} \times \vec{c}) = \vec{b}(\vec{a} \cdot \vec{c}) - \vec{c}(\vec{a} \cdot \vec{b})\)
  3. Formal Proof (Component-wise): A formal proof involves defining each vector by its components (e.g., \(\vec{a} = (a_1, a_2, a_3)\)) and calculating both sides of the equation. This process is algebraically intensive but confirms the identity. Let’s demonstrate for the x-component:
    • Left Hand Side (LHS):
      • First, \(\vec{b} \times \vec{c} = (b_yc_z - b_zc_y, b_zc_x - b_xc_z, b_xc_y - b_yc_x)\).
      • Then the x-component of \(\vec{a} \times (\vec{b} \times \vec{c})\) is \(a_y(b_xc_y - b_yc_x) - a_z(b_zc_x - b_xc_z)\).
      • LHS\(_x = a_yb_xc_y - a_yb_yc_x - a_zb_zc_x + a_zb_xc_z\).
    • Right Hand Side (RHS):
      • First, \(\vec{a} \cdot \vec{c} = a_xc_x + a_yc_y + a_zc_z\) and \(\vec{a} \cdot \vec{b} = a_xb_x + a_yb_y + a_zb_z\).
      • The x-component of \((\vec{a} \cdot \vec{c})\vec{b} - (\vec{a} \cdot \vec{b})\vec{c}\) is \(b_x(a_xc_x + a_yc_y + a_zc_z) - c_x(a_xb_x + a_yb_y + a_zb_z)\).
      • RHS\(_x = a_xb_xc_x + a_yb_xc_y + a_zb_xc_z - a_xb_xc_x - a_yb_yc_x - a_zb_zc_x\).
      • RHS\(_x = a_yb_xc_y - a_yb_yc_x - a_zb_zc_x + a_zb_xc_z\).
    • The x-components match. A similar calculation shows the y and z components also match.
Answer: The identity \(\vec{a} \times (\vec{b} \times \vec{c}) = (\vec{a} \cdot \vec{c})\vec{b} - (\vec{a} \cdot \vec{b})\vec{c}\) is a fundamental property of vectors.
4.29. Prove the Jacobi Identity (Homework 3, Task 2)

Prove that for any vectors \(\vec{a}\), \(\vec{b}\), and \(\vec{c}\), we have: \(\vec{a} \times (\vec{b} \times \vec{c}) + \vec{b} \times (\vec{c} \times \vec{a}) + \vec{c} \times (\vec{a} \times \vec{b}) = \vec{0}\). This is known as the Jacobi identity for vector triple products.

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  1. Apply the vector triple product identity to each term: We use the identity \(\vec{X} \times (\vec{Y} \times \vec{Z}) = (\vec{X} \cdot \vec{Z})\vec{Y} - (\vec{X} \cdot \vec{Y})\vec{Z}\).
  2. Expand the first term:
    • \(\vec{a} \times (\vec{b} \times \vec{c}) = (\vec{a} \cdot \vec{c})\vec{b} - (\vec{a} \cdot \vec{b})\vec{c}\).
  3. Expand the second term:
    • \(\vec{b} \times (\vec{c} \times \vec{a}) = (\vec{b} \cdot \vec{a})\vec{c} - (\vec{b} \cdot \vec{c})\vec{a}\).
  4. Expand the third term:
    • \(\vec{c} \times (\vec{a} \times \vec{b}) = (\vec{c} \cdot \vec{b})\vec{a} - (\vec{c} \cdot \vec{a})\vec{b}\).
  5. Sum the three expanded expressions:
    • Sum = \([(\vec{a} \cdot \vec{c})\vec{b} - (\vec{a} \cdot \vec{b})\vec{c}] + [(\vec{b} \cdot \vec{a})\vec{c} - (\vec{b} \cdot \vec{c})\vec{a}] + [(\vec{c} \cdot \vec{b})\vec{a} - (\vec{c} \cdot \vec{a})\vec{b}]\).
  6. Group terms by vector and simplify: We use the commutative property of the dot product (e.g., \(\vec{a} \cdot \vec{b} = \vec{b} \cdot \vec{a}\)).
    • Terms with \(\vec{a}\): \(-(\vec{b} \cdot \vec{c})\vec{a} + (\vec{c} \cdot \vec{b})\vec{a} = \vec{0}\).
    • Terms with \(\vec{b}\): \((\vec{a} \cdot \vec{c})\vec{b} - (\vec{c} \cdot \vec{a})\vec{b} = \vec{0}\).
    • Terms with \(\vec{c}\): \(-(\vec{a} \cdot \vec{b})\vec{c} + (\vec{b} \cdot \vec{a})\vec{c} = \vec{0}\).
  7. Conclusion: All terms cancel out.
    • Sum = \(\vec{0} + \vec{0} + \vec{0} = \vec{0}\).
Answer: The sum of the three vector triple products is the zero vector, which proves the Jacobi identity.
4.30. Effect of Row/Column Swap on Determinant (Homework 3, Task 3)

What is the effect on a matrix’s determinant when two of its rows or columns are swapped?

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  1. State the property: Swapping any two rows or any two columns of a square matrix multiplies its determinant by \(-1\).
  2. Formal statement: If \(A\) is a square matrix and \(B\) is the matrix that results from swapping two rows (or columns) of \(A\), then \(\det(B) = -\det(A)\).
  3. Illustrate with a \(2 \times 2\) example:
    • Let matrix \(A = \begin{pmatrix} a & b \\ c & d \end{pmatrix}\). Its determinant is \(\det(A) = ad - bc\).
    • Now, swap the two rows to get matrix \(B = \begin{pmatrix} c & d \\ a & b \end{pmatrix}\).
    • The determinant of the new matrix is \(\det(B) = cb - da = -(ad - bc)\).
    • Therefore, \(\det(B) = -\det(A)\).
  4. Conclusion: The effect is that the determinant changes its sign.
Answer: When two rows or columns of a matrix are swapped, the determinant of the new matrix is the negative of the determinant of the original matrix.