Engineering projects in quantity surveying often involve systems of linear equations to model costs, resources, and schedules. Efficiently solving these systems requires mastery of matrix operations, particularly determinants and inverses. This chapter focuses on the calculation of determinants and inverses of 2x2 and 3x3 matrices, foundational skills essential for analyzing project data, cost allocations, and optimization in Kenyan construction and infrastructure projects.

2.1 Determinant and inverse of 2x2 matrix

Calculating the determinant and inverse of a 2x2 matrix is a fundamental skill in quantity surveying for solving simple linear systems such as budgeting constraints or resource allocations. These calculations enable quick evaluation of matrix properties and facilitate solving equations related to cost distribution or material quantities in small-scale projects.

2.1.1 Determinant of a 2x2 matrix

The determinant of a 2x2 matrix gives a scalar value that indicates whether the matrix is invertible. For a matrix

$$ A = \begin{bmatrix} a & b \\ c & d \end{bmatrix}, $$

the determinant is calculated as

$$ \det(A) = ad, bc. $$

Worked Examples

Example 1:
Given a matrix representing cost coefficients in a small project

$$ A = \begin{bmatrix} 4 & 3 \\ 2 & 5 \end{bmatrix}, $$

Calculate the determinant.

$$ \det(A) = (4)(5) - (3)(2) $$

$$ = 20 - 6 $$

$$ = 14 $$

Answer: \(\det(A) = 14\)

Example 2:
For matrix

$$ B = \begin{bmatrix} 7 & 2 \\ 6 & 7 \end{bmatrix}, $$

find the determinant.

$$ \det(B) = (7)(7) - (2)(6) $$

$$ = 49 - 12 $$

$$ = 37 $$

Answer: \(\det(B) = 37\)

Example 3:
Calculate the determinant of

$$ C = \begin{bmatrix} 1 & 4 \\ 5 & 1 \end{bmatrix}. $$

$$ \det(C) = (1)(1) - (4)(5) $$

$$ = 1 - 20 $$

$$ = -19 $$

Answer: \(\det(C) = -19\)

2.1.2 Inverse of a 2x2 matrix

The inverse of a 2x2 matrix exists only if its determinant is non-zero. The inverse matrix \(A^{-1}\) satisfies

$$ A A^{-1} = I, $$

where \(I\) is the 2x2 identity matrix. For matrix

$$ A = \begin{bmatrix} a & b \\ c & d \end{bmatrix}, $$

the inverse is given by

$$ A^{-1} = \frac{1}{\det(A)} \begin{bmatrix} d & -b \\ -c & a \end{bmatrix}. $$

Worked Examples

Example 1:
Find the inverse of

$$ A = \begin{bmatrix} 3 & 2 \\ 1 & 4 \end{bmatrix}. $$

Calculate determinant:

$$ \det(A) = (3)(4) - (2)(1) = 12 - 2 = 10. $$

Calculate inverse:

$$ A^{-1} = \frac{1}{10} \begin{bmatrix} 4 & -2 \\ -1 & 3 \end{bmatrix} = \begin{bmatrix} 0.4 & -0.2 \\ -0.1 & 0.3 \end{bmatrix}. $$

Answer:

$$ A^{-1} = \begin{bmatrix} 0.4 & -0.2 \\ -0.1 & 0.3 \end{bmatrix}. $$

Example 2:
Calculate the inverse of

$$ B = \begin{bmatrix} 5 & 1 \\ 2 & 3 \end{bmatrix}. $$

Determinant:

$$ \det(B) = (5)(3) - (1)(2) = 15 - 2 = 13. $$

Inverse:

$$ B^{-1} = \frac{1}{13} \begin{bmatrix} 3 & -1 \\ -2 & 5 \end{bmatrix} = \begin{bmatrix} \frac{3}{13} & -\frac{1}{13} \\ -\frac{2}{13} & \frac{5}{13} \end{bmatrix}. $$

Answer:

$$ B^{-1} = \begin{bmatrix} 0.2308 & -0.0769 \\ -0.1538 & 0.3846 \end{bmatrix}. $$

Example 3:
Find inverse of

$$ C = \begin{bmatrix} 6 & 4 \\ 3 & 2 \end{bmatrix}. $$

Determinant:

$$ \det(C) = (6)(2) - (4)(3) = 12 - 12 = 0. $$

Since determinant is zero, inverse does not exist.

Answer: Inverse does not exist because \(\det(C) = 0\).

2.1.3 Application of 2x2 inverse matrix in solving linear systems

Inverse matrices solve linear systems of two equations in two unknowns, common in quantity surveying for cost and resource allocation.

Given system:

$$ a x + b y = e $$

$$ c x + d y = f, $$

expressed as

$$ A \mathbf{x} = \mathbf{b}, $$

where

$$ A = \begin{bmatrix} a & b \\ c & d \end{bmatrix}, \quad \mathbf{x} = \begin{bmatrix} x \\ y \end{bmatrix}, \quad \mathbf{b} = \begin{bmatrix} e \\ f \end{bmatrix}. $$

Solution:

$$ \mathbf{x} = A^{-1} \mathbf{b}. $$

Worked Examples

Example 1:
Solve the system

$$ 2x + 3y = 8, $$

$$ 4x + y = 7. $$

Matrix form:

$$ A = \begin{bmatrix} 2 & 3 \\ 4 & 1 \end{bmatrix}, \quad \mathbf{b} = \begin{bmatrix} 8 \\ 7 \end{bmatrix}. $$

Determinant:

$$ \det(A) = (2)(1) - (3)(4) = 2 - 12 = -10. $$

Inverse:

$$ A^{-1} = \frac{1}{-10} \begin{bmatrix} 1 & -3 \\ -4 & 2 \end{bmatrix} = \begin{bmatrix} -0.1 & 0.3 \\ 0.4 & -0.2 \end{bmatrix}. $$

Calculate:

$$ \mathbf{x} = A^{-1} \mathbf{b} = \begin{bmatrix} -0.1 & 0.3 \\ 0.4 & -0.2 \end{bmatrix} \begin{bmatrix} 8 \\ 7 \end{bmatrix} $$

$$ = \begin{bmatrix} (-0.1)(8) + (0.3)(7) \\ (0.4)(8) + (-0.2)(7) \end{bmatrix} = \begin{bmatrix} -0.8 + 2.1 \\ 3.2 - 1.4 \end{bmatrix} = \begin{bmatrix} 1.3 \\ 1.8 \end{bmatrix}. $$

Answer:

$$ x = 1.3, \quad y = 1.8. $$

Example 2:
Solve

$$ 3x, y = 5, $$

$$ 2x + 4y = 12. $$

Matrix:

$$ A = \begin{bmatrix} 3 & -1 \\ 2 & 4 \end{bmatrix}, \quad \mathbf{b} = \begin{bmatrix} 5 \\ 12 \end{bmatrix}. $$

Determinant:

$$ \det(A) = (3)(4) - (-1)(2) = 12 + 2 = 14. $$

Inverse:

$$ A^{-1} = \frac{1}{14} \begin{bmatrix} 4 & 1 \\ -2 & 3 \end{bmatrix} = \begin{bmatrix} 0.2857 & 0.0714 \\ -0.1429 & 0.2143 \end{bmatrix}. $$

Calculate:

$$ \mathbf{x} = A^{-1} \mathbf{b} = \begin{bmatrix} 0.2857 & 0.0714 \\ -0.1429 & 0.2143 \end{bmatrix} \begin{bmatrix} 5 \\ 12 \end{bmatrix} $$

$$ = \begin{bmatrix} (0.2857)(5) + (0.0714)(12) \\ (-0.1429)(5) + (0.2143)(12) \end{bmatrix} = \begin{bmatrix} 1.4285 + 0.857 \\ -0.7145 + 2.5716 \end{bmatrix} = \begin{bmatrix} 2.2855 \\ 1.8571 \end{bmatrix}. $$

Answer:

$$ x = 2.29, \quad y = 1.86. $$

Example 3:
Solve

$$ 5x + 2y = 14, $$

$$ -3x + 4y = 1. $$

Matrix:

$$ A = \begin{bmatrix} 5 & 2 \\ -3 & 4 \end{bmatrix}, \quad \mathbf{b} = \begin{bmatrix} 14 \\ 1 \end{bmatrix}. $$

Determinant:

$$ \det(A) = (5)(4) - (2)(-3) = 20 + 6 = 26. $$

Inverse:

$$ A^{-1} = \frac{1}{26} \begin{bmatrix} 4 & -2 \\ 3 & 5 \end{bmatrix} = \begin{bmatrix} 0.1538 & -0.0769 \\ 0.1154 & 0.1923 \end{bmatrix}. $$

Calculate:

$$ \mathbf{x} = A^{-1} \mathbf{b} = \begin{bmatrix} 0.1538 & -0.0769 \\ 0.1154 & 0.1923 \end{bmatrix} \begin{bmatrix} 14 \\ 1 \end{bmatrix} $$

$$ = \begin{bmatrix} (0.1538)(14) + (-0.0769)(1) \\ (0.1154)(14) + (0.1923)(1) \end{bmatrix} = \begin{bmatrix} 2.1532 - 0.0769 \\ 1.6156 + 0.1923 \end{bmatrix} = \begin{bmatrix} 2.0763 \\ 1.8079 \end{bmatrix}. $$

Answer:

$$ x = 2.08, \quad y = 1.81. $$

2.1.4 Properties of 2x2 determinants and inverses

Determinants and inverses of 2x2 matrices have properties useful in quantity surveying calculations for checking matrix behavior.

• The determinant changes sign if rows or columns are interchanged.
• The determinant of a product of matrices equals the product of their determinants.
• The inverse of an inverse matrix equals the original matrix.
• If the determinant is zero, the matrix is singular and not invertible.
• The transpose of an inverse equals the inverse of the transpose.

Worked Examples

Example 1:
Given

$$ A = \begin{bmatrix} 1 & 2 \\ 3 & 4 \end{bmatrix}, $$

compute determinant of \(A\) and \(A^T\).

$$ \det(A) = (1)(4) - (2)(3) = 4 - 6 = -2. $$

Transpose:

$$ A^T = \begin{bmatrix} 1 & 3 \\ 2 & 4 \end{bmatrix}. $$

Determinant:

$$ \det(A^T) = (1)(4) - (3)(2) = 4 - 6 = -2. $$

Answer: Determinants of \(A\) and \(A^T\) are equal: \(-2\).

Example 2:
Verify that determinant of product equals product of determinants for

$$ A = \begin{bmatrix} 2 & 0 \\ 1 & 3 \end{bmatrix}, \quad B = \begin{bmatrix} 1 & 4 \\ 2 & 5 \end{bmatrix}. $$

Calculate:

$$ \det(A) = (2)(3) - (0)(1) = 6 - 0 = 6, $$

$$ \det(B) = (1)(5) - (4)(2) = 5 - 8 = -3. $$

Product:

$$ AB = \begin{bmatrix} (2)(1)+(0)(2) & (2)(4)+(0)(5) \\ (1)(1)+(3)(2) & (1)(4)+(3)(5) \end{bmatrix} = \begin{bmatrix} 2 & 8 \\ 7 & 19 \end{bmatrix}. $$

Determinant:

$$ \det(AB) = (2)(19) - (8)(7) = 38 - 56 = -18. $$

Product of determinants:

$$ \det(A) \times \det(B) = 6 \times (-3) = -18. $$

Answer: \(\det(AB) = \det(A) \times \det(B) = -18\).

Chapter Summary

This chapter focused on performing matrix calculations essential for engineering applications. It began with methods to find the determinant and inverse of a 2x2 matrix, providing the foundation for understanding matrix operations. The study then extended these concepts to 3x3 matrices, explaining the more complex procedures for calculating determinants and inverses. Following this, techniques for solving simultaneous linear equations using matrix methods were explored, demonstrating how matrices can simplify solving systems of equations. The chapter concluded with an introduction to eigenvalues and eigenvectors, illustrating their calculation and significance in matrix analysis. Together, these topics equip learners with practical skills for handling matrices in various engineering problems. Mastery of these concepts enables effective problem-solving in structural analysis, electrical circuits, and other technical fields.