Engineering Mathematics III is essential for Land Survey professionals in Kenya, providing mathematical tools to solve spatial and measurement problems using matrices. Matrices simplify complex systems of linear equations that frequently arise in coordinate transformations, least squares adjustments, and network computations in surveying. Mastery of determinants and matrix inverses is fundamental for solving simultaneous equations and transforming coordinate data efficiently.

3.1 Determinant and inverse of 2x2

Determinants and inverses of 2x2 matrices are fundamental concepts in matrix algebra, widely used in surveying computations such as coordinate transformations and solving linear systems. The determinant helps determine if a matrix is invertible, while the inverse matrix is crucial for solving matrix equations. This section covers the calculation of determinants and inverses of 2x2 matrices with worked examples related to land survey problems.

3.1.1 Determinant of a 2x2 matrix

The determinant of a 2x2 matrix is a scalar value that provides information about the matrix, such as whether it is invertible. For a matrix
\[ A = \begin{bmatrix} a & b \\ c & d \end{bmatrix} \]
the determinant is calculated by the formula:

$$\det(A) = ad - bc$$

Worked Examples

Example 1: Given a matrix representing a simple transformation in a survey coordinate system:

\[ A = \begin{bmatrix} 3 & 5 \\ 2 & 7 \end{bmatrix} \]

Calculate the determinant.

Given: \(a=3\), \(b=5\), \(c=2\), \(d=7\)

$$\det(A) = ad - bc$$

$$= (3)(7) - (5)(2)$$

$$= 21 - 10$$

Answer: \(\det(A) = 11\)

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Example 2: The determinant of the matrix

\[ B = \begin{bmatrix} -4 & 6 \\ 3 & 2 \end{bmatrix} \]

Given: \(a=-4\), \(b=6\), \(c=3\), \(d=2\)

$$\det(B) = ad - bc$$

$$= (-4)(2) - (6)(3)$$

$$= -8 - 18$$

Answer: \(\det(B) = -26\)

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Example 3: Calculate the determinant of the matrix

\[ C = \begin{bmatrix} 0 & 7 \\ -3 & 4 \end{bmatrix} \]

Given: \(a=0\), \(b=7\), \(c=-3\), \(d=4\)

$$\det(C) = ad - bc$$

$$= (0)(4) - (7)(-3)$$

$$= 0 + 21$$

Answer: \(\det(C) = 21\)

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Example 4: A surveyor has the matrix

\[ D = \begin{bmatrix} 1.5 & 0.5 \\ 2 & 3 \end{bmatrix} \]

Calculate the determinant.

Given: \(a=1.5\), \(b=0.5\), \(c=2\), \(d=3\)

$$\det(D) = ad - bc$$

$$= (1.5)(3) - (0.5)(2)$$

$$= 4.5 - 1$$

Answer: \(\det(D) = 3.5\)

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Example 5: For the matrix

\[ E = \begin{bmatrix} 8 & -2 \\ -4 & 5 \end{bmatrix} \]

Calculate the determinant.

Given: \(a=8\), \(b=-2\), \(c=-4\), \(d=5\)

$$\det(E) = ad - bc$$

$$= (8)(5) - (-2)(-4)$$

$$= 40 - 8$$

Answer: \(\det(E) = 32\)

3.1.2 Conditions for invertibility of a 2x2 matrix

A 2x2 matrix is invertible if and only if its determinant is nonzero. This condition ensures that the matrix has a unique inverse, which is critical in solving survey equations involving coordinate transformations or network adjustments.

The invertibility condition is:

$$\det(A) \neq 0$$

If \(\det(A) = 0\), the matrix is singular and does not have an inverse.

Worked Examples

Example 1: Determine if the matrix

\[ A = \begin{bmatrix} 2 & 3 \\ 4 & 6 \end{bmatrix} \]

is invertible.

Given: \(a=2\), \(b=3\), \(c=4\), \(d=6\)

$$\det(A) = ad - bc$$

$$= (2)(6) - (3)(4)$$

$$= 12 - 12$$

$$= 0$$

Answer: Matrix A is not invertible (singular).

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Example 2: Check invertibility of

\[ B = \begin{bmatrix} 1 & 0 \\ 0 & 1 \end{bmatrix} \]

Given \(a=1\), \(b=0\), \(c=0\), \(d=1\)

$$\det(B) = (1)(1) - (0)(0) = 1$$

Answer: Matrix B is invertible.

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Example 3: For the matrix

\[ C = \begin{bmatrix} 0 & 5 \\ 2 & 1 \end{bmatrix} \]

Calculate determinant.

$$\det(C) = (0)(1) - (5)(2) = 0 - 10 = -10$$

Answer: Matrix C is invertible.

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Example 4: Matrix

\[ D = \begin{bmatrix} 7 & 14 \\ 2 & 4 \end{bmatrix} \]

Calculate determinant.

$$\det(D) = (7)(4) - (14)(2) = 28 - 28 = 0$$

Answer: Matrix D is not invertible.

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Example 5: Matrix

\[ E = \begin{bmatrix} 3 & 1 \\ 5 & 2 \end{bmatrix} \]

Calculate determinant.

$$\det(E) = (3)(2) - (1)(5) = 6 - 5 = 1$$

Answer: Matrix E is invertible.

3.1.3 Inverse of a 2x2 matrix

The inverse of a 2x2 matrix \(A\) is another matrix \(A^{-1}\) such that:

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

where \(I\) is the 2x2 identity matrix. The formula for the inverse is:

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

This formula applies only if \(\det(A) \neq 0\).

Worked Examples

Example 1: Find the inverse of

\[ A = \begin{bmatrix} 1 & 2 \\ 3 & 4 \end{bmatrix} \]

Given: \(a=1\), \(b=2\), \(c=3\), \(d=4\)

Calculate determinant:

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

Calculate inverse:

$$A^{-1} = \frac{1}{-2} \begin{bmatrix} 4 & -2 \\ -3 & 1 \end{bmatrix}$$

$$= \begin{bmatrix} -2 & 1 \\ 1.5 & -0.5 \end{bmatrix}$$

Answer:

\[ A^{-1} = \begin{bmatrix} -2 & 1 \\ 1.5 & -0.5 \end{bmatrix} \]

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Example 2: Find the inverse of

\[ B = \begin{bmatrix} 5 & 1 \\ 2 & 3 \end{bmatrix} \]

Calculate determinant:

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

Calculate 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:

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

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Example 3: Find the inverse of

\[ C = \begin{bmatrix} 4 & 7 \\ 2 & 6 \end{bmatrix} \]

Calculate determinant:

$$\det(C) = (4)(6) - (7)(2) = 24 - 14 = 10$$

Calculate inverse:

$$C^{-1} = \frac{1}{10} \begin{bmatrix} 6 & -7 \\ -2 & 4 \end{bmatrix}$$

$$= \begin{bmatrix} 0.6 & -0.7 \\ -0.2 & 0.4 \end{bmatrix}$$

Answer:

\[ C^{-1} = \begin{bmatrix} 0.6 & -0.7 \\ -0.2 & 0.4 \end{bmatrix} \]

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Example 4: Find the inverse of

\[ D = \begin{bmatrix} 2 & 3 \\ 1 & 4 \end{bmatrix} \]

Calculate determinant:

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

Calculate inverse:

$$D^{-1} = \frac{1}{5} \begin{bmatrix} 4 & -3 \\ -1 & 2 \end{bmatrix}$$

$$= \begin{bmatrix} 0.8 & -0.6 \\ -0.2 & 0.4 \end{bmatrix}$$

Answer:

\[ D^{-1} = \begin{bmatrix} 0.8 & -0.6 \\ -0.2 & 0.4 \end{bmatrix} \]

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Example 5: Find the inverse of

\[ E = \begin{bmatrix} 1.5 & 0.5 \\ 2 & 3 \end{bmatrix} \]

Calculate determinant:

$$\det(E) = (1.5)(3) - (0.5)(2) = 4.5 - 1 = 3.5$$

Calculate inverse:

$$E^{-1} = \frac{1}{3.5} \begin{bmatrix} 3 & -0.5 \\ -2 & 1.5 \end{bmatrix}$$

$$= \begin{bmatrix} \frac{3}{3.5} & -\frac{0.5}{3.5} \\ -\frac{2}{3.5} & \frac{1.5}{3.5} \end{bmatrix}$$

$$= \begin{bmatrix} 0.8571 & -0.1429 \\ -0.5714 & 0.4286 \end{bmatrix}$$

Answer:

\[ E^{-1} = \begin{bmatrix} 0.8571 & -0.1429 \\ -0.5714 & 0.4286 \end{bmatrix} \]

3.1.4 Application of inverse matrices in solving linear survey equations

In land surveying, systems of linear equations often arise when adjusting coordinates or solving transformation parameters. The inverse of a 2x2 matrix can be used to solve a system of equations of the form:

$$AX = B$$

where \(A\) is a 2x2 coefficient matrix, \(X\) is the unknown vector, and \(B\) is the constants vector. The solution is:

$$X = A^{-1} B$$

Worked Examples

Example 1: Solve the system

\[ \begin{cases} 2x + 3y = 8 \\ 5x + 4y = 13 \end{cases} \]

Matrix form:

\[ A = \begin{bmatrix} 2 & 3 \\ 5 & 4 \end{bmatrix}, \quad B = \begin{bmatrix} 8 \\ 13 \end{bmatrix} \]

Calculate determinant:

$$\det(A) = (2)(4) - (3)(5) = 8 - 15 = -7$$

Calculate inverse:

$$A^{-1} = \frac{1}{-7} \begin{bmatrix} 4 & -3 \\ -5 & 2 \end{bmatrix} = \begin{bmatrix} -\frac{4}{7} & \frac{3}{7} \\ \frac{5}{7} & -\frac{2}{7} \end{bmatrix}$$

Calculate solution:

$$X = A^{-1} B = \begin{bmatrix} -\frac{4}{7} & \frac{3}{7} \\ \frac{5}{7} & -\frac{2}{7} \end{bmatrix} \begin{bmatrix} 8 \\ 13 \end{bmatrix}$$

Calculate components:

$$x = -\frac{4}{7} \times 8 + \frac{3}{7} \times 13 = -\frac{32}{7} + \frac{39}{7} = \frac{7}{7} = 1$$

$$y = \frac{5}{7} \times 8 - \frac{2}{7} \times 13 = \frac{40}{7} - \frac{26}{7} = \frac{14}{7} = 2$$

Answer:

\[ x = 1, \quad y = 2 \]

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Example 2: Solve

\[ \begin{cases} 4x + y = 9 \\ 2x + 3y = 13 \end{cases} \]

Matrix form:

\[ A = \begin{bmatrix} 4 & 1 \\ 2 & 3 \end{bmatrix}, \quad B = \begin{bmatrix} 9 \\ 13 \end{bmatrix} \]

Calculate determinant:

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

Calculate inverse:

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

Calculate solution:

$$X = A^{-1} B = \begin{bmatrix} 0.3 & -0.1 \\ -0.2 & 0.4 \end{bmatrix} \begin{bmatrix} 9 \\ 13 \end{bmatrix}$$

Calculate components:

$$x = 0.3 \times 9 - 0.1 \times 13 = 2.7 - 1.3 = 1.4$$

$$y = -0.2 \times 9 + 0.4 \times 13 = -1.8 + 5.2 = 3.4$$

Answer:

\[ x = 1.4, \quad y = 3.4 \]

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Example 3: Solve the system:

\[ \begin{cases} x + 2y = 7 \\ 3x + 4y = 18 \end{cases} \]

Matrix form:

\[ A = \begin{bmatrix} 1 & 2 \\ 3 & 4 \end{bmatrix}, \quad B = \begin{bmatrix} 7 \\ 18 \end{bmatrix} \]

Calculate determinant:

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

Calculate inverse:

$$A^{-1} = \frac{1}{-2} \begin{bmatrix} 4 & -2 \\ -3 & 1 \end{bmatrix} = \begin{bmatrix} -2 & 1 \\ 1.5 & -0.5 \end{bmatrix}$$

Calculate solution:

$$X = A^{-1} B = \begin{bmatrix} -2 & 1 \\ 1.5 & -0.5 \end{bmatrix} \begin{bmatrix} 7 \\ 18 \end{bmatrix}$$

Calculate components:

$$x = (-2)(7) + (1)(18) = -14 + 18 = 4$$

$$y = (1.5)(7) + (-0.5)(18) = 10.5 - 9 = 1.5$$

Answer:

\[ x = 4, \quad y = 1.5 \]

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Example 4: Solve the system:

\[ \begin{cases} 3x + 5y = 23 \\ 4x + 2y = 16 \end{cases} \]

Matrix form:

\[ A = \begin{bmatrix} 3 & 5 \\ 4 & 2 \end{bmatrix}, \quad B = \begin{bmatrix} 23 \\ 16 \end{bmatrix} \]

Calculate determinant:

$$\det(A) = (3)(2) - (5)(4) = 6 - 20 = -14$$

Calculate inverse:

$$A^{-1} = \frac{1}{-14} \begin{bmatrix} 2 & -5 \\ -4 & 3 \end{bmatrix} = \begin{bmatrix} -\frac{1}{7} & \frac{5}{14} \\ \frac{2}{7} & -\frac{3}{14} \end{bmatrix}$$

Calculate solution:

$$X = A^{-1} B = \begin{bmatrix} -\frac{1}{7} & \frac{5}{14} \\ \frac{2}{7} & -\frac{3}{14} \end{bmatrix} \begin{bmatrix} 23 \\ 16 \end{bmatrix}$$

Calculate components:

$$x = -\frac{1}{7} \times 23 + \frac{5}{14} \times 16 = -\frac{23}{7} + \frac{80}{14} = -3.2857 + 5.7143 = 2.4286$$

$$y = \frac{2}{7} \times 23 - \frac{3}{14} \times 16 = \frac{46}{7} - \frac{48}{14} = 6.5714 - 3.4286 = 3.1428$$

Answer:

\[ x \approx 2.43, \quad y \approx 3.14 \]

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Example 5: Solve the system:

\[ \begin{cases} 1.5x + 0.5y = 4 \\ 2x + 3y = 12 \end{cases} \]

Matrix form:

\[ A = \begin{bmatrix} 1.5 & 0.5 \\ 2 & 3 \end{bmatrix}, \quad B = \begin{bmatrix} 4 \\ 12 \end{bmatrix} \]

Calculate determinant:

$$\det(A) = (1.5)(3) - (0.5)(2) = 4.5 - 1 = 3.5$$

Calculate inverse:

$$A^{-1} = \frac{1}{3.5} \begin{bmatrix} 3 & -0.5 \\ -2 & 1.5 \end{bmatrix} = \begin{bmatrix} 0.8571 & -0.1429 \\ -0.5714 & 0.4286 \end{bmatrix}$$

Calculate solution:

$$X = A^{-1} B = \begin{bmatrix} 0.8571 & -0.1429 \\ -0.5714 & 0.4286 \end{bmatrix} \begin{bmatrix} 4 \\ 12 \end{bmatrix}$$

Calculate components:

$$x = 0.8571 \times 4 - 0.1429 \times 12 = 3.4284 - 1.7148 = 1.7136$$

$$y = -0.5714 \times 4 + 0.4286 \times 12 = -2.2856 + 5.1432 = 2.8576$$

Answer:

\[ x \approx 1.71, \quad y \approx 2.86 \]

Chapter Summary

This chapter focused on the application of matrices in solving engineering problems, beginning with the calculation of the determinant and inverse of a 2x2 matrix. Understanding how to find the determinant is essential as it determines whether the matrix is invertible. The chapter then advanced to the more complex 3x3 matrices, where methods for computing both the determinant and the inverse were explored in detail. Mastery of these concepts enables the handling of larger systems of equations. Finally, the chapter demonstrated how to use these matrix operations to solve simultaneous linear equations, a critical skill for analyzing engineering systems. Through systematic calculation of inverses and determinants, students can efficiently find unique solutions to systems represented in matrix form. This foundation supports more advanced studies in linear algebra and its applications in engineering design and analysis.