Matrices are fundamental in land surveying for organizing and solving systems of linear equations that arise in coordinate transformations, least squares adjustments, and network computations. The determinant and inverse of a matrix are essential tools to assess matrix properties and to find solutions to linear systems critical in survey computations. Mastery of these concepts enables surveyors to process measurement data accurately and efficiently, ensuring precise positioning and mapping in various Kenyan surveying projects.

3.1 Determinant and inverse of 2x2

Determinants and inverses of 2x2 matrices play a pivotal role in solving simultaneous linear equations commonly encountered in coordinate geometry and transformation tasks in land surveying. The determinant helps to determine if a matrix is invertible, which is crucial when solving for unknown coordinates or parameters. The inverse matrix, when it exists, provides a direct method to solve linear systems, making it indispensable for survey computations such as adjusting traverse coordinates or converting between coordinate systems.

3.1.1 Determinant of a 2x2 matrix

The determinant of a 2x2 matrix is a scalar value that provides important information about the matrix, including whether it is singular (non-invertible) or invertible. For a matrix

\[ A = \begin{bmatrix} a & b \\ c & d \end{bmatrix} \]

the determinant is calculated by

\[ \det(A) = ad - bc \]

The determinant's value indicates the matrix's behavior in transformations: a zero determinant means the matrix compresses space into a lower dimension, making it non-invertible.

Worked Examples

Example 1: Calculate the determinant of the matrix

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

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

\[ \det(A) = ad - bc \]

\[ = (2)(4) - (3)(1) \]

\[ = 8 - 3 \]

Answer: 5

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Example 2: Find the determinant of the matrix

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

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

\[ \det(B) = ad - bc \]

\[ = (5)(3) - (-2)(7) \]

\[ = 15 + 14 \]

Answer: 29

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Example 3: Compute the determinant of matrix

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

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

\[ \det(C) = ad - bc \]

\[ = (0)(0) - (4)(-3) \]

\[ = 0 + 12 \]

Answer: 12

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Example 4: Calculate the determinant of

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

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

\[ \det(D) = ad - bc \]

\[ = (1)(4) - (2)(2) \]

\[ = 4 - 4 \]

Answer: 0

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Example 5: Determine the determinant of

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

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

\[ \det(E) = ad - bc \]

\[ = (-1)(-2) - (3)(4) \]

\[ = 2 - 12 \]

Answer: -10

3.1.2 Conditions for invertibility of a 2x2 matrix

A 2x2 matrix is invertible if and only if its determinant is non-zero. In land surveying, this condition guarantees that systems of linear equations used for coordinate calculations have unique solutions. If the determinant is zero, the matrix is singular and the system either has no solutions or infinitely many, which complicates survey computations.

The inverse of a 2x2 matrix \( A = \begin{bmatrix} a & b \\ c & d \end{bmatrix} \) exists only when

\[ \det(A) = ad - bc \neq 0 \]

Worked Examples

Example 1: Check if the matrix

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

is invertible.

Given: \(a=3\), \(b=1\), \(c=2\), \(d=5\)

\[ \det(A) = ad - bc \]

\[ = (3)(5) - (1)(2) \]

\[ = 15 - 2 = 13 \]

Since 13 ≠ 0, matrix \(A\) is invertible.

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Example 2: Determine if matrix

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

is invertible.

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

\[ \det(B) = ad - bc \]

\[ = (4)(3) - (2)(6) \]

\[ = 12 - 12 = 0 \]

Since determinant is 0, matrix \(B\) is not invertible.

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Example 3: Verify invertibility of matrix

\[ C = \begin{bmatrix} -2 & 7 \\ 5 & 1 \end{bmatrix} \]

Given: \(a=-2\), \(b=7\), \(c=5\), \(d=1\)

\[ \det(C) = ad - bc \]

\[ = (-2)(1) - (7)(5) \]

\[ = -2 - 35 = -37 \]

Since -37 ≠ 0, matrix \(C\) is invertible.

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Example 4: Is the matrix

\[ D = \begin{bmatrix} 0 & 0 \\ 1 & 4 \end{bmatrix} \]

invertible?

Given: \(a=0\), \(b=0\), \(c=1\), \(d=4\)

\[ \det(D) = ad - bc \]

\[ = (0)(4) - (0)(1) \]

\[ = 0 - 0 = 0 \]

Matrix \(D\) is not invertible.

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Example 5: Check invertibility of

\[ E = \begin{bmatrix} 1 & -1 \\ 1 & 1 \end{bmatrix} \]

Given: \(a=1\), \(b=-1\), \(c=1\), \(d=1\)

\[ \det(E) = ad - bc \]

\[ = (1)(1) - (-1)(1) \]

\[ = 1 + 1 = 2 \]

Matrix \(E\) is invertible.

3.1.3 Inverse of a 2x2 matrix

The inverse of an invertible 2x2 matrix \( A = \begin{bmatrix} a & b \\ c & d \end{bmatrix} \) is given by

\[ A^{-1} = \frac{1}{ad - bc} \begin{bmatrix} d & -b \\ -c & a \end{bmatrix} \]

This formula is essential in land surveying for solving linear systems involving coordinate transformations or least squares adjustments. The inverse matrix, when multiplied by the constant vector, yields the solution vector.

Worked Examples

Example 1: Find the inverse of

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

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

Calculate determinant:

\[ \det(A) = (2)(4) - (3)(1) = 8 - 3 = 5 \]

Calculate inverse:

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

\[ = \begin{bmatrix} \frac{4}{5} & -\frac{3}{5} \\ -\frac{1}{5} & \frac{2}{5} \end{bmatrix} \]

Answer:

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

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Example 2: Compute the inverse of

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

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

Calculate determinant:

\[ \det(B) = (5)(3) - (-2)(7) = 15 + 14 = 29 \]

Calculate inverse:

\[ B^{-1} = \frac{1}{29} \begin{bmatrix} 3 & 2 \\ -7 & 5 \end{bmatrix} \]

\[ = \begin{bmatrix} \frac{3}{29} & \frac{2}{29} \\ -\frac{7}{29} & \frac{5}{29} \end{bmatrix} \]

Answer:

\[ B^{-1} = \begin{bmatrix} 0.1034 & 0.0690 \\ -0.2414 & 0.1724 \end{bmatrix} \]

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Example 3: Find the inverse of

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

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

Calculate determinant:

\[ \det(C) = (1)(4) - (2)(3) = 4 - 6 = -2 \]

Calculate inverse:

\[ C^{-1} = \frac{1}{-2} \begin{bmatrix} 4 & -2 \\ -3 & 1 \end{bmatrix} \]

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

Answer:

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

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Example 4: Calculate the inverse of

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

Given: \(a=4\), \(b=7\), \(c=2\), \(d=6\)

Calculate determinant:

\[ \det(D) = (4)(6) - (7)(2) = 24 - 14 = 10 \]

Calculate inverse:

\[ D^{-1} = \frac{1}{10} \begin{bmatrix} 6 & -7 \\ -2 & 4 \end{bmatrix} \]

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

Answer:

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

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Example 5: Find the inverse of

\[ E = \begin{bmatrix} 3 & 0 \\ 0 & 2 \end{bmatrix} \]

Given: \(a=3\), \(b=0\), \(c=0\), \(d=2\)

Calculate determinant:

\[ \det(E) = (3)(2) - (0)(0) = 6 \]

Calculate inverse:

\[ E^{-1} = \frac{1}{6} \begin{bmatrix} 2 & 0 \\ 0 & 3 \end{bmatrix} \]

\[ = \begin{bmatrix} \frac{2}{6} & 0 \\ 0 & \frac{3}{6} \end{bmatrix} \]

\[ = \begin{bmatrix} 0.3333 & 0 \\ 0 & 0.5 \end{bmatrix} \]

Answer:

\[ E^{-1} = \begin{bmatrix} 0.3333 & 0 \\ 0 & 0.5 \end{bmatrix} \]

3.1.4 Application of inverse matrix in solving linear systems

In land surveying, solving simultaneous linear equations is common when adjusting coordinates or transforming between coordinate systems. Given a system

\[ AX = B \]

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

\[ X = A^{-1} B \]

provided \(A\) is invertible. This method provides an efficient way to compute unknowns such as coordinates or correction factors.

Worked Examples

Example 1: Solve the system

\[ \begin{cases} 2x + 3y = 8 \\ x + 4y = 7 \end{cases} \]

Given:

\[ A = \begin{bmatrix} 2 & 3 \\ 1 & 4 \end{bmatrix}, \quad B = \begin{bmatrix} 8 \\ 7 \end{bmatrix} \]

Calculate determinant:

\[ \det(A) = (2)(4) - (3)(1) = 8 - 3 = 5 \]

Calculate inverse:

\[ A^{-1} = \frac{1}{5} \begin{bmatrix} 4 & -3 \\ -1 & 2 \end{bmatrix} = \begin{bmatrix} 0.8 & -0.6 \\ -0.2 & 0.4 \end{bmatrix} \]

Calculate \(X\):

\[ X = A^{-1} B = \begin{bmatrix} 0.8 & -0.6 \\ -0.2 & 0.4 \end{bmatrix} \begin{bmatrix} 8 \\ 7 \end{bmatrix} \]

Calculate each element:

\[ x = (0.8)(8) + (-0.6)(7) = 6.4 - 4.2 = 2.2 \]

\[ y = (-0.2)(8) + (0.4)(7) = -1.6 + 2.8 = 1.2 \]

Answer:

\[ x = 2.2, \quad y = 1.2 \]

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Example 2: Solve

\[ \begin{cases} 5x - 2y = 1 \\ 7x + 3y = 13 \end{cases} \]

Given:

\[ A = \begin{bmatrix} 5 & -2 \\ 7 & 3 \end{bmatrix}, \quad B = \begin{bmatrix} 1 \\ 13 \end{bmatrix} \]

Calculate determinant:

\[ \det(A) = (5)(3) - (-2)(7) = 15 + 14 = 29 \]

Calculate inverse:

\[ A^{-1} = \frac{1}{29} \begin{bmatrix} 3 & 2 \\ -7 & 5 \end{bmatrix} = \begin{bmatrix} 0.1034 & 0.0690 \\ -0.2414 & 0.1724 \end{bmatrix} \]

Calculate \(X\):

\[ X = A^{-1} B = \begin{bmatrix} 0.1034 & 0.0690 \\ -0.2414 & 0.1724 \end{bmatrix} \begin{bmatrix} 1 \\ 13 \end{bmatrix} \]

Calculate each element:

\[ x = (0.1034)(1) + (0.0690)(13) = 0.1034 + 0.897 = 1.0004 \approx 1.0 \]

\[ y = (-0.2414)(1) + (0.1724)(13) = -0.2414 + 2.2412 = 1.9998 \approx 2.0 \]

Answer:

\[ x = 1.0, \quad y = 2.0 \]

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Example 3: Solve the system

\[ \begin{cases} 1x + 2y = 5 \\ 3x + 4y = 11 \end{cases} \]

Given:

\[ A = \begin{bmatrix} 1 & 2 \\ 3 & 4 \end{bmatrix}, \quad B = \begin{bmatrix} 5 \\ 11 \end{bmatrix} \]

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 \(X\):

\[ X = A^{-1} B = \begin{bmatrix} -2 & 1 \\ 1.5 & -0.5 \end{bmatrix} \begin{bmatrix} 5 \\ 11 \end{bmatrix} \]

Calculate each element:

\[ x = (-2)(5) + (1)(11) = -10 + 11 = 1 \]

\[ y = (1.5)(5) + (-0.5)(11) = 7.5 - 5.5 = 2 \]

Answer:

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

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Example 4: Solve

\[ \begin{cases} 4x + 7y = 31 \\ 2x + 6y = 26 \end{cases} \]

Given:

\[ A = \begin{bmatrix} 4 & 7 \\ 2 & 6 \end{bmatrix}, \quad B = \begin{bmatrix} 31 \\ 26 \end{bmatrix} \]

Calculate determinant:

\[ \det(A) = (4)(6) - (7)(2) = 24 - 14 = 10 \]

Calculate inverse:

\[ A^{-1} = \frac{1}{10} \begin{bmatrix} 6 & -7 \\ -2 & 4 \end{bmatrix} = \begin{bmatrix} 0.6 & -0.7 \\ -0.2 & 0.4 \end{bmatrix} \]

Calculate \(X\):

\[ X = A^{-1} B = \begin{bmatrix} 0.6 & -0.7 \\ -0.2 & 0.4 \end{bmatrix} \begin{bmatrix} 31 \\ 26 \end{bmatrix} \]

Calculate each element:

\[ x = (0.6)(31) + (-0.7)(26) = 18.6 - 18.2 = 0.4 \]

\[ y = (-0.2)(31) + (0.4)(26) = -6.2 + 10.4 = 4.2 \]

Answer:

\[ x = 0.4, \quad y = 4.2 \]

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Example 5: Solve

\[ \begin{cases} 3x + 0y = 9 \\ 0x + 2y = 8 \end{cases} \]

Given:

\[ A = \begin{bmatrix} 3 & 0 \\ 0 & 2 \end{bmatrix}, \quad B = \begin{bmatrix} 9 \\ 8 \end{bmatrix} \]

Calculate determinant:

\[ \det(A) = (3)(2) - (0)(0) = 6 \]

Calculate inverse:

\[ A^{-1} = \frac{1}{6} \begin{bmatrix} 2 & 0 \\ 0 & 3 \end{bmatrix} = \begin{bmatrix} 0.3333 & 0 \\ 0 & 0.5 \end{bmatrix} \]

Calculate \(X\):

\[ X = A^{-1} B = \begin{bmatrix} 0.3333 & 0 \\ 0 & 0.5 \end{bmatrix} \begin{bmatrix} 9 \\ 8 \end{bmatrix} \]

Calculate each element:

\[ x = (0.3333)(9) + (0)(8) = 3 + 0 = 3 \]

\[ y = (0)(9) + (0.5)(8) = 0 + 4 = 4 \]

Answer:

\[ x = 3, \quad y = 4 \]

Chapter Summary

This chapter focused on the application of matrices in solving engineering problems, starting with the calculation of the determinant and inverse of a 2x2 matrix. Understanding how to find the determinant is crucial as it determines whether the matrix is invertible. The chapter then extended these concepts to 3x3 matrices, where the process of calculating the determinant and inverse becomes more involved but follows similar principles. With these tools, the chapter demonstrated how to apply matrix methods to find solutions to simultaneous linear equations, a common requirement in engineering contexts. Techniques such as using the inverse matrix to solve systems of equations were emphasized, providing a systematic approach to problems that would otherwise be difficult to handle. The progression from simple 2x2 matrices to more complex 3x3 matrices illustrated the increasing computational effort while reinforcing the underlying concepts. Overall, the chapter equipped students with essential matrix operations and their practical use in engineering mathematics.