Matrix calculations are essential tools in civil engineering for solving systems of linear equations, transforming coordinate systems, and modeling structural behavior. Determinants and inverses of matrices allow engineers to analyze and manipulate these systems efficiently. This chapter focuses on calculating determinants and inverses for matrices commonly used in civil engineering contexts in Kenya, providing thorough worked examples for practical understanding.

2.1 Determinants and Inverses

Determinants provide a scalar value associated with a square matrix that helps determine whether the matrix is invertible. In civil engineering, this is crucial when solving simultaneous equations for forces in structures or when performing coordinate transformations. The inverse of a matrix, when it exists, is used to solve matrix equations such as \(AX = B\), where \(A\) and \(B\) are known matrices, and \(X\) is the unknown matrix to be found. Understanding how to compute determinants and inverses is fundamental for structural analysis, geotechnical modeling, and other engineering computations.

2.1.1 Determinant Definition and Calculation for Square Matrices

The determinant is a scalar value calculated from a square matrix that reflects properties such as matrix invertibility and the scaling factor of linear transformations. For a matrix \(A\), the determinant is denoted \(\det(A)\) or \(|A|\), and it is defined only for square matrices.

For a \(2 \times 2\) matrix
\[ A = \begin{bmatrix} a & b \\ c & d \end{bmatrix}, \]
the determinant is
\[ \det(A) = ad - bc. \]

For higher-order matrices, the determinant is calculated using expansion by minors or cofactors.

$$\text{Determinant of } A = \sum_{j=1}^n (-1)^{1+j} a_{1j} \det(M_{1j})$$

where \(M_{1j}\) is the \((n-1) \times (n-1)\) submatrix formed by deleting the first row and \(j^{th}\) column.

Worked Examples

Example 1: Calculate the determinant of a \(2 \times 2\) matrix representing a transformation of coordinates on a small construction site.

Given:
\[ A = \begin{bmatrix} 4 & 2 \\ 3 & 5 \end{bmatrix} \]

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

$$= 20 - 6$$

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

Example 2: Calculate the determinant of a \(3 \times 3\) matrix representing stiffness coefficients in a structural analysis problem.

Given:
\[ B = \begin{bmatrix} 2 & 1 & 3 \\ 0 & -1 & 4 \\ 5 & 2 & 0 \end{bmatrix} \]

$$\det(B) = 2 \times \det\begin{bmatrix} -1 & 4 \\ 2 & 0 \end{bmatrix} - 1 \times \det\begin{bmatrix} 0 & 4 \\ 5 & 0 \end{bmatrix} + 3 \times \det\begin{bmatrix} 0 & -1 \\ 5 & 2 \end{bmatrix}$$

Calculate minors:

$$\det\begin{bmatrix} -1 & 4 \\ 2 & 0 \end{bmatrix} = (-1)(0) - (4)(2) = 0 - 8 = -8$$

$$\det\begin{bmatrix} 0 & 4 \\ 5 & 0 \end{bmatrix} = (0)(0) - (4)(5) = 0 - 20 = -20$$

$$\det\begin{bmatrix} 0 & -1 \\ 5 & 2 \end{bmatrix} = (0)(2) - (-1)(5) = 0 + 5 = 5$$

Substitute:

$$= 2 \times (-8) - 1 \times (-20) + 3 \times 5$$

$$= -16 + 20 + 15$$

$$= 19$$

Answer: \(\det(B) = 19\)

Example 3: Calculate the determinant of a \(3 \times 3\) matrix representing load distribution coefficients on a bridge deck.

Given:
\[ C = \begin{bmatrix} 1 & 2 & 3 \\ 4 & 5 & 6 \\ 7 & 8 & 9 \end{bmatrix} \]

$$\det(C) = 1 \times \det\begin{bmatrix} 5 & 6 \\ 8 & 9 \end{bmatrix} - 2 \times \det\begin{bmatrix} 4 & 6 \\ 7 & 9 \end{bmatrix} + 3 \times \det\begin{bmatrix} 4 & 5 \\ 7 & 8 \end{bmatrix}$$

Calculate minors:

$$\det\begin{bmatrix} 5 & 6 \\ 8 & 9 \end{bmatrix} = (5)(9) - (6)(8) = 45 - 48 = -3$$

$$\det\begin{bmatrix} 4 & 6 \\ 7 & 9 \end{bmatrix} = (4)(9) - (6)(7) = 36 - 42 = -6$$

$$\det\begin{bmatrix} 4 & 5 \\ 7 & 8 \end{bmatrix} = (4)(8) - (5)(7) = 32 - 35 = -3$$

Substitute:

$$= 1 \times (-3) - 2 \times (-6) + 3 \times (-3)$$

$$= -3 + 12 - 9$$

$$= 0$$

Answer: \(\det(C) = 0\)

2.1.2 Matrix Inverse Definition and Calculation

The inverse of a square matrix \(A\), denoted \(A^{-1}\), is a matrix such that

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

where \(I\) is the identity matrix of the same order. The inverse exists only if \(\det(A) \neq 0\). In civil engineering, inverses are used to solve matrix equations, such as systems of forces or displacements.

For a \(2 \times 2\) matrix

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

if \(\det(A) \neq 0\), then

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

For larger matrices, the inverse is calculated using cofactors and adjoint matrices or by row reduction techniques.

Worked Examples

Example 1: Find the inverse of a \(2 \times 2\) matrix representing a transformation matrix for site coordinates.

Given:
\[ A = \begin{bmatrix} 3 & 4 \\ 2 & 5 \end{bmatrix} \]

Calculate determinant:

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

Calculate inverse:

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

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

Answer:

\[ A^{-1} = \begin{bmatrix} 0.714 & -0.571 \\ -0.286 & 0.429 \end{bmatrix} \]

Example 2: Find the inverse of a \(3 \times 3\) matrix representing coefficients of a load distribution system.

Given:
\[ B = \begin{bmatrix} 1 & 2 & 3 \\ 0 & 1 & 4 \\ 5 & 6 & 0 \end{bmatrix} \]

Calculate determinant (using expansion):

$$\det(B) = 1 \times \det\begin{bmatrix} 1 & 4 \\ 6 & 0 \end{bmatrix} - 2 \times \det\begin{bmatrix} 0 & 4 \\ 5 & 0 \end{bmatrix} + 3 \times \det\begin{bmatrix} 0 & 1 \\ 5 & 6 \end{bmatrix}$$

Calculate minors:

$$\det\begin{bmatrix} 1 & 4 \\ 6 & 0 \end{bmatrix} = (1)(0) - (4)(6) = -24$$

$$\det\begin{bmatrix} 0 & 4 \\ 5 & 0 \end{bmatrix} = (0)(0) - (4)(5) = -20$$

$$\det\begin{bmatrix} 0 & 1 \\ 5 & 6 \end{bmatrix} = (0)(6) - (1)(5) = -5$$

Substitute:

$$= 1 \times (-24) - 2 \times (-20) + 3 \times (-5)$$

$$= -24 + 40 - 15 = 1$$

Since determinant = 1, inverse exists.

Calculate cofactors matrix \(C\):

\[ C = \begin{bmatrix} \det\begin{bmatrix} 1 & 4 \\ 6 & 0 \end{bmatrix} & -\det\begin{bmatrix} 0 & 4 \\ 5 & 0 \end{bmatrix} & \det\begin{bmatrix} 0 & 1 \\ 5 & 6 \end{bmatrix} \\ -\det\begin{bmatrix} 2 & 3 \\ 6 & 0 \end{bmatrix} & \det\begin{bmatrix} 1 & 3 \\ 5 & 0 \end{bmatrix} & -\det\begin{bmatrix} 1 & 2 \\ 5 & 6 \end{bmatrix} \\ \det\begin{bmatrix} 2 & 3 \\ 1 & 4 \end{bmatrix} & -\det\begin{bmatrix} 1 & 3 \\ 0 & 4 \end{bmatrix} & \det\begin{bmatrix} 1 & 2 \\ 0 & 1 \end{bmatrix} \end{bmatrix} \]

Calculate each minor:

• \(C_{11} = -24\)
• \(C_{12} = -(-20) = 20\)
• \(C_{13} = -5\)
• \(C_{21} = -\det\begin{bmatrix} 2 & 3 \\ 6 & 0 \end{bmatrix} = -((2)(0)-(3)(6)) = - (0 - 18) = 18\)
• \(C_{22} = \det\begin{bmatrix} 1 & 3 \\ 5 & 0 \end{bmatrix} = (1)(0) - (3)(5) = -15\)
• \(C_{23} = -\det\begin{bmatrix} 1 & 2 \\ 5 & 6 \end{bmatrix} = -((1)(6) - (2)(5)) = - (6 - 10) = 4\)
• \(C_{31} = \det\begin{bmatrix} 2 & 3 \\ 1 & 4 \end{bmatrix} = (2)(4) - (3)(1) = 8 - 3 = 5\)
• \(C_{32} = -\det\begin{bmatrix} 1 & 3 \\ 0 & 4 \end{bmatrix} = -((1)(4) - (3)(0)) = - (4 - 0) = -4\)
• \(C_{33} = \det\begin{bmatrix} 1 & 2 \\ 0 & 1 \end{bmatrix} = (1)(1) - (2)(0) = 1\)

Cofactor matrix:

\[ C = \begin{bmatrix} -24 & 20 & -5 \\ 18 & -15 & 4 \\ 5 & -4 & 1 \end{bmatrix} \]

Transpose \(C\) to get adjoint \(adj(B)\):

\[ adj(B) = \begin{bmatrix} -24 & 18 & 5 \\ 20 & -15 & -4 \\ -5 & 4 & 1 \end{bmatrix} \]

Calculate inverse:

$$B^{-1} = \frac{1}{\det(B)} adj(B) = 1 \times adj(B) = adj(B)$$

Answer:

\[ B^{-1} = \begin{bmatrix} -24 & 18 & 5 \\ 20 & -15 & -4 \\ -5 & 4 & 1 \end{bmatrix} \]

Example 3: Determine if the matrix below is invertible and find its inverse if possible:

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

Calculate determinant:

$$\det(C) = (2)(2) - (4)(1) = 4 - 4 = 0$$

Since determinant is zero, inverse does not exist.

Answer: The matrix \(C\) is singular and has no inverse.

2.1.3 Using Inverses to Solve Matrix Equations

Solving systems of linear equations is a common task in civil engineering, such as determining forces in a truss or displacements in a frame. Given a matrix equation

$$A X = B,$$

where \(A\) is a known square matrix, \(B\) is a known vector or matrix, and \(X\) is unknown, if \(A^{-1}\) exists, the solution is

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

Worked Examples

Example 1: Solve the system of equations representing forces in a beam:

\[ \begin{cases} 3x + 4y = 10 \\ 2x + 5y = 13 \end{cases} \]

Matrix form:

\[ A = \begin{bmatrix} 3 & 4 \\ 2 & 5 \end{bmatrix}, \quad B = \begin{bmatrix} 10 \\ 13 \end{bmatrix} \]

Calculate determinant:

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

Calculate inverse:

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

Calculate \(X\):

$$X = A^{-1} B = \frac{1}{7} \begin{bmatrix} 5 & -4 \\ -2 & 3 \end{bmatrix} \begin{bmatrix} 10 \\ 13 \end{bmatrix}$$

Compute multiplication:

$$\begin{bmatrix} 5 \times 10 + (-4) \times 13 \\ -2 \times 10 + 3 \times 13 \end{bmatrix} = \begin{bmatrix} 50 - 52 \\ -20 + 39 \end{bmatrix} = \begin{bmatrix} -2 \\ 19 \end{bmatrix}$$

Multiply by \(1/7\):

$$X = \begin{bmatrix} -\frac{2}{7} \\ \frac{19}{7} \end{bmatrix} = \begin{bmatrix} -0.286 \\ 2.714 \end{bmatrix}$$

Answer: \(x = -0.286\), \(y = 2.714\)

Example 2: Solve for nodal displacements in a truss represented by

\[ A = \begin{bmatrix} 4 & 1 & 0 \\ 1 & 3 & 1 \\ 0 & 1 & 2 \end{bmatrix}, \quad B = \begin{bmatrix} 7 \\ 8 \\ 5 \end{bmatrix} \]

Calculate determinant (from previous methods or calculator):

\[ \det(A) = 4 \times (3 \times 2 - 1 \times 1) - 1 \times (1 \times 2 - 1 \times 0) + 0 = 4 \times (6 - 1) - 1 \times (2 - 0) + 0 = 4 \times 5 - 2 = 20 - 2 = 18 \]

Calculate inverse (using cofactors and adjoint):

Cofactors matrix \(C\):

\[ C = \begin{bmatrix} \det\begin{bmatrix} 3 & 1 \\ 1 & 2 \end{bmatrix} & -\det\begin{bmatrix} 1 & 1 \\ 0 & 2 \end{bmatrix} & \det\begin{bmatrix} 1 & 3 \\ 0 & 1 \end{bmatrix} \\ -\det\begin{bmatrix} 1 & 0 \\ 1 & 2 \end{bmatrix} & \det\begin{bmatrix} 4 & 0 \\ 0 & 2 \end{bmatrix} & -\det\begin{bmatrix} 4 & 1 \\ 0 & 1 \end{bmatrix} \\ \det\begin{bmatrix} 1 & 0 \\ 3 & 1 \end{bmatrix} & -\det\begin{bmatrix} 4 & 1 \\ 1 & 3 \end{bmatrix} & \det\begin{bmatrix} 4 & 1 \\ 1 & 3 \end{bmatrix} \end{bmatrix} \]

Calculate each minor:

• \(C_{11} = (3)(2) - (1)(1) = 6 - 1 = 5\)
• \(C_{12} = -((1)(2) - (1)(0)) = -(2 - 0) = -2\)
• \(C_{13} = (1)(1) - (3)(0) = 1 - 0 = 1\)
• \(C_{21} = -((1)(2) - (0)(1)) = - (2 - 0) = -2\)
• \(C_{22} = (4)(2) - (0)(0) = 8 - 0 = 8\)
• \(C_{23} = -((4)(1) - (0)(1)) = - (4 - 0) = -4\)
• \(C_{31} = (1)(1) - (0)(3) = 1 - 0 = 1\)
• \(C_{32} = -((4)(3) - (1)(1)) = - (12 - 1) = -11\)
• \(C_{33} = (4)(3) - (1)(1) = 12 - 1 = 11\)

Cofactor matrix:

\[ C = \begin{bmatrix} 5 & -2 & 1 \\ -2 & 8 & -4 \\ 1 & -11 & 11 \end{bmatrix} \]

Transpose to get adjoint:

\[ adj(A) = \begin{bmatrix} 5 & -2 & 1 \\ -2 & 8 & -11 \\ 1 & -4 & 11 \end{bmatrix} \]

Calculate inverse:

$$A^{-1} = \frac{1}{18} adj(A) = \frac{1}{18} \begin{bmatrix} 5 & -2 & 1 \\ -2 & 8 & -11 \\ 1 & -4 & 11 \end{bmatrix}$$

Calculate \(X = A^{-1} B\):

\[ X = \frac{1}{18} \begin{bmatrix} 5 & -2 & 1 \\ -2 & 8 & -11 \\ 1 & -4 & 11 \end{bmatrix} \begin{bmatrix} 7 \\ 8 \\ 5 \end{bmatrix} \]

Multiply:

• First row: \(5 \times 7 + (-2) \times 8 + 1 \times 5 = 35 - 16 + 5 = 24\)
• Second row: \(-2 \times 7 + 8 \times 8 + (-11) \times 5 = -14 + 64 - 55 = -5\)
• Third row: \(1 \times 7 + (-4) \times 8 + 11 \times 5 = 7 - 32 + 55 = 30\)

Multiply by \(1/18\):

\[ X = \begin{bmatrix} \frac{24}{18} \\ \frac{-5}{18} \\ \frac{30}{18} \end{bmatrix} = \begin{bmatrix} 1.333 \\ -0.278 \\ 1.667 \end{bmatrix} \]

Answer:

\[ x = 1.333, \quad y = -0.278, \quad z = 1.667 \]

Example 3: Solve the system

\[ \begin{cases} x + y + z = 6 \\ 2x + 3y + 5z = 14 \\ 4x + 6y + 8z = 24 \end{cases} \]

Matrix form:

\[ A = \begin{bmatrix} 1 & 1 & 1 \\ 2 & 3 & 5 \\ 4 & 6 & 8 \end{bmatrix}, \quad B = \begin{bmatrix} 6 \\ 14 \\ 24 \end{bmatrix} \]

Calculate determinant:

\[ \det(A) = 1 \times (3 \times 8 - 5 \times 6) - 1 \times (2 \times 8 - 5 \times 4) + 1 \times (2 \times 6 - 3 \times 4) \]

Calculate minors:

• \(3 \times 8 - 5 \times 6 = 24 - 30 = -6\)
• \(2 \times 8 - 5 \times 4 = 16 - 20 = -4\)
• \(2 \times 6 - 3 \times 4 = 12 - 12 = 0\)

Substitute:

\[ = 1 \times (-6) - 1 \times (-4) + 1 \times 0 = -6 + 4 + 0 = -2 \]

Since \(\det(A) \neq 0\), inverse exists.

Calculate cofactors matrix \(C\):

\[ C = \begin{bmatrix} \det\begin{bmatrix} 3 & 5 \\ 6 & 8 \end{bmatrix} & -\det\begin{bmatrix} 2 & 5 \\ 4 & 8 \end{bmatrix} & \det\begin{bmatrix} 2 & 3 \\ 4 & 6 \end{bmatrix} \\ -\det\begin{bmatrix} 1 & 1 \\ 6 & 8 \end{bmatrix} & \det\begin{bmatrix} 1 & 1 \\ 4 & 8 \end{bmatrix} & -\det\begin{bmatrix} 1 & 1 \\ 4 & 6 \end{bmatrix} \\ \det\begin{bmatrix} 1 & 1 \\ 3 & 5 \end{bmatrix} & -\det\begin{bmatrix} 1 & 1 \\ 2 & 5 \end{bmatrix} & \det\begin{bmatrix} 1 & 1 \\ 2 & 3 \end{bmatrix} \end{bmatrix} \]

Calculate each minor:

• \(C_{11} = (3)(8) - (5)(6) = 24 - 30 = -6\)
• \(C_{12} = -((2)(8) - (5)(4)) = - (16 - 20) = 4\)
• \(C_{13} = (2)(6) - (3)(4) = 12 - 12 = 0\)
• \(C_{21} = -((1)(8) - (1)(6)) = - (8 - 6) = -2\)
• \(C_{22} = (1)(8) - (1)(4) = 8 - 4 = 4\)
• \(C_{23} = -((1)(6) - (1)(4)) = - (6 - 4) = -2\)
• \(C_{31} = (1)(5) - (1)(3) = 5 - 3 = 2\)
• \(C_{32} = -((1)(5) - (1)(2)) = - (5 - 2) = -3\)
• \(C_{33} = (1)(3) - (1)(2) = 3 - 2 = 1\)

Cofactor matrix:

\[ C = \begin{bmatrix} -6 & 4 & 0 \\ -2 & 4 & -2 \\ 2 & -3 & 1 \end{bmatrix} \]

Transpose to get adjoint:

\[ adj(A) = \begin{bmatrix} -6 & -2 & 2 \\ 4 & 4 & -3 \\ 0 & -2 & 1 \end{bmatrix} \]

Calculate inverse:

$$A^{-1} = \frac{1}{-2} adj(A) = -\frac{1}{2} \begin{bmatrix} -6 & -2 & 2 \\ 4 & 4 & -3 \\ 0 & -2 & 1 \end{bmatrix} = \begin{bmatrix} 3 & 1 & -1 \\ -2 & -2 & \frac{3}{2} \\ 0 & 1 & -\frac{1}{2} \end{bmatrix}$$

Calculate \(X = A^{-1} B\):

\[ X = \begin{bmatrix} 3 & 1 & -1 \\ -2 & -2 & \frac{3}{2} \\ 0 & 1 & -\frac{1}{2} \end{bmatrix} \begin{bmatrix} 6 \\ 14 \\ 24 \end{bmatrix} \]

Multiply:

• First row: \(3 \times 6 + 1 \times 14 + (-1) \times 24 = 18 + 14 - 24 = 8\)
• Second row: \(-2 \times 6 + (-2) \times 14 + \frac{3}{2} \times 24 = -12 - 28 + 36 = -4\)
• Third row: \(0 \times 6 + 1 \times 14 + (-\frac{1}{2}) \times 24 = 0 + 14 - 12 = 2\)

Answer:

\[ x = 8, \quad y = -4, \quad z = 2 \]

2.1.4 Conditions for Invertibility and Properties of Inverses

A square matrix \(A\) is invertible only if its determinant is nonzero. This condition is vital in civil engineering applications where matrix inversion solves equilibrium equations or transforms data.

Key properties of inverses include:

• \((A^{-1})^{-1} = A\)
• \((A B)^{-1} = B^{-1} A^{-1}\)
• \( (A^T)^{-1} = (A^{-1})^T \)
• \(A A^{-1} = A^{-1} A = I\)

Worked Examples

Example 1: Verify invertibility of the matrix

\[ A = \begin{bmatrix} 1 & 2 \\ 3 & 6 \end{bmatrix} \]

Calculate determinant:

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

Matrix is singular, no inverse.

Example 2: For matrices

\[ A = \begin{bmatrix} 1 & 0 \\ 0 & 1 \end{bmatrix}, \quad B = \begin{bmatrix} 2 & 3 \\ 1 & 4 \end{bmatrix} \]

Calculate \((AB)^{-1}\) and verify \((AB)^{-1} = B^{-1} A^{-1}\).

Calculate \(AB\):

\[ AB = \begin{bmatrix} 1 & 0 \\ 0 & 1 \end{bmatrix} \begin{bmatrix} 2 & 3 \\ 1 & 4 \end{bmatrix} = \begin{bmatrix} 2 & 3 \\ 1 & 4 \end{bmatrix} = B \]

Calculate \(B^{-1}\):

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

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

Calculate \(A^{-1}\):

\[ A^{-1} = A = \begin{bmatrix} 1 & 0 \\ 0 & 1 \end{bmatrix} \]

Calculate \(B^{-1} A^{-1} = B^{-1} \times A = B^{-1}\)

Calculate \((AB)^{-1} = B^{-1}\)

Property verified.

Example 3: Show that \((A^T)^{-1} = (A^{-1})^T\) for

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

Calculate \(A^{-1}\):

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

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

Calculate transpose of \(A\):

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

Calculate \((A^T)^{-1}\):

\[ \det(A^T) = \det(A) = -2 \]

\[ (A^T)^{-1} = \frac{1}{-2} \begin{bmatrix} 4 & -3 \\ -2 & 1 \end{bmatrix} = \begin{bmatrix} -2 & 1.5 \\ 1 & -0.5 \end{bmatrix} \]

Calculate \((A^{-1})^T\):

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

They are equal.

Chapter Summary

This chapter explored key matrix operations starting with determinants and inverses, focusing on how to calculate determinants for 2x2 and 3x3 matrices. The process of finding the inverse of a matrix was detailed, highlighting its importance in solving matrix equations. Methods to solve simultaneous linear equations using matrix techniques were covered, particularly Cramer's rule for efficient computation. The concept of eigenvalues and eigenvectors was introduced, explaining their role in characterizing matrix properties. Finally, the chapter examined matrix transformations and their practical applications in two-dimensional and three-dimensional geometric transformations, demonstrating how matrices can be used to manipulate shapes and coordinate systems effectively.