Industrial Control and Installation professionals regularly encounter matrices when designing, analyzing, and troubleshooting control systems for Kenyan industries. Matrices are essential for handling multiple variables, such as electrical currents, voltages, and signals, and for solving systems of equations in automation and process control. This chapter equips you with practical skills to perform matrix operations, vital for tasks like calibrating sensors, managing PLC inputs and outputs, and optimizing industrial processes.

3.1 Matrix Operation

Matrix operations provide a structured way to handle complex calculations involving multiple variables, which are common in industrial control and installation. Understanding addition, subtraction, multiplication, and scalar multiplication of matrices is critical for professionals working with automation systems, electrical circuits, and networked control devices in Kenyan manufacturing, processing, and utility sectors.

3.1.1 Matrix Addition and Subtraction

Matrix addition and subtraction are performed element-wise, meaning corresponding elements are added or subtracted. Both matrices must be of the same size.

$$ \text{If } A = [a_{ij}],\ B = [b_{ij}],\ \text{then:} $$$$ A + B = [a_{ij} + b_{ij}] $$$$ A - B = [a_{ij} - b_{ij}] $$

Worked Examples

Example 1 (Easy):
A PLC system records sensor readings as matrices:
\(A = \begin{bmatrix} 2 & 5 \\ 4 & 7 \end{bmatrix}\) and \(B = \begin{bmatrix} 1 & 3 \\ 6 & 2 \end{bmatrix}\).
Find \(A + B\).

Given:
\(A = \begin{bmatrix} 2 & 5 \\ 4 & 7 \end{bmatrix}\), \(B = \begin{bmatrix} 1 & 3 \\ 6 & 2 \end{bmatrix}\)

$$ A + B = \begin{bmatrix} 2+1 & 5+3 \\ 4+6 & 7+2 \end{bmatrix} $$$$ = \begin{bmatrix} 3 & 8 \\ 10 & 9 \end{bmatrix} $$ Answer: \(\begin{bmatrix} 3 & 8 \\ 10 & 9 \end{bmatrix}\)

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Example 2 (Medium):
Subtract matrix \(B\) from matrix \(A\) for the same system.

Given:
\(A = \begin{bmatrix} 2 & 5 \\ 4 & 7 \end{bmatrix}\), \(B = \begin{bmatrix} 1 & 3 \\ 6 & 2 \end{bmatrix}\)

$$ A - B = \begin{bmatrix} 2-1 & 5-3 \\ 4-6 & 7-2 \end{bmatrix} $$$$ = \begin{bmatrix} 1 & 2 \\ -2 & 5 \end{bmatrix} $$ Answer: \(\begin{bmatrix} 1 & 2 \\ -2 & 5 \end{bmatrix}\)

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Example 3 (Hard):
A control panel has voltage readings stored as \(C = \begin{bmatrix} 9 & 4 & 7 \\ 3 & 8 & 5 \end{bmatrix}\) and \(D = \begin{bmatrix} 6 & 2 & 4 \\ 1 & 6 & 3 \end{bmatrix}\).
Find \(C + D\).

Given:
\(C = \begin{bmatrix} 9 & 4 & 7 \\ 3 & 8 & 5 \end{bmatrix}\), \(D = \begin{bmatrix} 6 & 2 & 4 \\ 1 & 6 & 3 \end{bmatrix}\)

$$ C + D = \begin{bmatrix} 9+6 & 4+2 & 7+4 \\ 3+1 & 8+6 & 5+3 \end{bmatrix} $$$$ = \begin{bmatrix} 15 & 6 & 11 \\ 4 & 14 & 8 \end{bmatrix} $$ Answer: \(\begin{bmatrix} 15 & 6 & 11 \\ 4 & 14 & 8 \end{bmatrix}\)

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Example 4 (Applied):
A water treatment plant records inflow and outflow rates as matrices.
Inflow: \(E = \begin{bmatrix} 12 & 7 \\ 15 & 9 \end{bmatrix}\), Outflow: \(F = \begin{bmatrix} 5 & 3 \\ 6 & 2 \end{bmatrix}\).
Find the net flow matrix \(E - F\).

Given:
\(E = \begin{bmatrix} 12 & 7 \\ 15 & 9 \end{bmatrix}\), \(F = \begin{bmatrix} 5 & 3 \\ 6 & 2 \end{bmatrix}\)

$$ E - F = \begin{bmatrix} 12-5 & 7-3 \\ 15-6 & 9-2 \end{bmatrix} $$$$ = \begin{bmatrix} 7 & 4 \\ 9 & 7 \end{bmatrix} $$ Answer: \(\begin{bmatrix} 7 & 4 \\ 9 & 7 \end{bmatrix}\)

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Example 5 (Challenge):
A manufacturing line records daily output in two shifts as \(G = \begin{bmatrix} 18 & 12 & 9 \\ 11 & 20 & 15 \end{bmatrix}\) and \(H = \begin{bmatrix} 8 & 6 & 4 \\ 5 & 10 & 7 \end{bmatrix}\).
Compute total output \(G + H\).

Given:
\(G = \begin{bmatrix} 18 & 12 & 9 \\ 11 & 20 & 15 \end{bmatrix}\), \(H = \begin{bmatrix} 8 & 6 & 4 \\ 5 & 10 & 7 \end{bmatrix}\)

$$ G + H = \begin{bmatrix} 18+8 & 12+6 & 9+4 \\ 11+5 & 20+10 & 15+7 \end{bmatrix} $$$$ = \begin{bmatrix} 26 & 18 & 13 \\ 16 & 30 & 22 \end{bmatrix} $$ Answer: \(\begin{bmatrix} 26 & 18 & 13 \\ 16 & 30 & 22 \end{bmatrix}\)

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3.1.2 Scalar Multiplication of Matrices

Scalar multiplication involves multiplying every element of a matrix by a constant (scalar). This operation is used to scale measurements, adjust signal strengths, or model proportional changes in industrial control systems.

$$ \text{If } k \text{ is a scalar and } A = [a_{ij}],\ \text{then:} $$$$ kA = [k \times a_{ij}] $$

Worked Examples

Example 1 (Easy):
Multiply matrix \(A = \begin{bmatrix} 2 & 3 \\ 4 & 5 \end{bmatrix}\) by scalar \(k = 3\).

Given:
\(A = \begin{bmatrix} 2 & 3 \\ 4 & 5 \end{bmatrix}\), \(k = 3\)

$$ 3A = \begin{bmatrix} 3 \times 2 & 3 \times 3 \\ 3 \times 4 & 3 \times 5 \end{bmatrix} $$$$ = \begin{bmatrix} 6 & 9 \\ 12 & 15 \end{bmatrix} $$ Answer: \(\begin{bmatrix} 6 & 9 \\ 12 & 15 \end{bmatrix}\)

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Example 2 (Medium):
For \(B = \begin{bmatrix} 7 & 1 \\ 3 & 6 \end{bmatrix}\), multiply by scalar \(k = 2\).

Given:
\(B = \begin{bmatrix} 7 & 1 \\ 3 & 6 \end{bmatrix}\), \(k = 2\)

$$ 2B = \begin{bmatrix} 2 \times 7 & 2 \times 1 \\ 2 \times 3 & 2 \times 6 \end{bmatrix} $$$$ = \begin{bmatrix} 14 & 2 \\ 6 & 12 \end{bmatrix} $$ Answer: \(\begin{bmatrix} 14 & 2 \\ 6 & 12 \end{bmatrix}\)

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Example 3 (Hard):
A control system adjusts readings in \(C = \begin{bmatrix} 5 & 8 & 2 \\ 4 & 7 & 9 \end{bmatrix}\) by a factor of \(k = 4\).

Given:
\(C = \begin{bmatrix} 5 & 8 & 2 \\ 4 & 7 & 9 \end{bmatrix}\), \(k = 4\)

$$ 4C = \begin{bmatrix} 4 \times 5 & 4 \times 8 & 4 \times 2 \\ 4 \times 4 & 4 \times 7 & 4 \times 9 \end{bmatrix} $$$$ = \begin{bmatrix} 20 & 32 & 8 \\ 16 & 28 & 36 \end{bmatrix} $$ Answer: \(\begin{bmatrix} 20 & 32 & 8 \\ 16 & 28 & 36 \end{bmatrix}\)

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Example 4 (Applied):
A water treatment process needs to double all sensor readings recorded in \(D = \begin{bmatrix} 12 & 9 \\ 7 & 14 \end{bmatrix}\).

Given:
\(D = \begin{bmatrix} 12 & 9 \\ 7 & 14 \end{bmatrix}\), \(k = 2\)

$$ 2D = \begin{bmatrix} 2 \times 12 & 2 \times 9 \\ 2 \times 7 & 2 \times 14 \end{bmatrix} $$$$ = \begin{bmatrix} 24 & 18 \\ 14 & 28 \end{bmatrix} $$ Answer: \(\begin{bmatrix} 24 & 18 \\ 14 & 28 \end{bmatrix}\)

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Example 5 (Challenge):
A manufacturing plant multiplies the output matrix \(E = \begin{bmatrix} 10 & 15 & 20 \\ 8 & 12 & 16 \end{bmatrix}\) by \(k = 5\).

Given:
\(E = \begin{bmatrix} 10 & 15 & 20 \\ 8 & 12 & 16 \end{bmatrix}\), \(k = 5\)

$$ 5E = \begin{bmatrix} 5 \times 10 & 5 \times 15 & 5 \times 20 \\ 5 \times 8 & 5 \times 12 & 5 \times 16 \end{bmatrix} $$$$ = \begin{bmatrix} 50 & 75 & 100 \\ 40 & 60 & 80 \end{bmatrix} $$ Answer: \(\begin{bmatrix} 50 & 75 & 100 \\ 40 & 60 & 80 \end{bmatrix}\)

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3.1.3 Matrix Multiplication

Matrix multiplication combines two matrices to produce a new matrix, used extensively in industrial process modeling, signal processing, and systems analysis. The number of columns in the first matrix must equal the number of rows in the second.

$$ \text{If } A \text{ is } m \times n,\ B \text{ is } n \times p,\ \text{then } AB \text{ is } m \times p $$$$ (AB)_{ij} = \sum_{k=1}^{n} a_{ik} b_{kj} $$

Worked Examples

Example 1 (Easy):
Multiply \(A = \begin{bmatrix} 2 & 3 \\ 1 & 4 \end{bmatrix}\) by \(B = \begin{bmatrix} 5 & 6 \\ 7 & 8 \end{bmatrix}\).

Given:
\(A = \begin{bmatrix} 2 & 3 \\ 1 & 4 \end{bmatrix}\), \(B = \begin{bmatrix} 5 & 6 \\ 7 & 8 \end{bmatrix}\)

$$ AB = \begin{bmatrix} (2 \times 5 + 3 \times 7) & (2 \times 6 + 3 \times 8) \\ (1 \times 5 + 4 \times 7) & (1 \times 6 + 4 \times 8) \end{bmatrix} $$$$ = \begin{bmatrix} (10 + 21) & (12 + 24) \\ (5 + 28) & (6 + 32) \end{bmatrix} $$$$ = \begin{bmatrix} 31 & 36 \\ 33 & 38 \end{bmatrix} $$ Answer: \(\begin{bmatrix} 31 & 36 \\ 33 & 38 \end{bmatrix}\)

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Example 2 (Medium):
Multiply \(C = \begin{bmatrix} 4 & 1 \\ 2 & 5 \end{bmatrix}\) by \(D = \begin{bmatrix} 7 & 3 \\ 6 & 2 \end{bmatrix}\).

Given:
\(C = \begin{bmatrix} 4 & 1 \\ 2 & 5 \end{bmatrix}\), \(D = \begin{bmatrix} 7 & 3 \\ 6 & 2 \end{bmatrix}\)

$$ CD = \begin{bmatrix} (4 \times 7 + 1 \times 6) & (4 \times 3 + 1 \times 2) \\ (2 \times 7 + 5 \times 6) & (2 \times 3 + 5 \times 2) \end{bmatrix} $$$$ = \begin{bmatrix} (28 + 6) & (12 + 2) \\ (14 + 30) & (6 + 10) \end{bmatrix} $$$$ = \begin{bmatrix} 34 & 14 \\ 44 & 16 \end{bmatrix} $$ Answer: \(\begin{bmatrix} 34 & 14 \\ 44 & 16 \end{bmatrix}\)

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Example 3 (Hard):
Multiply \(E = \begin{bmatrix} 3 & 2 & 1 \\ 4 & 0 & 5 \end{bmatrix}\) by \(F = \begin{bmatrix} 1 & 2 \\ 0 & 3 \\ 4 & 5 \end{bmatrix}\).

Given:
\(E = \begin{bmatrix} 3 & 2 & 1 \\ 4 & 0 & 5 \end{bmatrix}\), \(F = \begin{bmatrix} 1 & 2 \\ 0 & 3 \\ 4 & 5 \end{bmatrix}\)

$$ EF = \begin{bmatrix} (3 \times 1 + 2 \times 0 + 1 \times 4) & (3 \times 2 + 2 \times 3 + 1 \times 5) \\ (4 \times 1 + 0 \times 0 + 5 \times 4) & (4 \times 2 + 0 \times 3 + 5 \times 5) \end{bmatrix} $$$$ = \begin{bmatrix} (3 + 0 + 4) & (6 + 6 + 5) \\ (4 + 0 + 20) & (8 + 0 + 25) \end{bmatrix} $$$$ = \begin{bmatrix} 7 & 17 \\ 24 & 33 \end{bmatrix} $$ Answer: \(\begin{bmatrix} 7 & 17 \\ 24 & 33 \end{bmatrix}\)

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Example 4 (Applied):
A control system matrix \(G = \begin{bmatrix} 2 & 4 \\ 3 & 1 \end{bmatrix}\) multiplies a settings matrix \(H = \begin{bmatrix} 5 & 7 \\ 6 & 8 \end{bmatrix}\).

Given:
\(G = \begin{bmatrix} 2 & 4 \\ 3 & 1 \end{bmatrix}\), \(H = \begin{bmatrix} 5 & 7 \\ 6 & 8 \end{bmatrix}\)

$$ GH = \begin{bmatrix} (2 \times 5 + 4 \times 6) & (2 \times 7 + 4 \times 8) \\ (3 \times 5 + 1 \times 6) & (3 \times 7 + 1 \times 8) \end{bmatrix} $$$$ = \begin{bmatrix} (10 + 24) & (14 + 32) \\ (15 + 6) & (21 + 8) \end{bmatrix} $$$$ = \begin{bmatrix} 34 & 46 \\ 21 & 29 \end{bmatrix} $$ Answer: \(\begin{bmatrix} 34 & 46 \\ 21 & 29 \end{bmatrix}\)

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Example 5 (Challenge):
Multiply \(J = \begin{bmatrix} 1 & 2 & 3 \\ 4 & 5 & 6 \end{bmatrix}\) by \(K = \begin{bmatrix} 7 & 8 \\ 9 & 10 \\ 11 & 12 \end{bmatrix}\).

Given:
\(J = \begin{bmatrix} 1 & 2 & 3 \\ 4 & 5 & 6 \end{bmatrix}\), \(K = \begin{bmatrix} 7 & 8 \\ 9 & 10 \\ 11 & 12 \end{bmatrix}\)

$$ JK = \begin{bmatrix} (1 \times 7 + 2 \times 9 + 3 \times 11) & (1 \times 8 + 2 \times 10 + 3 \times 12) \\ (4 \times 7 + 5 \times 9 + 6 \times 11) & (4 \times 8 + 5 \times 10 + 6 \times 12) \end{bmatrix} $$$$ = \begin{bmatrix} (7 + 18 + 33) & (8 + 20 + 36) \\ (28 + 45 + 66) & (32 + 50 + 72) \end{bmatrix} $$$$ = \begin{bmatrix} 58 & 64 \\ 139 & 154 \end{bmatrix} $$ Answer: \(\begin{bmatrix} 58 & 64 \\ 139 & 154 \end{bmatrix}\)

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3.1.4 Determinant of a 2x2 Matrix

The determinant of a \(2 \times 2\) matrix is a single value that can be used to analyze system stability, invert matrices, and solve equations in industrial control systems.

$$ \text{For } M = \begin{bmatrix} a & b \\ c & d \end{bmatrix},\ \text{det}(M) = ad, bc $$

Worked Examples

Example 1 (Easy):
Find the determinant of \(A = \begin{bmatrix} 3 & 4 \\ 2 & 5 \end{bmatrix}\).

Given:
\(A = \begin{bmatrix} 3 & 4 \\ 2 & 5 \end{bmatrix}\)

$$ \text{det}(A) = 3 \times 5 - 4 \times 2 $$$$ = 15 - 8 $$$$ = 7 $$ Answer: 7

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Example 2 (Medium):
Calculate the determinant of \(B = \begin{bmatrix} 6 & 1 \\ 3 & 8 \end{bmatrix}\).

Given:
\(B = \begin{bmatrix} 6 & 1 \\ 3 & 8 \end{bmatrix}\)

$$ \text{det}(B) = 6 \times 8 - 1 \times 3 $$$$ = 48 - 3 $$$$ = 45 $$ Answer: 45

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Example 3 (Hard):
Determine the determinant of \(C = \begin{bmatrix} 9 & 5 \\ 7 & 2 \end{bmatrix}\).

Given:
\(C = \begin{bmatrix} 9 & 5 \\ 7 & 2 \end{bmatrix}\)

$$ \text{det}(C) = 9 \times 2 - 5 \times 7 $$$$ = 18 - 35 $$$$ = -17 $$ Answer: -17

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Example 4 (Applied):
A sensor matrix \(D = \begin{bmatrix} 12 & 4 \\ 3 & 10 \end{bmatrix}\), find its determinant.

Given:
\(D = \begin{bmatrix} 12 & 4 \\ 3 & 10 \end{bmatrix}\)

$$ \text{det}(D) = 12 \times 10 - 4 \times 3 $$$$ = 120 - 12 $$$$ = 108 $$ Answer: 108

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Example 5 (Challenge):
Compute the determinant of \(E = \begin{bmatrix} 15 & 8 \\ 6 & 13 \end{bmatrix}\).

Given:
\(E = \begin{bmatrix} 15 & 8 \\ 6 & 13 \end{bmatrix}\)

$$ \text{det}(E) = 15 \times 13 - 8 \times 6 $$$$ = 195 - 48 $$$$ = 147 $$ Answer: 147

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3.1.5 Unit Matrix (Identity Matrix)

The unit matrix (identity matrix) is a square matrix with ones on the diagonal and zeros elsewhere. Multiplying any matrix by the identity matrix leaves it unchanged, analogous to multiplying by 1.

$$ I = \begin{bmatrix} 1 & 0 \\ 0 & 1 \end{bmatrix} $$

Worked Examples

Example 1 (Easy):
Multiply \(A = \begin{bmatrix} 2 & 5 \\ 3 & 4 \end{bmatrix}\) by the identity matrix \(I = \begin{bmatrix} 1 & 0 \\ 0 & 1 \end{bmatrix}\).

Given:
\(A = \begin{bmatrix} 2 & 5 \\ 3 & 4 \end{bmatrix}\), \(I = \begin{bmatrix} 1 & 0 \\ 0 & 1 \end{bmatrix}\)

$$ AI = \begin{bmatrix} 2 \times 1 + 5 \times 0 & 2 \times 0 + 5 \times 1 \\ 3 \times 1 + 4 \times 0 & 3 \times 0 + 4 \times 1 \end{bmatrix} $$$$ = \begin{bmatrix} 2 & 5 \\ 3 & 4 \end{bmatrix} $$ Answer: \(\begin{bmatrix} 2 & 5 \\ 3 & 4 \end{bmatrix}\)

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Example 2 (Medium):
Multiply \(B = \begin{bmatrix} 7 & 8 \\ 9 & 6 \end{bmatrix}\) by identity matrix \(I\).

Given:
\(B = \begin{bmatrix} 7 & 8 \\ 9 & 6 \end{bmatrix}\), \(I = \begin{bmatrix} 1 & 0 \\ 0 & 1 \end{bmatrix}\)

$$ BI = \begin{bmatrix} 7 \times 1 + 8 \times 0 & 7 \times 0 + 8 \times 1 \\ 9 \times 1 + 6 \times 0 & 9 \times 0 + 6 \times 1 \end{bmatrix} $$$$ = \begin{bmatrix} 7 & 8 \\ 9 & 6 \end{bmatrix} $$ Answer: \(\begin{bmatrix} 7 & 8 \\ 9 & 6 \end{bmatrix}\)

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Example 3 (Hard):
Multiply \(C = \begin{bmatrix} 3 & 1 \\ 2 & 5 \end{bmatrix}\) by identity matrix \(I\).

Given:
\(C = \begin{bmatrix} 3 & 1 \\ 2 & 5 \end{bmatrix}\), \(I = \begin{bmatrix} 1 & 0 \\ 0 & 1 \end{bmatrix}\)

$$ CI = \begin{bmatrix} 3 \times 1 + 1 \times 0 & 3 \times 0 + 1 \times 1 \\ 2 \times 1 + 5 \times 0 & 2 \times 0 + 5 \times 1 \end{bmatrix} $$$$ = \begin{bmatrix} 3 & 1 \\ 2 & 5 \end{bmatrix} $$ Answer: \(\begin{bmatrix} 3 & 1 \\ 2 & 5 \end{bmatrix}\)

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Example 4 (Applied):
Multiply \(D = \begin{bmatrix} 15 & 7 \\ 8 & 12 \end{bmatrix}\) by identity matrix \(I\).

Given:
\(D = \begin{bmatrix} 15 & 7 \\ 8 & 12 \end{bmatrix}\), \(I = \begin{bmatrix} 1 & 0 \\ 0 & 1 \end{bmatrix}\)

$$ DI = \begin{bmatrix} 15 \times 1 + 7 \times 0 & 15 \times 0 + 7 \times 1 \\ 8 \times 1 + 12 \times 0 & 8 \times 0 + 12 \times 1 \end{bmatrix} $$$$ = \begin{bmatrix} 15 & 7 \\ 8 & 12 \end{bmatrix} $$ Answer: \(\begin{bmatrix} 15 & 7 \\ 8 & 12 \end{bmatrix}\)

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Example 5 (Challenge):
Multiply \(E = \begin{bmatrix} 6 & 9 \\ 2 & 13 \end{bmatrix}\) by identity matrix \(I\).

Given:
\(E = \begin{bmatrix} 6 & 9 \\ 2 & 13 \end{bmatrix}\), \(I = \begin{bmatrix} 1 & 0 \\ 0 & 1 \end{bmatrix}\)

$$ EI = \begin{bmatrix} 6 \times 1 + 9 \times 0 & 6 \times 0 + 9 \times 1 \\ 2 \times 1 + 13 \times 0 & 2 \times 0 + 13 \times 1 \end{bmatrix} $$$$ = \begin{bmatrix} 6 & 9 \\ 2 & 13 \end{bmatrix} $$ Answer: \(\begin{bmatrix} 6 & 9 \\ 2 & 13 \end{bmatrix}\)

3.2.1 Determinant of a 3x3 Matrix: Concept and Formula

The determinant of a \( 3 \times 3 \) matrix is a single value calculated from its elements, and it is essential in industrial control for solving systems of equations, checking if a matrix is invertible, and analyzing control system stability. For a matrix
\[ A = \begin{bmatrix} a & b & c \\ d & e & f \\ g & h & i \end{bmatrix} \]
the determinant, written as \( |A| \) or \( \det(A) \), is calculated using:$$ |A| = a(ei, fh) - b(di, fg) + c(dh, eg) $$

Worked Examples

Example 1:
A PLC control panel uses a transformation matrix
\[ M = \begin{bmatrix} 2 & 1 & 3 \\ 0 & 4 & 5 \\ 1 & 0 & 6 \end{bmatrix} \]
Find the determinant of \( M \).

Given:
\( a = 2, b = 1, c = 3, d = 0, e = 4, f = 5, g = 1, h = 0, i = 6 \)

$$ |M| = 2(4 \times 6 - 5 \times 0) - 1(0 \times 6 - 5 \times 1) + 3(0 \times 0 - 4 \times 1) $$$$ = 2(24 - 0) - 1(0 - 5) + 3(0 - 4) $$$$ = 2 \times 24 + 1 \times 5 + 3 \times (-4) $$$$ = 48 + 5 - 12 $$$$ = 41 $$ Answer: \( 41 \)

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Example 2:
A control engineer records sensor calibration values in a matrix:
\[ N = \begin{bmatrix} 1 & -2 & 0 \\ 3 & 5 & 7 \\ 4 & 1 & -1 \end{bmatrix} \]
Find \( |N| \).

Given:
\( a = 1, b = -2, c = 0, d = 3, e = 5, f = 7, g = 4, h = 1, i = -1 \)

$$ |N| = 1(5 \times -1 - 7 \times 1) - (-2)(3 \times -1 - 7 \times 4) + 0(3 \times 1 - 5 \times 4) $$$$ = 1(-5 - 7) + 2(-3 - 28) + 0(-17) $$$$ = 1 \times (-12) + 2 \times (-31) + 0 $$$$ = -12 - 62 + 0 $$$$ = -74 $$ Answer: \( -74 \)

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Example 3:
A three-phase motor control system uses the matrix:
\[ P = \begin{bmatrix} 3 & 2 & 1 \\ 4 & 0 & 5 \\ 6 & 7 & 8 \end{bmatrix} \]
Find the determinant of \( P \).

Given:
\( a = 3, b = 2, c = 1, d = 4, e = 0, f = 5, g = 6, h = 7, i = 8 \)

$$ |P| = 3(0 \times 8 - 5 \times 7) - 2(4 \times 8 - 5 \times 6) + 1(4 \times 7 - 0 \times 6) $$$$ = 3(0 - 35) - 2(32 - 30) + 1(28 - 0) $$$$ = 3 \times (-35) - 2 \times 2 + 1 \times 28 $$$$ = -105 - 4 + 28 $$$$ = -81 $$ Answer: \( -81 \)

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Example 4:
A temperature control matrix is:
\[ Q = \begin{bmatrix} 5 & 0 & 2 \\ 1 & 4 & 3 \\ 2 & 1 & 0 \end{bmatrix} \]
Find \( |Q| \).

Given:
\( a = 5, b = 0, c = 2, d = 1, e = 4, f = 3, g = 2, h = 1, i = 0 \)

$$ |Q| = 5(4 \times 0 - 3 \times 1) - 0(1 \times 0 - 3 \times 2) + 2(1 \times 1 - 4 \times 2) $$$$ = 5(0 - 3) - 0(0 - 6) + 2(1 - 8) $$$$ = 5 \times (-3) - 0 + 2 \times (-7) $$$$ = -15 + 0 - 14 $$$$ = -29 $$ Answer: \( -29 \)

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Example 5:
A lighting automation matrix is:
\[ R = \begin{bmatrix} 7 & 2 & 5 \\ 1 & 3 & 4 \\ 6 & 8 & 9 \end{bmatrix} \]
Find \( |R| \).

Given:
\( a = 7, b = 2, c = 5, d = 1, e = 3, f = 4, g = 6, h = 8, i = 9 \)

$$ |R| = 7(3 \times 9 - 4 \times 8) - 2(1 \times 9 - 4 \times 6) + 5(1 \times 8 - 3 \times 6) $$$$ = 7(27 - 32) - 2(9 - 24) + 5(8 - 18) $$$$ = 7 \times (-5) - 2 \times (-15) + 5 \times (-10) $$$$ = -35 + 30 - 50 $$$$ = -55 $$ Answer: \( -55 \)

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3.3.1 Concept of Matrix Inverse

The inverse of a matrix is a key tool in industrial control and installation, enabling engineers to solve systems of equations for circuit analysis, automation, and system modeling. For a square matrix \(A\), the inverse, denoted \(A^{-1}\), exists only if the determinant of \(A\) is non-zero. The inverse matrix is used to find unknown variables and to check system controllability in Kenyan industrial plants.

For a \(3 \times 3\) matrix:$$ A = \begin{pmatrix} a_{11} & a_{12} & a_{13} \\ a_{21} & a_{22} & a_{23} \\ a_{31} & a_{32} & a_{33} \end{pmatrix} $$

The inverse is given by:$$ A^{-1} = \frac{1}{\det(A)} \cdot \text{adj}(A) $$ where \(\det(A)\) is the determinant of \(A\) and \(\text{adj}(A)\) is the adjugate matrix.

Worked Examples

Example 1:
A control system in a Nairobi factory uses the matrix:$$ A = \begin{pmatrix} 2 & 1 & 3 \\ 0 & -1 & 2 \\ 1 & 4 & 0 \end{pmatrix} $$ Find \(A^{-1}\).

Given:
\(A\) as above.

Step 1: Find the determinant \(\det(A)\).
\[ \det(A) = 2 ((-1) \times 0 - 2 \times 4) - 1 (0 \times 0 - 2 \times 1) + 3 (0 \times 4 - (-1) \times 1) \]
\[ = 2 (0 - 8) - 1 (0 - 2) + 3 (0 + 1) \]
\[ = 2 \times (-8) - 1 \times (-2) + 3 \times 1 \]
\[ = -16 + 2 + 3 \]
\[ = -11 \]

Step 2: Find the matrix of cofactors and adjugate.

Calculate cofactors:

• \(C_{11}\): \((-1) \times 0 - 2 \times 4 = -8\)
• \(C_{12}\): \((0) \times 0 - 2 \times 1 = -2\), sign change: \(+2\)
• \(C_{13}\): \((0) \times 4 - (-1) \times 1 = 1\)
• \(C_{21}\): \((1) \times 0 - 3 \times 4 = -12\), sign change: \(+12\)
• \(C_{22}\): \((2) \times 0 - 3 \times 1 = -3\)
• \(C_{23}\): \((2) \times 4 - 1 \times 1 = 7\), sign change: \(-7\)
• \(C_{31}\): \((1) \times 2 - 3 \times (-1) = 2 + 3 = 5\)
• \(C_{32}\): \((2) \times 2 - 3 \times 0 = 4\), sign change: \(-4\)
• \(C_{33}\): \((2) \times (-1) - 1 \times 0 = -2\)

So, cofactor matrix:
\[ \begin{pmatrix} -8 & 2 & 1 \\ 12 & -3 & -7 \\ 5 & -4 & -2 \end{pmatrix} \]

Adjugate (transpose):
\[ \text{adj}(A) = \begin{pmatrix} -8 & 12 & 5 \\ 2 & -3 & -4 \\ 1 & -7 & -2 \end{pmatrix} \]

Step 3: Find \(A^{-1}\):
\[ A^{-1} = \frac{1}{-11} \begin{pmatrix} -8 & 12 & 5 \\ 2 & -3 & -4 \\ 1 & -7 & -2 \end{pmatrix} \]

Step 4: Multiply each element:
\[ A^{-1} = \begin{pmatrix} \frac{-8}{-11} & \frac{12}{-11} & \frac{5}{-11} \\ \frac{2}{-11} & \frac{-3}{-11} & \frac{-4}{-11} \\ \frac{1}{-11} & \frac{-7}{-11} & \frac{-2}{-11} \end{pmatrix} = \begin{pmatrix} 0.727 & -1.091 & -0.455 \\ -0.182 & 0.273 & 0.364 \\ -0.091 & 0.636 & 0.182 \end{pmatrix} \]

Answer:
\[ A^{-1} = \begin{pmatrix} 0.727 & -1.091 & -0.455 \\ -0.182 & 0.273 & 0.364 \\ -0.091 & 0.636 & 0.182 \end{pmatrix} \]

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Example 2:
A PLC system in a Mombasa hotel uses matrix:$$ B = \begin{pmatrix} 1 & 0 & 2 \\ 3 & 1 & 0 \\ 0 & 4 & 1 \end{pmatrix} $$ Find \(B^{-1}\).

Given:
\(B\) as above.

Step 1: Find determinant:
\[ \det(B) = 1 (1 \times 1 - 0 \times 4) - 0 (3 \times 1 - 0 \times 0) + 2 (3 \times 4 - 1 \times 0) \]
\[ = 1 (1 - 0) - 0 (3 - 0) + 2 (12 - 0) \]
\[ = 1 - 0 + 24 \]
\[ = 25 \]

Step 2: Cofactors:

• \(C_{11}\): \(1 \times 1 - 0 \times 4 = 1\)
• \(C_{12}\): \(3 \times 1 - 0 \times 0 = 3\), sign change: \(-3\)
• \(C_{13}\): \(3 \times 4 - 1 \times 0 = 12\)
• \(C_{21}\): \(0 \times 1 - 2 \times 4 = -8\), sign change: \(+8\)
• \(C_{22}\): \(1 \times 1 - 2 \times 0 = 1\)
• \(C_{23}\): \(1 \times 4 - 0 \times 0 = 4\), sign change: \(-4\)
• \(C_{31}\): \(0 \times 0 - 2 \times 1 = -2\)
• \(C_{32}\): \(1 \times 0 - 2 \times 3 = -6\), sign change: \(+6\)
• \(C_{33}\): \(1 \times 1 - 0 \times 3 = 1\)

Cofactor matrix:
\[ \begin{pmatrix} 1 & -3 & 12 \\ 8 & 1 & -4 \\ -2 & 6 & 1 \end{pmatrix} \]

Adjugate:
\[ \text{adj}(B) = \begin{pmatrix} 1 & 8 & -2 \\ -3 & 1 & 6 \\ 12 & -4 & 1 \end{pmatrix} \]

Step 3: Inverse:
\[ B^{-1} = \frac{1}{25} \begin{pmatrix} 1 & 8 & -2 \\ -3 & 1 & 6 \\ 12 & -4 & 1 \end{pmatrix} \]

Step 4: Multiply out:
\[ B^{-1} = \begin{pmatrix} 0.04 & 0.32 & -0.08 \\ -0.12 & 0.04 & 0.24 \\ 0.48 & -0.16 & 0.04 \end{pmatrix} \]

Answer:
\[ B^{-1} = \begin{pmatrix} 0.04 & 0.32 & -0.08 \\ -0.12 & 0.04 & 0.24 \\ 0.48 & -0.16 & 0.04 \end{pmatrix} \]

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Example 3:
A SACCO security system uses:$$ C = \begin{pmatrix} 4 & 2 & 1 \\ 3 & 5 & 2 \\ 2 & 1 & 3 \end{pmatrix} $$ Find \(C^{-1}\).

Given:
\(C\) as above.

Step 1: Determinant:
\[ \det(C) = 4(5 \times 3 - 2 \times 1) - 2(3 \times 3 - 2 \times 2) + 1(3 \times 1 - 5 \times 2) \]
\[ = 4(15 - 2) - 2(9 - 4) + 1(3 - 10) \]
\[ = 4 \times 13 - 2 \times 5 + 1 \times (-7) \]
\[ = 52 - 10 - 7 \]
\[ = 35 \]

Step 2: Cofactors:

• \(C_{11}\): \(5 \times 3 - 2 \times 1 = 15 - 2 = 13\)
• \(C_{12}\): \(3 \times 3 - 2 \times 2 = 9 - 4 = 5\), sign change: \(-5\)
• \(C_{13}\): \(3 \times 1 - 5 \times 2 = 3 - 10 = -7\)
• \(C_{21}\): \(2 \times 3 - 1 \times 1 = 6 - 1 = 5\), sign change: \(-5\)
• \(C_{22}\): \(4 \times 3 - 1 \times 2 = 12 - 2 = 10\)
• \(C_{23}\): \(4 \times 1 - 2 \times 2 = 4 - 4 = 0\), sign change: \(0\)
• \(C_{31}\): \(2 \times 2 - 1 \times 5 = 4 - 5 = -1\)
• \(C_{32}\): \(4 \times 2 - 1 \times 3 = 8 - 3 = 5\), sign change: \(-5\)
• \(C_{33}\): \(4 \times 5 - 2 \times 3 = 20 - 6 = 14\)

Cofactor matrix:
\[ \begin{pmatrix} 13 & -5 & -7 \\ -5 & 10 & 0 \\ -1 & -5 & 14 \end{pmatrix} \]

Adjugate:
\[ \text{adj}(C) = \begin{pmatrix} 13 & -5 & -1 \\ -5 & 10 & -5 \\ -7 & 0 & 14 \end{pmatrix} \]

Step 3: Inverse:
\[ C^{-1} = \frac{1}{35} \begin{pmatrix} 13 & -5 & -1 \\ -5 & 10 & -5 \\ -7 & 0 & 14 \end{pmatrix} \]

Step 4: Multiply:
\[ C^{-1} = \begin{pmatrix} 0.371 & -0.143 & -0.029 \\ -0.143 & 0.286 & -0.143 \\ -0.200 & 0 & 0.400 \end{pmatrix} \]

Answer:
\[ C^{-1} = \begin{pmatrix} 0.371 & -0.143 & -0.029 \\ -0.143 & 0.286 & -0.143 \\ -0.200 & 0 & 0.400 \end{pmatrix} \]

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Example 4:
A university automation system uses:$$ D = \begin{pmatrix} 3 & 0 & 2 \\ 2 & 0 & -2 \\ 0 & 1 & 1 \end{pmatrix} $$ Find \(D^{-1}\).

Given:
\(D\) as above.

Step 1: Determinant:
\[ \det(D) = 3(0 \times 1 - (-2) \times 1) - 0(2 \times 1 - (-2) \times 0) + 2(2 \times 1 - 0 \times 0) \]
\[ = 3(0 + 2) - 0(2 - 0) + 2(2 - 0) \]
\[ = 3 \times 2 + 2 \times 2 \]
\[ = 6 + 4 \]
\[ = 10 \]

Step 2: Cofactors:

• \(C_{11}\): \(0 \times 1 - (-2) \times 1 = 0 + 2 = 2\)
• \(C_{12}\): \(2 \times 1 - (-2) \times 0 = 2 - 0 = 2\), sign change: \(-2\)
• \(C_{13}\): \(2 \times 1 - 0 \times 0 = 2 - 0 = 2\)
• \(C_{21}\): \(0 \times 1 - 2 \times 1 = 0 - 2 = -2\), sign change: \(+2\)
• \(C_{22}\): \(3 \times 1 - 2 \times 0 = 3 - 0 = 3\)
• \(C_{23}\): \(3 \times 1 - 0 \times 0 = 3 - 0 = 3\), sign change: \(-3\)
• \(C_{31}\): \(0 \times -2 - 0 \times 0 = 0 - 0 = 0\)
• \(C_{32}\): \(3 \times -2 - 2 \times 0 = -6 - 0 = -6\), sign change: \(+6\)
• \(C_{33}\): \(3 \times 0 - 2 \times 0 = 0 - 0 = 0\)

Cofactor matrix:
\[ \begin{pmatrix} 2 & -2 & 2 \\ 2 & 3 & -3 \\ 0 & 6 & 0 \end{pmatrix} \]

Adjugate:
\[ \text{adj}(D) = \begin{pmatrix} 2 & 2 & 0 \\ -2 & 3 & 6 \\ 2 & -3 & 0 \end{pmatrix} \]

Step 3: Inverse:
\[ D^{-1} = \frac{1}{10} \begin{pmatrix} 2 & 2 & 0 \\ -2 & 3 & 6 \\ 2 & -3 & 0 \end{pmatrix} \]

Step 4: Multiply:
\[ D^{-1} = \begin{pmatrix} 0.2 & 0.2 & 0 \\ -0.2 & 0.3 & 0.6 \\ 0.2 & -0.3 & 0 \end{pmatrix} \]

Answer:
\[ D^{-1} = \begin{pmatrix} 0.2 & 0.2 & 0 \\ -0.2 & 0.3 & 0.6 \\ 0.2 & -0.3 & 0 \end{pmatrix} \]

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Example 5:
A county government office uses:$$ E = \begin{pmatrix} 1 & 2 & 3 \\ 0 & 1 & 4 \\ 5 & 6 & 0 \end{pmatrix} $$ Find \(E^{-1}\).

Given:
\(E\) as above.

Step 1: Determinant:
\[ \det(E) = 1(1 \times 0 - 4 \times 6) - 2(0 \times 0 - 4 \times 5) + 3(0 \times 6 - 1 \times 5) \]
\[ = 1(0 - 24) - 2(0 - 20) + 3(0 - 5) \]
\[ = -24 + 40 - 15 \]
\[ = 1 \]

Step 2: Cofactors:

• \(C_{11}\): \(1 \times 0 - 4 \times 6 = 0 - 24 = -24\)
• \(C_{12}\): \(0 \times 0 - 4 \times 5 = 0 - 20 = -20\), sign change: \(+20\)
• \(C_{13}\): \(0 \times 6 - 1 \times 5 = 0 - 5 = -5\)
• \(C_{21}\): \(2 \times 0 - 3 \times 6 = 0 - 18 = -18\), sign change: \(+18\)
• \(C_{22}\): \(1 \times 0 - 3 \times 5 = 0 - 15 = -15\)
• \(C_{23}\): \(1 \times 6 - 2 \times 5 = 6 - 10 = -4\), sign change: \(+4\)
• \(C_{31}\): \(2 \times 4 - 3 \times 1 = 8 - 3 = 5\)
• \(C_{32}\): \(1 \times 4 - 3 \times 0 = 4 - 0 = 4\), sign change: \(-4\)
• \(C_{33}\): \(1 \times 1 - 2 \times 0 = 1 - 0 = 1\)

Cofactor matrix:
\[ \begin{pmatrix} -24 & 20 & -5 \\ 18 & -15 & 4 \\ 5 & -4 & 1 \end{pmatrix} \]

Adjugate:
\[ \text{adj}(E) = \begin{pmatrix} -24 & 18 & 5 \\ 20 & -15 & -4 \\ -5 & 4 & 1 \end{pmatrix} \]

Step 3: Inverse:
\[ E^{-1} = \frac{1}{1} \begin{pmatrix} -24 & 18 & 5 \\ 20 & -15 & -4 \\ -5 & 4 & 1 \end{pmatrix} \]

Step 4: Multiply:
\[ E^{-1} = \begin{pmatrix} -24 & 18 & 5 \\ 20 & -15 & -4 \\ -5 & 4 & 1 \end{pmatrix} \]

Answer:
\[ E^{-1} = \begin{pmatrix} -24 & 18 & 5 \\ 20 & -15 & -4 \\ -5 & 4 & 1 \end{pmatrix} \]

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Chapter Summary

This chapter provided a comprehensive exploration of matrix theory and its practical use in engineering contexts. Beginning with matrix operations, students learned how to perform addition, subtraction, and multiplication, as well as calculate the determinant of a \(3 \times 3\) matrix. The process of finding the inverse of a \(3 \times 3\) matrix was covered in detail, laying the groundwork for solving linear simultaneous equations involving three unknowns using matrix methods. The calculation and interpretation of eigenvalues and eigenvectors were introduced, demonstrating their significance in structural analysis and systems modeling. Finally, the chapter concluded with applications of matrices in probability, showing how these mathematical tools can be used to model and solve real-world engineering problems. Throughout, emphasis was placed on step-by-step calculations and logical reasoning, enabling students to master these essential skills. The chapter equips learners with the mathematical foundation necessary for advanced engineering analysis and decision-making.