Matrices and Linear Transformations

Matrix Operations

Notation

An $m \times n$ matrix $A$ has $m$ rows and $n$ columns. The entry in row $i$, column $j$ is $a_{ij}$.

Addition and Scalar Multiplication

For $A, B \in \mathcal{'\{'}M{'\}'}_{m \times n}(\mathbb{'\{'}R{'\}'})$ and $\lambda \in \mathbb{'\{'}R{'\}'}$:

$(A + B)_{ij} = a_{ij} + b_{ij}, \qquad (\lambda A)_{ij} = \lambda a_{ij}$

Matrix Multiplication

If $A$ is $m \times p$ and $B$ is $p \times n$, the product $C = AB$ is $m \times n$ with:

$c_{ij} = \sum_{k=1}^{p} a_{ik}b_{kj}$

Key properties:

• $(AB)C = A(BC)$ (associative)
• $A(B + C) = AB + AC$ (distributive)
• $AB \ne BA$ in general (not commutative)
• $(AB)^T = B^T A^T$

Example. If $A = \begin{pmatrix} 1 & 2 \\ 3 & 4 \end{pmatrix}$ and $B = \begin{pmatrix} 5 & 6 \\ 7 & 8 \end{pmatrix}$:

$AB = \begin{pmatrix} 1 \cdot 5 + 2 \cdot 7 & 1 \cdot 6 + 2 \cdot 8 \\ 3 \cdot 5 + 4 \cdot 7 & 3 \cdot 6 + 4 \cdot 8 \end{pmatrix} = \begin{pmatrix} 19 & 22 \\ 43 & 50 \end{pmatrix}$

$BA = \begin{pmatrix} 23 & 34 \\ 31 & 46 \end{pmatrix} \ne AB$

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Determinants

$2 \times 2$ Determinant

$\det\begin{pmatrix} a & b \\ c & d \end{pmatrix} = ad - bc$

$3 \times 3$ Determinant (Cofactor Expansion)

$\det A = a_{11}\begin{vmatrix} a_{22} & a_{23} \\ a_{32} & a_{33} \end{vmatrix} - a_{12}\begin{vmatrix} a_{21} & a_{23} \\ a_{31} & a_{33} \end{vmatrix} + a_{13}\begin{vmatrix} a_{21} & a_{22} \\ a_{31} & a_{32} \end{vmatrix}$

Example. $\det\begin{pmatrix} 1 & 0 & 2 \\ 3 & 1 & -1 \\ 2 & 4 & 1 \end{pmatrix}$

$= 1\begin{vmatrix} 1 & -1 \\ 4 & 1 \end{vmatrix} - 0 + 2\begin{vmatrix} 3 & 1 \\ 2 & 4 \end{vmatrix} = 1(1 + 4) + 2(12 - 2) = 5 + 20 = 25$

Properties

• $\det(AB) = \det(A)\det(B)$
• $\det(A^T) = \det(A)$
• $\det(\lambda A) = \lambda^n \det(A)$ for $n \times n$ matrix $A$
• Swapping two rows multiplies the determinant by $-1$
• If two rows are equal, $\det(A) = 0$
• $\det(A) = 0$ if and only if $A$ is singular (non-invertible)

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Matrix Inverses

Definition

If $A$ is a square matrix, its inverse $A^{-1}$ satisfies:

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

$A$ is invertible if and only if $\det(A) \ne 0$.

$2 \times 2$ Inverse

$A^{-1} = \frac{1}{ad - bc}\begin{pmatrix} d & -b \\ -c & a \end{pmatrix}$

Example. Find the inverse of $A = \begin{pmatrix} 3 & 1 \\ 5 & 2 \end{pmatrix}$.

$\det(A) = 6 - 5 = 1 \ne 0$, so $A$ is invertible.

$A^{-1} = \begin{pmatrix} 2 & -1 \\ -5 & 3 \end{pmatrix}$

Inverse via Row Reduction (Gauss-Jordan)

To find $A^{-1}$, augment $A$ with $I$ and apply row operations:

$\left[\,A \mid I\,\right] \xrightarrow{\mathrm{row\ ops}} \left[\,I \mid A^{-1}\,\right]$

If the left half reduces to $I$, the right half is $A^{-1}$. If the left half has a zero row, $A$ is singular.

Example. Find $A^{-1}$ for $A = \begin{pmatrix} 1 & 2 & 3 \\ 0 & 1 & 4 \\ 0 & 0 & 1 \end{pmatrix}$.

Since $A$ is upper triangular, $\det(A) = 1 \cdot 1 \cdot 1 = 1$.

$\left[\begin{array}{ccc|ccc} 1 & 2 & 3 & 1 & 0 & 0 \\ 0 & 1 & 4 & 0 & 1 & 0 \\ 0 & 0 & 1 & 0 & 0 & 1 \end{array}\right]$

$R_1 \to R_1 - 2R_2$:

$\left[\begin{array}{ccc|ccc} 1 & 0 & -5 & 1 & -2 & 0 \\ 0 & 1 & 4 & 0 & 1 & 0 \\ 0 & 0 & 1 & 0 & 0 & 1 \end{array}\right]$

$R_1 \to R_1 + 5R_3$, $R_2 \to R_2 - 4R_3$:

$\left[\begin{array}{ccc|ccc} 1 & 0 & 0 & 1 & -2 & 5 \\ 0 & 1 & 0 & 0 & 1 & -4 \\ 0 & 0 & 1 & 0 & 0 & 1 \end{array}\right]$

$A^{-1} = \begin{pmatrix} 1 & -2 & 5 \\ 0 & 1 & -4 \\ 0 & 0 & 1 \end{pmatrix}$

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Solving Systems of Linear Equations

Matrix Form

The system $A\mathbf{'\{'}x{'\}'} = \mathbf{'\{'}b{'\}'}$ where $A$ is $n \times n$:

• Unique solution: $\det(A) \ne 0$, giving $\mathbf{'\{'}x{'\}'} = A^{-1}\mathbf{'\{'}b{'\}'}$
• No solution or infinitely many: $\det(A) = 0$

Cramer's Rule

If $\det(A) \ne 0$, the $i$-th component of the solution is:

$x_i = \frac{\det(A_i)}{\det(A)}$

where $A_i$ is $A$ with the $i$-th column replaced by $\mathbf{'\{'}b{'\}'}$.

Example. Solve $\begin{cases} 2x + y = 5 \\ x - y = 1 \end{cases}$.

$A = \begin{pmatrix} 2 & 1 \\ 1 & -1 \end{pmatrix}, \quad \det(A) = -2 - 1 = -3$

$x = \frac{\det\begin{pmatrix} 5 & 1 \\ 1 & -1 \end{pmatrix}}{-3} = \frac{-6}{-3} = 2, \qquad y = \frac{\det\begin{pmatrix} 2 & 5 \\ 1 & 1 \end{pmatrix}}{-3} = \frac{-3}{-3} = 1$

Gaussian Elimination

For systems of any size, use row reduction to transform the augmented matrix to row echelon form, then back-substitute.

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Geometric Transformations

$2 \times 2$ Transformation Matrices

Each linear transformation of $\mathbb{'\{'}R{'\}'}^2$ can be represented by a $2 \times 2$ matrix.

Transformation	Matrix	Effect on Area
Reflection in $x$-axis	$\begin{pmatrix} 1 & 0 \\ 0 & -1 \end{pmatrix}$	Preserved
Reflection in $y$-axis	$\begin{pmatrix} -1 & 0 \\ 0 & 1 \end{pmatrix}$	Preserved
Reflection in $y = x$	$\begin{pmatrix} 0 & 1 \\ 1 & 0 \end{pmatrix}$	Preserved
Rotation $\theta$ anticlockwise	$\begin{pmatrix} \cos\theta & -\sin\theta \\ \sin\theta & \cos\theta \end{pmatrix}$	Preserved
Enlargement by $k$	$\begin{pmatrix} k & 0 \\ 0 & k \end{pmatrix}$	Scaled by $k^2$
Stretch parallel to $x$-axis by $k$	$\begin{pmatrix} k & 0 \\ 0 & 1 \end{pmatrix}$	Scaled by $
Shear parallel to $x$-axis by $k$	$\begin{pmatrix} 1 & k \\ 0 & 1 \end{pmatrix}$	Preserved

Composition of Transformations

If transformation $A$ is followed by transformation $B$, the combined transformation is represented by the matrix product $BA$ (note the order: apply $A$ first, then $B$).

Example. Find the matrix for a rotation of $90\,{}^{\circ}$ anticlockwise about the origin, followed by a reflection in the $x$-axis.

$A = \begin{pmatrix} 0 & -1 \\ 1 & 0 \end{pmatrix} \quad \mathrm{(rotation)}, \qquad B = \begin{pmatrix} 1 & 0 \\ 0 & -1 \end{pmatrix} \quad \mathrm{(reflection)}$

$BA = \begin{pmatrix} 0 & -1 \\ -1 & 0 \end{pmatrix}$

This is equivalent to a reflection in the line $y = -x$.

Invariant Points and Lines

An invariant point satisfies $A\mathbf{'\{'}x{'\}'} = \mathbf{'\{'}x{'\}'}$, i.e. $(A - I)\mathbf{'\{'}x{'\}'} = \mathbf{'\{'}0{'\}'}$. An invariant line is a line mapped to itself (points on the line may move along it).

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Eigenvalues and Eigenvectors

Definition

For a square matrix $A$, a scalar $\lambda$ is an eigenvalue if there exists a non-zero vector $\mathbf{'\{'}v{'\}'}$ such that:

$A\mathbf{'\{'}v{'\}'} = \lambda\mathbf{'\{'}v{'\}'}$

The vector $\mathbf{'\{'}v{'\}'}$ is the corresponding eigenvector.

Characteristic Equation

$\det(A - \lambda I) = 0$

This is a polynomial equation of degree $n$ in $\lambda$ whose roots are the eigenvalues.

Finding Eigenvectors

For each eigenvalue $\lambda_i$, solve $(A - \lambda_i I)\mathbf{'\{'}v{'\}'} = \mathbf{'\{'}0{'\}'}$.

Worked Examples

Example. Find the eigenvalues and eigenvectors of $A = \begin{pmatrix} 4 & 1 \\ 2 & 3 \end{pmatrix}$.

Characteristic equation:

$\det\begin{pmatrix} 4 - \lambda & 1 \\ 2 & 3 - \lambda \end{pmatrix} = (4 - \lambda)(3 - \lambda) - 2 = \lambda^2 - 7\lambda + 10 = 0$

$(\lambda - 5)(\lambda - 2) = 0 \implies \lambda_1 = 5, \quad \lambda_2 = 2$

For $\lambda_1 = 5$: $(A - 5I)\mathbf{'\{'}v{'\}'} = \mathbf{'\{'}0{'\}'}$:

$\begin{pmatrix} -1 & 1 \\ 2 & -2 \end{pmatrix}\begin{pmatrix} x \\ y \end{pmatrix} = \begin{pmatrix} 0 \\ 0 \end{pmatrix} \implies -x + y = 0 \implies y = x$

Eigenvector: $\mathbf{'\{'}v{'\}'}_1 = \begin{pmatrix} 1 \\ 1 \end{pmatrix}$ (or any scalar multiple).

For $\lambda_2 = 2$: $(A - 2I)\mathbf{'\{'}v{'\}'} = \mathbf{'\{'}0{'\}'}$:

$\begin{pmatrix} 2 & 1 \\ 2 & 1 \end{pmatrix}\begin{pmatrix} x \\ y \end{pmatrix} = \begin{pmatrix} 0 \\ 0 \end{pmatrix} \implies 2x + y = 0 \implies y = -2x$

Eigenvector: $\mathbf{'\{'}v{'\}'}_2 = \begin{pmatrix} 1 \\ -2 \end{pmatrix}$.

Diagonalisation

If $A$ has $n$ linearly independent eigenvectors, it can be diagonalised:

$A = PDP^{-1}$

where $D$ is the diagonal matrix of eigenvalues and $P$ has eigenvectors as columns.

Applications:

• Computing $A^n$ efficiently: $A^n = PD^nP^{-1}$
• Solving systems of differential equations
• Principal component analysis (statistics)

Trace and Determinant

For a $2 \times 2$ matrix with eigenvalues $\lambda_1, \lambda_2$:

$\mathrm{tr}(A) = \lambda_1 + \lambda_2 = a_{11} + a_{22}$

$\det(A) = \lambda_1 \lambda_2$

$3 \times 3$ Eigenvalue Example

Example. Find the eigenvalues and eigenvectors of $A = \begin{pmatrix} 2 & 0 & 0 \\ 0 & 3 & -1 \\ 0 & -1 & 3 \end{pmatrix}$.

Characteristic equation:

$\det\begin{pmatrix} 2 - \lambda & 0 & 0 \\ 0 & 3 - \lambda & -1 \\ 0 & -1 & 3 - \lambda \end{pmatrix} = (2 - \lambda)\bigl[(3 - \lambda)^2 - 1\bigr] = 0$

$(2 - \lambda)(\lambda^2 - 6\lambda + 8) = (2 - \lambda)(\lambda - 2)(\lambda - 4) = 0$

$\lambda_1 = 2$ (algebraic multiplicity 2), $\lambda_2 = 4$.

For $\lambda = 2$: $(A - 2I)\mathbf{'\{'}v{'\}'} = \mathbf{'\{'}0{'\}'}$ gives $\begin{pmatrix} 0 & 0 & 0 \\ 0 & 1 & -1 \\ 0 & -1 & 1 \end{pmatrix}\mathbf{'\{'}v{'\}'} = \mathbf{'\{'}0{'\}'}$, so $v_2 = v_3$ with $v_1$ free. Two linearly independent eigenvectors: $\mathbf{'\{'}v{'\}'}_1 = (1, 0, 0)$ and $\mathbf{'\{'}v{'\}'}_2 = (0, 1, 1)$.

For $\lambda = 4$: $(A - 4I)\mathbf{'\{'}v{'\}'} = \mathbf{'\{'}0{'\}'}$ gives $-v_2 - v_3 = 0$ and $-v_1 = 0$, so $\mathbf{'\{'}v{'\}'}_3 = (0, 1, -1)$.

The matrix is diagonalisable: $P = \begin{pmatrix} 1 & 0 & 0 \\ 0 & 1 & 1 \\ 0 & 1 & -1 \end{pmatrix}$, $D = \begin{pmatrix} 2 & 0 & 0 \\ 0 & 2 & 0 \\ 0 & 0 & 4 \end{pmatrix}$.

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Solving Systems Using Row Reduction

Augmented Matrix Method

For a system of $m$ equations in $n$ unknowns, write the augmented matrix $[A | \mathbf{'\{'}b{'\}'}]$ and apply elementary row operations to reach row echelon form.

Elementary row operations:

1. Swap two rows ($R_i \leftrightarrow R_j$).
2. Multiply a row by a non-zero scalar ($R_i \to kR_i$).
3. Add a multiple of one row to another ($R_i \to R_i + kR_j$).

Example: Three Equations in Three Unknowns

Solve the system:

$x + 2y - z = 3$ $2x - y + z = 1$ $3x + y + 2z = 10$

Augmented matrix:

$\left[\begin{array}{ccc|c} 1 & 2 & -1 & 3 \\ 2 & -1 & 1 & 1 \\ 3 & 1 & 2 & 10 \end{array}\right]$

$R_2 \to R_2 - 2R_1$, $R_3 \to R_3 - 3R_1$:

$\left[\begin{array}{ccc|c} 1 & 2 & -1 & 3 \\ 0 & -5 & 3 & -5 \\ 0 & -5 & 5 & 1 \end{array}\right]$

$R_3 \to R_3 - R_2$:

$\left[\begin{array}{ccc|c} 1 & 2 & -1 & 3 \\ 0 & -5 & 3 & -5 \\ 0 & 0 & 2 & 6 \end{array}\right]$

Back-substitute: $2z = 6 \implies z = 3$. Then $-5y + 9 = -5 \implies y = 14/5 = 2.8$. Then $x + 5.6 - 3 = 3 \implies x = 0.4$.

Solution: $(x, y, z) = (2/5,\, 14/5,\, 3)$.

Nature of Solutions

For an $n \times n$ system $A\mathbf{'\{'}x{'\}'} = \mathbf{'\{'}b{'\}'}$:

$\det(A)$	Rank	Solutions
$\ne 0$	$n$	Unique solution
$= 0$ and consistent	$\lt n$	Infinitely many solutions
$= 0$ and inconsistent	—	No solution

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$3 \times 3$ Determinant and Inverse

Cofactor Expansion Along Any Row or Column

The determinant can be computed by expanding along any row $i$:

$\det(A) = \sum_{j=1}^{n} (-1)^{i+j}\, a_{ij}\, M_{ij}$

where $M_{ij}$ is the minor (determinant of the submatrix obtained by removing row $i$ and column $j$). Choosing a row or column with the most zeros minimises computation.

Adjugate Method for $3 \times 3$ Inverse

$A^{-1} = \frac{1}{\det(A)}\,\mathrm{adj}(A)$

where the adjugate (adjoint) matrix is the transpose of the cofactor matrix.

Example. Find $A^{-1}$ for $A = \begin{pmatrix} 1 & 2 & 0 \\ 0 & 1 & 3 \\ 1 & 0 & 1 \end{pmatrix}$.

$\det(A) = 1(1 - 0) - 2(0 - 3) + 0 = 1 + 6 = 7$.

Cofactor matrix: $C = \begin{pmatrix} 1 & 3 & -1 \\ -2 & 1 & 2 \\ 6 & -3 & 1 \end{pmatrix}$.

$\mathrm{adj}(A) = C^T = \begin{pmatrix} 1 & -2 & 6 \\ 3 & 1 & -3 \\ -1 & 2 & 1 \end{pmatrix}$.

$A^{-1} = \frac{1}{7}\begin{pmatrix} 1 & -2 & 6 \\ 3 & 1 & -3 \\ -1 & 2 & 1 \end{pmatrix}$

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Applications of Eigenvalues

Powers of Matrices

If $A = PDP^{-1}$, then $A^n = PD^nP^{-1}$, where $D^n$ is trivial to compute (just raise diagonal entries to the $n$-th power).

Example. Compute $A^{10}$ for $A = \begin{pmatrix} 3 & 1 \\ 2 & 4 \end{pmatrix}$.

From earlier, $\lambda_1 = 5$, $\lambda_2 = 2$, $P = \begin{pmatrix} 1 & 1 \\ 1 & -2 \end{pmatrix}$.

$D^{10} = \begin{pmatrix} 5^{10} & 0 \\ 0 & 2^{10} \end{pmatrix}$

$A^{10} = PD^{10}P^{-1}$

Systems of Differential Equations

The system $\dfrac{d\mathbf{'\{'}x{'\}'}}{dt} = A\mathbf{'\{'}x{'\}'}$ has solution $\mathbf{'\{'}x{'\}'}(t) = e^{At}\mathbf{'\{'}x{'\}'}(0)$, which can be evaluated using diagonalisation: $e^{At} = Pe^{Dt}P^{-1}$.

warning

Common Pitfall

A matrix is diagonalisable if and only if it has a full set of linearly independent eigenvectors. A matrix with repeated eigenvalues may or may not be diagonalisable. For example, $\begin{pmatrix} 1 & 1 \\ 0 & 1 \end{pmatrix}$ has eigenvalue $1$ with algebraic multiplicity $2$ but geometric multiplicity $1$; it is not diagonalisable.

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Additional Worked Examples

Worked Example: $3 \times 3$ System with No Unique Solution

Determine the nature of solutions for the system:

$x + 2y + 3z = 4$ $2x + 4y + 6z = 8$ $x - y + z = 1$

Solution

Augmented matrix:

$\left[\begin{array}{ccc|c} 1 & 2 & 3 & 4 \\ 2 & 4 & 6 & 8 \\ 1 & -1 & 1 & 1 \end{array}\right]$

$R_2 \to R_2 - 2R_1$, $R_3 \to R_3 - R_1$:

$\left[\begin{array}{ccc|c} 1 & 2 & 3 & 4 \\ 0 & 0 & 0 & 0 \\ 0 & -3 & -2 & -3 \end{array}\right]$

Swap $R_2$ and $R_3$:

$\left[\begin{array}{ccc|c} 1 & 2 & 3 & 4 \\ 0 & -3 & -2 & -3 \\ 0 & 0 & 0 & 0 \end{array}\right]$

Row 2 gives $-3y - 2z = -3$, so $y = \dfrac{3 - 2z}{3} = 1 - \dfrac{2z}{3}$.

Row 1 gives $x + 2\!\left(1 - \dfrac{2z}{3}\right) + 3z = 4$, so $x + 2 - \dfrac{4z}{3} + 3z = 4$, hence $x = 2 - \dfrac{5z}{3}$.

Let $z = 3t$ (to avoid fractions): $x = 2 - 5t$, $y = 1 - 2t$, $z = 3t$.

The system has infinitely many solutions parametrised by $t \in \mathbb{'\{'}R{'\}'}$.

Worked Example: Composition of Transformations

Find the matrix representing an enlargement by scale factor $2$ about the origin, followed by a reflection in the line $y = x$. Determine the image of the point $(3, 1)$.

Solution

Enlargement by $2$: $E = \begin{pmatrix} 2 & 0 \\ 0 & 2 \end{pmatrix}$.

Reflection in $y = x$: $R = \begin{pmatrix} 0 & 1 \\ 1 & 0 \end{pmatrix}$.

Combined transformation (enlargement first, then reflection): $T = RE$.

$T = \begin{pmatrix} 0 & 1 \\ 1 & 0 \end{pmatrix}\begin{pmatrix} 2 & 0 \\ 0 & 2 \end{pmatrix} = \begin{pmatrix} 0 & 2 \\ 2 & 0 \end{pmatrix}$

Image of $(3, 1)$:

$\begin{pmatrix} 0 & 2 \\ 2 & 0 \end{pmatrix}\begin{pmatrix} 3 \\ 1 \end{pmatrix} = \begin{pmatrix} 2 \\ 6 \end{pmatrix}$

The image is $(2, 6)$.

Worked Example: $3 \times 3$ Eigenvalues with Complex Roots

Find the eigenvalues of $A = \begin{pmatrix} 1 & 0 & -1 \\ 0 & 2 & 0 \\ 1 & 0 & 1 \end{pmatrix}$.

Solution

Characteristic equation:

$\det\begin{pmatrix} 1 - \lambda & 0 & -1 \\ 0 & 2 - \lambda & 0 \\ 1 & 0 & 1 - \lambda \end{pmatrix} = 0$

Expand along the second row:

$(2 - \lambda)\begin{vmatrix} 1 - \lambda & -1 \\ 1 & 1 - \lambda \end{vmatrix} = 0$

$(2 - \lambda)\bigl[(1 - \lambda)^2 + 1\bigr] = 0$

$(2 - \lambda)(\lambda^2 - 2\lambda + 2) = 0$

From $2 - \lambda = 0$: $\lambda_1 = 2$.

From $\lambda^2 - 2\lambda + 2 = 0$: $\lambda = \dfrac{2 \pm \sqrt{4 - 8}}{2} = 1 \pm i$.

The three eigenvalues are $\lambda = 2$, $\lambda = 1 + i$, $\lambda = 1 - i$.

Since complex eigenvalues of a real matrix come in conjugate pairs, this is consistent. The matrix is not diagonalisable over $\mathbb{'\{'}R{'\}'}$ but is diagonalisable over $\mathbb{'\{'}C{'\}'}$.

Worked Example: Adjugate Method for $3 \times 3$ Inverse

Find the inverse of $A = \begin{pmatrix} 2 & 1 & 0 \\ 1 & 3 & 1 \\ 0 & 1 & 2 \end{pmatrix}$ using the adjugate method.

Solution

First, compute $\det(A)$ by expanding along the first row:

$\det(A) = 2\begin{vmatrix} 3 & 1 \\ 1 & 2 \end{vmatrix} - 1\begin{vmatrix} 1 & 1 \\ 0 & 2 \end{vmatrix} + 0 = 2(6 - 1) - 1(2 - 0) = 10 - 2 = 8$

Now compute the cofactor matrix $C$:

$C_{11} = +\begin{vmatrix} 3 & 1 \\ 1 & 2 \end{vmatrix} = 5$, $C_{12} = -\begin{vmatrix} 1 & 1 \\ 0 & 2 \end{vmatrix} = -2$, $C_{13} = +\begin{vmatrix} 1 & 3 \\ 0 & 1 \end{vmatrix} = 1$

$C_{21} = -\begin{vmatrix} 1 & 0 \\ 1 & 2 \end{vmatrix} = -2$, $C_{22} = +\begin{vmatrix} 2 & 0 \\ 0 & 2 \end{vmatrix} = 4$, $C_{23} = -\begin{vmatrix} 2 & 1 \\ 0 & 1 \end{vmatrix} = -2$

$C_{31} = +\begin{vmatrix} 1 & 0 \\ 3 & 1 \end{vmatrix} = 1$, $C_{32} = -\begin{vmatrix} 2 & 0 \\ 1 & 1 \end{vmatrix} = -2$, $C_{33} = +\begin{vmatrix} 2 & 1 \\ 1 & 3 \end{vmatrix} = 5$

$C = \begin{pmatrix} 5 & -2 & 1 \\ -2 & 4 & -2 \\ 1 & -2 & 5 \end{pmatrix}$

$\mathrm{adj}(A) = C^T = \begin{pmatrix} 5 & -2 & 1 \\ -2 & 4 & -2 \\ 1 & -2 & 5 \end{pmatrix}$

(Since this matrix is symmetric, $C^T = C$.)

$A^{-1} = \frac{1}{8}\begin{pmatrix} 5 & -2 & 1 \\ -2 & 4 & -2 \\ 1 & -2 & 5 \end{pmatrix}$

Worked Example: Matrix Power via Diagonalisation

Compute $A^5$ where $A = \begin{pmatrix} 3 & -1 \\ 2 & 0 \end{pmatrix}$.

Solution

Characteristic equation: $\lambda(\lambda - 3) + 2 = \lambda^2 - 3\lambda + 2 = 0$, so $(\lambda - 1)(\lambda - 2) = 0$, giving $\lambda_1 = 1$, $\lambda_2 = 2$.

For $\lambda_1 = 1$: $\begin{pmatrix} 2 & -1 \\ 2 & -1 \end{pmatrix}\mathbf{'\{'}v{'\}'} = \mathbf{'\{'}0{'\}'} \implies 2x = y$. Eigenvector: $\mathbf{'\{'}v{'\}'}_1 = \begin{pmatrix} 1 \\ 2 \end{pmatrix}$.

For $\lambda_2 = 2$: $\begin{pmatrix} 1 & -1 \\ 2 & -2 \end{pmatrix}\mathbf{'\{'}v{'\}'} = \mathbf{'\{'}0{'\}'} \implies x = y$. Eigenvector: $\mathbf{'\{'}v{'\}'}_2 = \begin{pmatrix} 1 \\ 1 \end{pmatrix}$.

$P = \begin{pmatrix} 1 & 1 \\ 2 & 1 \end{pmatrix}, \quad D = \begin{pmatrix} 1 & 0 \\ 0 & 2 \end{pmatrix}$

$P^{-1} = \frac{1}{1 - 2}\begin{pmatrix} 1 & -1 \\ -2 & 1 \end{pmatrix} = \begin{pmatrix} -1 & 1 \\ 2 & -1 \end{pmatrix}$

$A^5 = PD^5P^{-1} = \begin{pmatrix} 1 & 1 \\ 2 & 1 \end{pmatrix}\begin{pmatrix} 1 & 0 \\ 0 & 32 \end{pmatrix}\begin{pmatrix} -1 & 1 \\ 2 & -1 \end{pmatrix}$

$= \begin{pmatrix} 1 & 32 \\ 2 & 32 \end{pmatrix}\begin{pmatrix} -1 & 1 \\ 2 & -1 \end{pmatrix} = \begin{pmatrix} -1 + 64 & 1 - 32 \\ -2 + 64 & 2 - 32 \end{pmatrix} = \begin{pmatrix} 63 & -31 \\ 62 & -30 \end{pmatrix}$

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Common Pitfalls

1. Matrix multiplication is not commutative. $AB \ne BA$ in general. When composing transformations, the order matters: the matrix for "transform $A$ then transform $B$" is $BA$, not $AB$.

2. Forgetting the factor of $\lambda^n$ in $\det(\lambda A)$. Scaling a matrix by $\lambda$ multiplies the determinant by $\lambda^n$, where $n$ is the dimension. A common mistake is to write $\det(\lambda A) = \lambda\det(A)$ instead of $\lambda^n\det(A)$.

3. Wrong sign in the $2 \times 2$ inverse. The inverse of $\begin{pmatrix} a & b \\ c & d \end{pmatrix}$ is $\dfrac{1}{ad - bc}\begin{pmatrix} d & -b \\ -c & a \end{pmatrix}$. The $-b$ and $-c$ are easily misplaced.

4. Confusing algebraic and geometric multiplicity. The algebraic multiplicity of an eigenvalue is its multiplicity as a root of the characteristic equation. The geometric multiplicity is the dimension of the corresponding eigenspace. The geometric multiplicity is always less than or equal to the algebraic multiplicity.

5. Assuming every matrix is diagonalisable. Only matrices with a full set of linearly independent eigenvectors can be diagonalised. Matrices with defective eigenvalues (geometric multiplicity less than algebraic multiplicity) require the Jordan normal form instead.

6. Row operation errors in Gauss-Jordan elimination. When finding the inverse via augmented matrices, remember that row operations must be applied to both sides simultaneously. A single arithmetic mistake propagates through all remaining steps.

7. Incorrect cofactor signs. The sign of the cofactor $C_{ij}$ is $(-1)^{i+j}$. This follows a checkerboard pattern starting with $+$ in the top-left corner. A common error is to use all positive signs.

8. Misidentifying invariant lines. An invariant line is mapped to itself, but points on the line may move. To find invariant lines, solve $(A - \lambda I)\mathbf{'\{'}v{'\}'} = \mathbf{'\{'}0{'\}'}$ where $\lambda$ is an eigenvalue. Every eigenvector lies on an invariant line through the origin.

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Exam-Style Problems

1. Compute $AB - BA$ where $A = \begin{pmatrix} 1 & 2 \\ 0 & 3 \end{pmatrix}$ and $B = \begin{pmatrix} 4 & -1 \\ 1 & 2 \end{pmatrix}$.

2. Find the inverse of $\begin{pmatrix} 1 & 0 & 2 \\ -1 & 3 & 1 \\ 2 & 1 & 0 \end{pmatrix}$ using row reduction, and verify your answer by multiplication.

3. Find the eigenvalues and eigenvectors of $A = \begin{pmatrix} 5 & 4 \\ 1 & 2 \end{pmatrix}$. Use diagonalisation to find $A^6$.

4. A triangle has vertices at $(0, 0)$, $(2, 0)$, and $(1, 3)$. Find the matrix of the transformation that reflects the triangle in the $y$-axis and then enlarges it by a factor of $2$ about the origin. Determine the area of the image.

5. Determine the values of $k$ for which the system $\begin{cases} x + ky + z = 1 \\ x + y + kz = k \\ kx + y + z = 1 \end{cases}$ has (a) a unique solution, (b) no solution, (c) infinitely many solutions.

6. Find the characteristic equation of $A = \begin{pmatrix} 2 & 1 & 0 \\ 0 & 2 & 1 \\ 0 & 0 & 2 \end{pmatrix}$ and explain why $A$ is not diagonalisable.

7. The matrix $T = \begin{pmatrix} \cos\theta & -\sin\theta \\ \sin\theta & \cos\theta \end{pmatrix}$ represents a rotation. Find the eigenvalues when $\theta = \dfrac{\pi}{2}$ and interpret them geometrically.

8. Use Cramer's rule to solve $\begin{cases} 3x + 2y - z = 4 \\ x - y + 2z = -1 \\ 2x + 3y + z = 7 \end{cases}$.

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Cross-References

• Differential equations and systems solved via matrix methods: see Differential Equations
• Proof and reasoning for properties of determinants: see Proof

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tip

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