Matrices

Matrix Fundamentals

Definition and Notation

An $m \times n$ matrix $A$ is a rectangular array of real numbers with $m$ rows and $n$ columns:

$A = \begin{pmatrix} a_{11} & a_{12} & \cdots & a_{1n} \\ a_{21} & a_{22} & \cdots & a_{2n} \\ \vdots & \vdots & \ddots & \vdots \\ a_{m1} & a_{m2} & \cdots & a_{mn} \end{pmatrix}$

The entry in row $i$ and column $j$ is $a_{ij}$. When $m = n$, $A$ is a square matrix of order $n$.

The identity matrix $I_n$ is the $n \times n$ matrix with $1$s on the diagonal and $0$s elsewhere. For any compatible matrix $A$: $AI = IA = A$.

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Matrix Operations

Addition and Scalar Multiplication

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

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

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:

• Associative: $(AB)C = A(BC)$
• Distributive: $A(B + C) = AB + AC$
• NOT commutative: $AB \ne BA$ in general
• Transpose of product: $(AB)^T = B^T A^T$

Transpose

The transpose $A^T$ is obtained by reflecting $A$ across its main diagonal: $(A^T)_{ij} = a_{ji}$.

Properties:

• $(A^T)^T = A$
• $(A + B)^T = A^T + B^T$
• $(\lambda A)^T = \lambda A^T$
• $(AB)^T = B^T A^T$

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Determinants

$2 \times 2$ Determinant

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

$3 \times 3$ Determinant

Expand along any row or column. Expanding along row 1:

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

The signs follow the checkerboard pattern $+ - +$ along row 1, $- + -$ along row 2, etc.

Properties of Determinants

• $\det(AB) = \det(A)\det(B)$
• $\det(A^T) = \det(A)$
• $\det(\lambda A) = \lambda^n \det(A)$ for an $n \times n$ matrix
• Swapping two rows multiplies the determinant by $-1$
• A matrix with a zero row (or column) has determinant $0$
• A triangular matrix has determinant equal to the product of its diagonal entries

Singular and Non-Singular Matrices

A square matrix $A$ is singular if $\det(A) = 0$ and non-singular (invertible) if $\det(A) \ne 0$.

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Inverse Matrices

Definition

The inverse of a square matrix $A$ is the matrix $A^{-1}$ satisfying:

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

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

$2 \times 2$ Inverse

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

Adjugate Method ($3 \times 3$)

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

where the adjugate (classical adjoint) is the transpose of the cofactor matrix. The $(i, j)$-cofactor is $C_{ij} = (-1)^{i+j} M_{ij}$, where $M_{ij}$ is the $(i, j)$-minor (determinant of the submatrix obtained by deleting row $i$ and column $j$).

Properties of Inverses

• $(A^{-1})^{-1} = A$
• $(AB)^{-1} = B^{-1}A^{-1}$
• $(A^T)^{-1} = (A^{-1})^T$
• $(\lambda A)^{-1} = \dfrac{1}{\lambda} A^{-1}$, $\lambda \ne 0$

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Solving Systems of Linear Equations

Matrix Form

A system of $n$ linear equations in $n$ unknowns:

$A\mathbf{'\{'}x{'\}'} = \mathbf{'\{'}b{'\}'}$

has the unique solution $\mathbf{'\{'}x{'\}'} = A^{-1}\mathbf{'\{'}b{'\}'}$ when $\det(A) \ne 0$.

Cramer's Rule

For the system $A\mathbf{'\{'}x{'\}'} = \mathbf{'\{'}b{'\}'}$ where $A$ is $n \times n$ and $\det(A) \ne 0$:

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

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

Gaussian Elimination

For larger systems, use elementary row operations to reduce the augmented matrix $[A \mid \mathbf{'\{'}b{'\}'}]$ to row echelon form, then solve by back-substitution. The three elementary operations are:

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

Existence and Uniqueness

• $\det(A) \ne 0$: unique solution
• $\det(A) = 0$ and system is consistent: infinitely many solutions
• $\det(A) = 0$ and system is inconsistent: no solution

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Geometric Transformations

$2 \times 2$ Transformation Matrices

Each transformation matrix $T$ maps a point $(x, y)$ to a new point by left-multiplication:

$\begin{pmatrix} x' \\ y' \end{pmatrix} = T\begin{pmatrix} x \\ y \end{pmatrix}$

Transformation	Matrix
Reflection in $x$-axis	$\begin{pmatrix} 1 & 0 \\ 0 & -1 \end{pmatrix}$
Reflection in $y$-axis	$\begin{pmatrix} -1 & 0 \\ 0 & 1 \end{pmatrix}$
Reflection in $y = x$	$\begin{pmatrix} 0 & 1 \\ 1 & 0 \end{pmatrix}$
Rotation by $\theta$ about origin	$\begin{pmatrix} \cos\theta & -\sin\theta \\ \sin\theta & \cos\theta \end{pmatrix}$
Enlargement by factor $k$	$\begin{pmatrix} k & 0 \\ 0 & k \end{pmatrix}$
Horizontal stretch by factor $k$	$\begin{pmatrix} k & 0 \\ 0 & 1 \end{pmatrix}$
Vertical stretch by factor $k$	$\begin{pmatrix} 1 & 0 \\ 0 & k \end{pmatrix}$
Shear parallel to $x$-axis, factor $k$	$\begin{pmatrix} 1 & k \\ 0 & 1 \end{pmatrix}$

Determinant and Area

For a $2 \times 2$ transformation matrix $T$:

• $|\det(T)|$ is the area scale factor
• $\det(T) \gt 0$: orientation preserved
• $\det(T) \lt 0$: orientation reversed

Composition of Transformations

Applying transformation $A$ followed by $B$ corresponds to the matrix product $BA$ (right to left order):

$\mathbf{'\{'}v{'\}'}' = B(A\mathbf{'\{'}v{'\}'}) = (BA)\mathbf{'\{'}v{'\}'}$

Invariant Points and Lines

An invariant point satisfies $T\mathbf{'\{'}v{'\}'} = \mathbf{'\{'}v{'\}'}$, i.e. $(T - I)\mathbf{'\{'}v{'\}'} = \mathbf{'\{'}0{'\}'}$. An invariant line is mapped to itself (points may move along the line).

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Eigenvalues and Eigenvectors (HL)

Definition

Let $A$ be an $n \times n$ matrix. A scalar $\lambda$ is an eigenvalue of $A$ if there exists a nonzero vector $\mathbf{'\{'}v{'\}'}$ (an eigenvector) such that:

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

Geometrically, $A$ stretches or compresses the eigenvector $\mathbf{'\{'}v{'\}'}$ by factor $\lambda$ without changing its direction.

Characteristic Equation

Rearranging $A\mathbf{'\{'}v{'\}'} = \lambda\mathbf{'\{'}v{'\}'}$ gives $(A - \lambda I)\mathbf{'\{'}v{'\}'} = \mathbf{'\{'}0{'\}'}$. For a nonzero solution $\mathbf{'\{'}v{'\}'}$, we require:

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

This is the characteristic equation. Its roots are the eigenvalues of $A$.

For a $2 \times 2$ matrix $A = \begin{pmatrix} a & b \\ c & d \end{pmatrix}$:

$\det\begin{pmatrix} a - \lambda & b \\ c & d - \lambda \end{pmatrix} = (a - \lambda)(d - \lambda) - bc = 0$

$\lambda^2 - (a + d)\lambda + (ad - bc) = 0$

Note: $a + d = \mathrm{tr}(A)$ (the trace) and $ad - bc = \det(A)$, so:

$\lambda^2 - \mathrm{tr}(A)\,\lambda + \det(A) = 0$

Finding Eigenvectors

For each eigenvalue $\lambda_i$, substitute into $(A - \lambda_i I)\mathbf{'\{'}v{'\}'} = \mathbf{'\{'}0{'\}'}$ and solve for $\mathbf{'\{'}v{'\}'}$. The solution space is the eigenspace corresponding to $\lambda_i$.

Properties

• The trace equals the sum of eigenvalues: $\mathrm{tr}(A) = \sum \lambda_i$
• The determinant equals the product of eigenvalues: $\det(A) = \prod \lambda_i$
• If $A$ is symmetric ($A = A^T$), all eigenvalues are real and eigenvectors corresponding to distinct eigenvalues are orthogonal.
• If $A$ has $n$ linearly independent eigenvectors, it is diagonalisable: $A = PDP^{-1}$ where $D$ is diagonal with eigenvalues on the diagonal.

Diagonalisation (HL Extension)

If $A$ has eigenvalues $\lambda_1, \ldots, \lambda_n$ and corresponding eigenvectors $\mathbf{'\{'}v{'\}'}_1, \ldots, \mathbf{'\{'}v{'\}'}_n$ forming a linearly independent set, then:

$A = PDP^{-1}$

where $P = \begin{pmatrix} \mathbf{'\{'}v{'\}'}_1 & \cdots & \mathbf{'\{'}v{'\}'}_n \end{pmatrix}$ and $D = \mathrm{diag}(\lambda_1, \ldots, \lambda_n)$.

This is useful for computing large powers: $A^n = PD^nP^{-1}$.

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Common Pitfalls

1. Matrix multiplication order. $AB \ne BA$ in general. When composing transformations, the rightmost matrix is applied first.

2. Transpose of a product. $(AB)^T = B^T A^T$, not $A^T B^T$. The order reverses.

3. Determinant of a scalar multiple. $\det(kA) = k^n\det(A)$ for an $n \times n$ matrix, not $k\det(A)$.

4. Forgetting to check invertibility. Only non-singular matrices have inverses. Always verify $\det(A) \ne 0$ before attempting to compute $A^{-1}$.

5. Eigenvector scaling. Eigenvectors are determined only up to a nonzero scalar multiple. Any nonzero multiple of an eigenvector is also an eigenvector.

6. Confusing invariant points with fixed points of composition. A point fixed by $A$ is not necessarily fixed by $BA$.

7. Cofactor sign errors. The $(i, j)$-cofactor sign is $(-1)^{i+j}$. Losing track of the sign pattern is the most common error in computing adjugates.

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Practice Problems

Problem 1

Let $A = \begin{pmatrix} 2 & 3 \\ 1 & 4 \end{pmatrix}$ and $B = \begin{pmatrix} 1 & -1 \\ 2 & 0 \end{pmatrix}$. Find $AB$ and $BA$. Is $AB = BA$?

Problem 2

Find $\det(A)$ and $A^{-1}$ where $A = \begin{pmatrix} 1 & 2 & 0 \\ 3 & 1 & -1 \\ 2 & 0 & 1 \end{pmatrix}$.

Problem 3

Solve the system using an inverse matrix: $x + 2y = 5$ $3x + 7y = 18$

Problem 4

Find the matrix representing a rotation of $90^\circ$ anticlockwise about the origin, and verify that applying it twice gives a rotation of $180^\circ$.

Problem 5

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

Problem 6

The transformation represented by matrix $T = \begin{pmatrix} 3 & 1 \\ 0 & 2 \end{pmatrix}$ is applied to the unit square. Find the area of the image.

Problem 7

Use Cramer's rule to solve: $2x - y + z = 3$ $x + 3y - z = 1$ $3x + y + 2z = 7$

Problem 8

For the matrix $A = \begin{pmatrix} 5 & -6 \\ 2 & -2 \end{pmatrix}$, find $A^{10}$ by diagonalisation.

Answers to Selected Problems

Problem 1: $AB = \begin{pmatrix} 8 & -2 \\ 9 & -1 \end{pmatrix}$, $BA = \begin{pmatrix} 1 & -1 \\ 4 & 6 \end{pmatrix}$. $AB \ne BA$, confirming non-commutativity.

Problem 2: Expanding along row 1: $\det(A) = 1(1 - 0) - 2(3 - (-2)) + 0 = 1 - 10 = -9$. Cofactors: $C_{11} = 1, C_{12} = -5, C_{13} = -2, C_{21} = -2, C_{22} = 1, C_{23} = 4, C_{31} = -2, C_{32} = 1, C_{33} = -5$. $A^{-1} = \dfrac{1}{-9}\begin{pmatrix} 1 & 2 & -2 \\ -5 & 1 & 1 \\ -2 & 4 & -5 \end{pmatrix} = \begin{pmatrix} -1/9 & -2/9 & 2/9 \\ 5/9 & -1/9 & -1/9 \\ 2/9 & -4/9 & 5/9 \end{pmatrix}$.

Problem 3: $A = \begin{pmatrix} 1 & 2 \\ 3 & 7 \end{pmatrix}$, $\det(A) = 7 - 6 = 1$. $A^{-1} = \begin{pmatrix} 7 & -2 \\ -3 & 1 \end{pmatrix}$. $\begin{pmatrix} x \\ y \end{pmatrix} = \begin{pmatrix} 7 & -2 \\ -3 & 1 \end{pmatrix}\begin{pmatrix} 5 \\ 18 \end{pmatrix} = \begin{pmatrix} 35 - 36 \\ -15 + 18 \end{pmatrix} = \begin{pmatrix} -1 \\ 3 \end{pmatrix}$. So $x = -1, y = 3$.

Problem 4: Rotation by $90^\circ$: $R = \begin{pmatrix} 0 & -1 \\ 1 & 0 \end{pmatrix}$. $R^2 = \begin{pmatrix} 0 & -1 \\ 1 & 0 \end{pmatrix}\begin{pmatrix} 0 & -1 \\ 1 & 0 \end{pmatrix} = \begin{pmatrix} -1 & 0 \\ 0 & -1 \end{pmatrix}$, which is a $180^\circ$ rotation.

Problem 5: Characteristic equation: $\lambda^2 - 7\lambda + 10 = 0$, so $\lambda = 2$ or $\lambda = 5$. For $\lambda = 2$: $(A - 2I)\mathbf{'\{'}v{'\}'} = \mathbf{'\{'}0{'\}'} \implies \begin{pmatrix} 2 & 1 \\ 2 & 1 \end{pmatrix}\mathbf{'\{'}v{'\}'} = \mathbf{'\{'}0{'\}'} \implies \mathbf{'\{'}v{'\}'} = \begin{pmatrix} 1 \\ -2 \end{pmatrix}$ (up to scalar). For $\lambda = 5$: $(A - 5I)\mathbf{'\{'}v{'\}'} = \mathbf{'\{'}0{'\}'} \implies \begin{pmatrix} -1 & 1 \\ 2 & -2 \end{pmatrix}\mathbf{'\{'}v{'\}'} = \mathbf{'\{'}0{'\}'} \implies \mathbf{'\{'}v{'\}'} = \begin{pmatrix} 1 \\ 1 \end{pmatrix}$.

Problem 6: Area $= |\det(T)| = |6 - 0| = 6$.

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Worked Examples

Worked Example: Finding the Inverse of a $3 \times 3$ Matrix

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

Solution

Step 1: Compute $\det(A)$. Expanding along row 1:

$\det(A) = 2\begin{vmatrix} 3 & -1 \\ 1 & 2 \end{vmatrix} - 1\begin{vmatrix} 1 & -1 \\ 0 & 2 \end{vmatrix} + 0 = 2(7) - 1(2) = 14 - 2 = 12$

Step 2: Compute the cofactor matrix.

$C_{11} = +\begin{vmatrix} 3 & -1 \\ 1 & 2 \end{vmatrix} = 7$, $C_{12} = -\begin{vmatrix} 1 & -1 \\ 0 & 2 \end{vmatrix} = -2$, $C_{13} = +\begin{vmatrix} 1 & 3 \\ 0 & 1 \end{vmatrix} = 1$

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

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

Cofactor matrix: $\begin{pmatrix} 7 & -2 & 1 \\ -2 & 4 & -2 \\ -1 & 2 & 5 \end{pmatrix}$.

Step 3: Transpose to get the adjugate.

$\mathrm{adj}(A) = \begin{pmatrix} 7 & -2 & -1 \\ -2 & 4 & 2 \\ 1 & -2 & 5 \end{pmatrix}$

Step 4: Divide by the determinant.

$A^{-1} = \frac{1}{12}\begin{pmatrix} 7 & -2 & -1 \\ -2 & 4 & 2 \\ 1 & -2 & 5 \end{pmatrix}$

Worked Example: Composition of Transformations

Find the matrix for the transformation that first reflects in the line $y = x$, then rotates $90^\circ$ anticlockwise about the origin. Verify the result by applying the combined transformation to the point $(3, 1)$.

Solution

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

Rotation $90^\circ$ anticlockwise: $R_\theta = \begin{pmatrix} 0 & -1 \\ 1 & 0 \end{pmatrix}$.

The combined transformation (reflect first, then rotate) is:

$T = R_\theta \cdot R_f = \begin{pmatrix} 0 & -1 \\ 1 & 0 \end{pmatrix}\begin{pmatrix} 0 & 1 \\ 1 & 0 \end{pmatrix} = \begin{pmatrix} -1 & 0 \\ 0 & 1 \end{pmatrix}$

This is a reflection in the $y$-axis.

Verification: $(3, 1)$ reflected in $y = x$ gives $(1, 3)$. Rotating $(1, 3)$ by $90^\circ$ anticlockwise: $R_\theta\begin{pmatrix} 1 \\ 3 \end{pmatrix} = \begin{pmatrix} -3 \\ 1 \end{pmatrix}$.

Direct application: $T\begin{pmatrix} 3 \\ 1 \end{pmatrix} = \begin{pmatrix} -3 \\ 1 \end{pmatrix}$. Confirmed.

Worked Example: Solving a System Using Gaussian Elimination

Solve the system using Gaussian elimination:

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

Solution

Augmented matrix:

$\begin{pmatrix} 2 & 1 & -1 & | & 3 \\ 1 & -1 & 2 & | & 6 \\ 3 & 2 & 1 & | & 1 \end{pmatrix}$

$R_1 \leftrightarrow R_2$:

$\begin{pmatrix} 1 & -1 & 2 & | & 6 \\ 2 & 1 & -1 & | & 3 \\ 3 & 2 & 1 & | & 1 \end{pmatrix}$

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

$\begin{pmatrix} 1 & -1 & 2 & | & 6 \\ 0 & 3 & -5 & | & -9 \\ 0 & 5 & -5 & | & -17 \end{pmatrix}$

$R_2 \to R_2/3$:

$\begin{pmatrix} 1 & -1 & 2 & | & 6 \\ 0 & 1 & -5/3 & | & -3 \\ 0 & 5 & -5 & | & -17 \end{pmatrix}$

$R_3 \to R_3 - 5R_2$:

$\begin{pmatrix} 1 & -1 & 2 & | & 6 \\ 0 & 1 & -5/3 & | & -3 \\ 0 & 0 & 20/3 & | & -2 \end{pmatrix}$

Back-substitution: $R_3 \implies z = -2 \cdot \dfrac{3}{20} = -\dfrac{3}{10}$.

$R_2$: $y - \dfrac{5}{3}\!\left(-\dfrac{3}{10}\right) = -3 \implies y + \dfrac{1}{2} = -3 \implies y = -\dfrac{7}{2}$.

$R_1$: $x - \left(-\dfrac{7}{2}\right) + 2\!\left(-\dfrac{3}{10}\right) = 6 \implies x + \dfrac{7}{2} - \dfrac{3}{5} = 6 \implies x = 6 - \dfrac{29}{10} = \dfrac{31}{10}$.

Solution: $x = \dfrac{31}{10}$, $y = -\dfrac{7}{2}$, $z = -\dfrac{3}{10}$.

Worked Example: Diagonalisation and Large Powers

For $A = \begin{pmatrix} 3 & 1 \\ 0 & 2 \end{pmatrix}$, find $A^5$ using diagonalisation.

Solution

The matrix is already upper triangular, so eigenvalues are $\lambda_1 = 3$ and $\lambda_2 = 2$.

For $\lambda_1 = 3$: $(A - 3I)\mathbf{'\{'}v{'\}'} = \mathbf{'\{'}0{'\}'} \implies \begin{pmatrix} 0 & 1 \\ 0 & -1 \end{pmatrix}\mathbf{'\{'}v{'\}'} = \mathbf{'\{'}0{'\}'} \implies \mathbf{'\{'}v{'\}'}_1 = \begin{pmatrix} 1 \\ 0 \end{pmatrix}$.

For $\lambda_2 = 2$: $(A - 2I)\mathbf{'\{'}v{'\}'} = \mathbf{'\{'}0{'\}'} \implies \begin{pmatrix} 1 & 1 \\ 0 & 0 \end{pmatrix}\mathbf{'\{'}v{'\}'} = \mathbf{'\{'}0{'\}'} \implies \mathbf{'\{'}v{'\}'}_2 = \begin{pmatrix} 1 \\ -1 \end{pmatrix}$.

$P = \begin{pmatrix} 1 & 1 \\ 0 & -1 \end{pmatrix}$, $D = \begin{pmatrix} 3 & 0 \\ 0 & 2 \end{pmatrix}$.

$P^{-1} = \dfrac{1}{-1}\begin{pmatrix} -1 & -1 \\ 0 & 1 \end{pmatrix} = \begin{pmatrix} 1 & 1 \\ 0 & -1 \end{pmatrix}$.

$A^5 = PD^5P^{-1} = \begin{pmatrix} 1 & 1 \\ 0 & -1 \end{pmatrix}\begin{pmatrix} 243 & 0 \\ 0 & 32 \end{pmatrix}\begin{pmatrix} 1 & 1 \\ 0 & -1 \end{pmatrix}$

$= \begin{pmatrix} 243 & 32 \\ 0 & -32 \end{pmatrix}\begin{pmatrix} 1 & 1 \\ 0 & -1 \end{pmatrix} = \begin{pmatrix} 243 & 211 \\ 0 & 32 \end{pmatrix}$.

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Additional Common Pitfalls

• Inverse of a product order. $(AB)^{-1} = B^{-1}A^{-1}$, not $A^{-1}B^{-1}$. The order reverses, just as with the transpose.

• Gaussian elimination arithmetic errors. A single arithmetic mistake in row operations propagates through all subsequent steps. Always verify the solution by substituting back into the original equations.

• Misidentifying eigenvectors. For each eigenvalue, solve $(A - \lambda I)\mathbf{'\{'}v{'\}'} = \mathbf{'\{'}0{'\}'}$. Substituting the eigenvalue into the original matrix $A$ instead of $A - \lambda I$ is a common error.

• Transformation composition order. The matrix $BA$ represents "apply $A$ first, then $B$." Writing $AB$ when $B$ should be applied first is the single most common transformation error.

• Determinant and area relationship for reflections. A reflection matrix has $\det(T) = -1$. The area scale factor is $|\det(T)| = 1$, but the negative sign indicates orientation reversal. The area is always positive; the sign encodes geometric information.

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Exam-Style Problems

Problem 9

Given $A = \begin{pmatrix} 1 & 2 \\ 3 & 4 \end{pmatrix}$ and $B = \begin{pmatrix} 0 & -1 \\ 1 & 2 \end{pmatrix}$, verify that $(AB)^{-1} = B^{-1}A^{-1}$.

Problem 10

Find the invariant points and invariant lines of the transformation represented by $T = \begin{pmatrix} 3 & 1 \\ 0 & 2 \end{pmatrix}$.

Problem 11

The matrix $A = \begin{pmatrix} 1 & 0 & 2 \\ 0 & 1 & -1 \\ 0 & 0 & 1 \end{pmatrix}$ represents a transformation in 3D. Describe the geometric effect of $A$ and find $A^n$ for any positive integer $n$.

Problem 12

Use Cramer's rule to solve: $3x + 2y - z = 4$, $x - y + 3z = 5$, $2x + y + z = 3$.

Problem 13

The transformation $T$ consists of an enlargement by factor $2$ followed by a rotation of $45^\circ$ anticlockwise. Find the matrix $T$ and the area scale factor.

Problem 14

For $A = \begin{pmatrix} 2 & 1 \\ 1 & 2 \end{pmatrix}$, find the eigenvalues and eigenvectors. Verify that the eigenvectors are orthogonal and diagonalise $A$.

Problem 15

Determine all values of $k$ for which the system has (a) a unique solution, (b) no solution, (c) infinitely many solutions:

$x + 2y + kz = 1$ $2x + ky + 2z = 2$ $kx + 2y + z = 3$

Answers to Additional Problems

Problem 9: $AB = \begin{pmatrix} 2 & 3 \\ 4 & 5 \end{pmatrix}$, $\det(AB) = 10 - 12 = -2$. $(AB)^{-1} = \dfrac{1}{-2}\begin{pmatrix} 5 & -3 \\ -4 & 2 \end{pmatrix} = \begin{pmatrix} -5/2 & 3/2 \\ 2 & -1 \end{pmatrix}$. $A^{-1} = \dfrac{1}{-2}\begin{pmatrix} 4 & -2 \\ -3 & 1 \end{pmatrix} = \begin{pmatrix} -2 & 1 \\ 3/2 & -1/2 \end{pmatrix}$. $B^{-1} = \dfrac{1}{1}\begin{pmatrix} 2 & 1 \\ -1 & 0 \end{pmatrix} = \begin{pmatrix} 2 & 1 \\ -1 & 0 \end{pmatrix}$. $B^{-1}A^{-1} = \begin{pmatrix} 2 & 1 \\ -1 & 0 \end{pmatrix}\begin{pmatrix} -2 & 1 \\ 3/2 & -1/2 \end{pmatrix} = \begin{pmatrix} -5/2 & 3/2 \\ 2 & -1 \end{pmatrix}$. Verified: $(AB)^{-1} = B^{-1}A^{-1}$.

Problem 10: Invariant points: $(T - I)\mathbf{'\{'}v{'\}'} = \mathbf{'\{'}0{'\}'} \implies \begin{pmatrix} 2 & 1 \\ 0 & 1 \end{pmatrix}\mathbf{'\{'}v{'\}'} = \mathbf{'\{'}0{'\}'} \implies 2x + y = 0$ and $y = 0$, so $x = 0$, $y = 0$. Only the origin is invariant. For invariant lines $y = mx + c$: $T\begin{pmatrix} x \\ mx + c \end{pmatrix}$ must lie on $y = mx + c$. This gives conditions leading to the invariant line $y = 0$ (the $x$-axis is mapped to itself).

Problem 11: $A$ is an upper triangular matrix with $1$s on the diagonal. This represents a shear transformation. For any $n$, $A^n = \begin{pmatrix} 1 & 0 & 2n \\ 0 & 1 & -n \\ 0 & 0 & 1 \end{pmatrix}$, which can be verified by induction.

Problem 12: $\det(A) = 3(-1 - 3) - 2(1 - 6) + (-1)(1 - (-2k)) = -12 + 10 - 1 + 2k = 2k - 3$. $A_x = \begin{pmatrix} 4 & 2 & -1 \\ 5 & -1 & 3 \\ 3 & 1 & 1 \end{pmatrix}$, $\det(A_x) = 4(-1-3) - 2(5-9) + (-1)(5+3) = -16 + 8 - 8 = -16$. $A_y = \begin{pmatrix} 3 & 4 & -1 \\ 1 & 5 & 3 \\ 2 & 3 & 1 \end{pmatrix}$, $\det(A_y) = 3(5-9) - 4(1-6) + (-1)(3-10) = -12 + 20 + 7 = 15$. $A_z = \begin{pmatrix} 3 & 2 & 4 \\ 1 & -1 & 5 \\ 2 & 1 & 3 \end{pmatrix}$, $\det(A_z) = 3(-3-5) - 2(3-10) + 4(1+2) = -24 + 14 + 12 = 2$. $x = -16/(2k-3)$, $y = 15/(2k-3)$, $z = 2/(2k-3)$.

Problem 13: Enlargement by $2$: $E = \begin{pmatrix} 2 & 0 \\ 0 & 2 \end{pmatrix}$. Rotation by $45^\circ$: $R = \begin{pmatrix} \cos 45^\circ & -\sin 45^\circ \\ \sin 45^\circ & \cos 45^\circ \end{pmatrix} = \dfrac{1}{\sqrt{2}}\begin{pmatrix} 1 & -1 \\ 1 & 1 \end{pmatrix}$. $T = R \cdot E = \dfrac{1}{\sqrt{2}}\begin{pmatrix} 1 & -1 \\ 1 & 1 \end{pmatrix}\begin{pmatrix} 2 & 0 \\ 0 & 2 \end{pmatrix} = \dfrac{2}{\sqrt{2}}\begin{pmatrix} 1 & -1 \\ 1 & 1 \end{pmatrix} = \sqrt{2}\begin{pmatrix} 1 & -1 \\ 1 & 1 \end{pmatrix}$. Area scale factor $= |\det(T)| = |2 - (-2)| = 4$.

Problem 14: Characteristic equation: $\lambda^2 - 4\lambda + 3 = 0 \implies \lambda = 1$ or $\lambda = 3$. For $\lambda = 1$: $\begin{pmatrix} 1 & 1 \\ 1 & 1 \end{pmatrix}\mathbf{'\{'}v{'\}'} = \mathbf{'\{'}0{'\}'} \implies \mathbf{'\{'}v{'\}'}_1 = \begin{pmatrix} 1 \\ -1 \end{pmatrix}$. For $\lambda = 3$: $\begin{pmatrix} -1 & 1 \\ 1 & -1 \end{pmatrix}\mathbf{'\{'}v{'\}'} = \mathbf{'\{'}0{'\}'} \implies \mathbf{'\{'}v{'\}'}_2 = \begin{pmatrix} 1 \\ 1 \end{pmatrix}$. Orthogonality check: $\mathbf{'\{'}v{'\}'}_1 \cdot \mathbf{'\{'}v{'\}'}_2 = 1 - 1 = 0$. Confirmed. $A = PDP^{-1}$ where $P = \begin{pmatrix} 1 & 1 \\ -1 & 1 \end{pmatrix}$, $D = \begin{pmatrix} 1 & 0 \\ 0 & 3 \end{pmatrix}$.

Problem 15: $\det(A) = 1(k - 4) - 2(2 - 2k) + k(4 - k^2) = k - 4 - 4 + 4k + 4k - k^3 = -k^3 + 9k - 8$. Factor: $-k^3 + 9k - 8 = -(k-1)(k^2 + k - 8)$. The discriminant of $k^2 + k - 8$ is $1 + 32 = 33$. (a) Unique solution when $\det(A) \ne 0$: $k \ne 1$ and $k \ne \dfrac{-1 \pm \sqrt{33}}{2}$. (b) No solution when $\det(A) = 0$ and the system is inconsistent. (c) Infinitely many solutions when $\det(A) = 0$ and the system is consistent.

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If You Get These Wrong, Revise:

• Solving simultaneous equations → Review ./matrices (sections on Solving Systems and Cramer's Rule)
• Trigonometry for rotation matrices → Review geometry and trigonometry topics
• Algebraic manipulation and factorisation → Review algebra fundamentals
• Vectors and linear independence → Review ./vectors for related concepts
• Determinants and their geometric meaning → Review ./matrices (section on Determinants and Area)

For the A-Level Further Maths treatment of this topic, see Matrices.