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A-Level Further Mathematics – Chapter 3: Matrices

Edexcel · AQA · OCR A-Level Further Mathematics · Updated March 2026

Matrices are rectangular arrays of numbers that encode linear transformations, systems of equations, and a vast range of algebraic structures. In Further Mathematics, you will learn to multiply matrices, compute determinants, find inverses, and use these tools to solve simultaneous equations and analyse geometric transformations. The algebraic richness of matrices — they do not commute in general, they can fail to be invertible, and their determinants carry geometric meaning — makes them both challenging and rewarding to master.

Specification Note

All content in this chapter is Further level. The 2×2 material (Sections 3.1–3.4) appears on all Further Pure papers; the 3×3 material (Section 3.5) is required for Edexcel FP1 and is also examined by AQA and OCR. Simultaneous equations and invariant lines (Sections 3.6–3.7) are high-priority examination topics.

Contents

  1. 3.1 Matrix Algebra
  2. 3.2 Matrix Multiplication
  3. 3.3 Determinant of a 2×2 Matrix
  4. 3.4 Inverse of a 2×2 Matrix
  5. 3.5 Determinant and Inverse of a 3×3 Matrix
  6. 3.6 Solving Simultaneous Equations
  7. 3.7 Invariant Points and Lines
  8. Practice Problems

3.1 Matrix Algebra Further

Definition 3.1 — Matrix

A matrix is a rectangular array of numbers. An $m \times n$ matrix has $m$ rows and $n$ columns; element $a_{ij}$ lies in row $i$ and column $j$:

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

The zero matrix $\mathbf{0}$ has all entries equal to zero. The $n \times n$ identity matrix $I_n$ has $1$s on the main diagonal and $0$s elsewhere.

Addition, Subtraction, Scalar Multiplication

Two matrices may be added or subtracted only if they have the same order. Addition is performed entry-wise: $(A + B)_{ij} = a_{ij} + b_{ij}$. Scalar multiplication: $(\lambda A)_{ij} = \lambda a_{ij}$.

Example 3.1 — Addition and Scalar Multiplication

Given $A = \begin{pmatrix}1 & 3\\-2 & 4\end{pmatrix}$ and $B = \begin{pmatrix}0 & -1\\5 & 2\end{pmatrix}$, find $2A - B$.

$2A = \begin{pmatrix}2 & 6\\-4 & 8\end{pmatrix}$, so $2A - B = \begin{pmatrix}2-0 & 6-(-1)\\-4-5 & 8-2\end{pmatrix} = \begin{pmatrix}2 & 7\\-9 & 6\end{pmatrix}$.

3.2 Matrix Multiplication Further

Definition 3.2 — Matrix Multiplication

The product $AB$ is defined if and only if the number of columns of $A$ equals the number of rows of $B$ (conformability condition). If $A$ is $m \times p$ and $B$ is $p \times n$, then $AB$ is $m \times n$ with:

$$(AB)_{ij} = \sum_{k=1}^{p} a_{ik}\, b_{kj}.$$

In words: the $(i,j)$ entry of $AB$ is the dot product of row $i$ of $A$ with column $j$ of $B$.

Key Property — Non-Commutativity

In general $AB \neq BA$. Always check the order of multiplication. However, matrix multiplication is associative: $(AB)C = A(BC)$, and $AI = IA = A$ for any matrix $A$ and identity matrix $I$ of appropriate size.

Example 3.2 — 2×2 Matrix Product

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

$AB = \begin{pmatrix}2(1)+1(2) & 2(4)+1(0)\\3(1)+(-1)(2) & 3(4)+(-1)(0)\end{pmatrix} = \begin{pmatrix}4 & 8\\1 & 12\end{pmatrix}$.

$BA = \begin{pmatrix}1(2)+4(3) & 1(1)+4(-1)\\2(2)+0(3) & 2(1)+0(-1)\end{pmatrix} = \begin{pmatrix}14 & -3\\4 & 2\end{pmatrix}$.

Since $AB \neq BA$, multiplication is non-commutative in this case.

Example 3.3 — Power of a Matrix

Given $A = \begin{pmatrix}1 & 1\\0 & 2\end{pmatrix}$, find $A^3$.

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

$A^3 = A^2 \cdot A = \begin{pmatrix}1 & 3\\0 & 4\end{pmatrix}\begin{pmatrix}1 & 1\\0 & 2\end{pmatrix} = \begin{pmatrix}1 & 7\\0 & 8\end{pmatrix}$.

3.3 Determinant of a 2×2 Matrix Further

Definition 3.3 — Determinant of a 2×2 Matrix

For $A = \begin{pmatrix}a & b\\c & d\end{pmatrix}$, the determinant is:

$$\det(A) = |A| = ad - bc.$$

A matrix is singular if $\det(A) = 0$, and non-singular if $\det(A) \neq 0$.

Geometric Meaning

The matrix $A$ represents a linear transformation of $\mathbb{R}^2$. The signed area of the image of the unit square under $A$ equals $\det(A)$. If $\det(A) < 0$, the transformation reverses orientation.

Theorem 3.1 — Multiplicativity of Determinant

For any two $n \times n$ matrices $A$ and $B$: $\det(AB) = \det(A)\det(B)$.

Corollary: if $A$ is non-singular, $\det(A^{-1}) = 1/\det(A)$.

Example 3.4 — Checking for Singularity and det(AB)

Let $A = \begin{pmatrix}3 & 1\\6 & 2\end{pmatrix}$ and $B = \begin{pmatrix}2 & -1\\3 & 4\end{pmatrix}$. Determine whether $A$ is singular, and find $\det(AB)$.

$\det(A) = 3(2) - 1(6) = 6 - 6 = 0$. So $A$ is singular.

$\det(B) = 2(4) - (-1)(3) = 8 + 3 = 11$.

$\det(AB) = \det(A)\det(B) = 0 \times 11 = 0$. The product $AB$ is also singular.

3.4 Inverse of a 2×2 Matrix Further

Theorem 3.2 — Inverse of a 2×2 Matrix

For a non-singular matrix $A = \begin{pmatrix}a & b\\c & d\end{pmatrix}$:

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

The matrix $\begin{pmatrix}d & -b\\-c & a\end{pmatrix}$ is called the adjugate (or adjoint) of $A$.

For products: $(AB)^{-1} = B^{-1}A^{-1}$ (note the reversed order).

Example 3.5 — Finding the Inverse

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

$\det(A) = 5(1) - 2(3) = 5 - 6 = -1$.

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

Verify: $AA^{-1} = \begin{pmatrix}5 & 2\\3 & 1\end{pmatrix}\begin{pmatrix}-1 & 2\\3 & -5\end{pmatrix} = \begin{pmatrix}-5+6 & 10-10\\-3+3 & 6-5\end{pmatrix} = \begin{pmatrix}1 & 0\\0 & 1\end{pmatrix} = I$. ✓

Example 3.6 — Solving $AX = B$ Using the Inverse

Solve for the matrix $X$ given $AX = B$ where $A = \begin{pmatrix}2 & 3\\1 & 2\end{pmatrix}$ and $B = \begin{pmatrix}7 & 0\\4 & 1\end{pmatrix}$.

$\det(A) = 4 - 3 = 1$, so $A^{-1} = \begin{pmatrix}2 & -3\\-1 & 2\end{pmatrix}$.

$X = A^{-1}B = \begin{pmatrix}2 & -3\\-1 & 2\end{pmatrix}\begin{pmatrix}7 & 0\\4 & 1\end{pmatrix} = \begin{pmatrix}14-12 & 0-3\\-7+8 & 0+2\end{pmatrix} = \begin{pmatrix}2 & -3\\1 & 2\end{pmatrix}$.

3.5 Determinant and Inverse of a 3×3 Matrix Further

Cofactor Expansion

The determinant of a $3 \times 3$ matrix is computed by expanding along any row or column using cofactors. Expanding along row 1:

$$\det(A) = a_{11}C_{11} + a_{12}C_{12} + a_{13}C_{13},$$

where the cofactor $C_{ij} = (-1)^{i+j} M_{ij}$ and $M_{ij}$ is the $2 \times 2$ minor obtained by deleting row $i$ and column $j$.

Definition 3.4 — Minor and Cofactor

The minor $M_{ij}$ of a matrix $A$ is the determinant of the submatrix formed by deleting row $i$ and column $j$. The cofactor is $C_{ij} = (-1)^{i+j}M_{ij}$. The sign pattern for a $3 \times 3$ matrix is:

$$\begin{pmatrix}+ & - & +\\- & + & -\\+ & - & +\end{pmatrix}.$$

Example 3.7 — 3×3 Determinant by Cofactor Expansion

Compute $\det(A)$ for $A = \begin{pmatrix}1 & 2 & 3\\4 & 5 & 6\\7 & 8 & 0\end{pmatrix}$.

Expanding along row 1:

$\det(A) = 1 \cdot \det\begin{pmatrix}5 & 6\\8 & 0\end{pmatrix} - 2 \cdot \det\begin{pmatrix}4 & 6\\7 & 0\end{pmatrix} + 3 \cdot \det\begin{pmatrix}4 & 5\\7 & 8\end{pmatrix}$

$= 1(0 - 48) - 2(0 - 42) + 3(32 - 35)$

$= -48 + 84 - 9 = 27$.

Inverse of a 3×3 Matrix via the Adjugate

The inverse is found using the cofactor matrix (matrix of cofactors), then taking its transpose to form the adjugate matrix:

$$A^{-1} = \frac{1}{\det(A)}\,\mathrm{adj}(A), \qquad \mathrm{adj}(A) = \left(C_{ij}\right)^{\mathrm{T}}.$$

Example 3.8 — 3×3 Inverse

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

Step 1 Compute $\det(A)$ by expansion along row 1:

$\det(A) = 2(3 \cdot 2 - 1 \cdot 1) - 1(1 \cdot 2 - 1 \cdot 0) + 0 = 2(5) - 1(2) = 10 - 2 = 8$.

Step 2 Compute all nine cofactors:

Step 3 Form the adjugate (transpose of cofactor matrix):

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

(In this case the cofactor matrix is symmetric, so the adjugate equals the cofactor matrix.)

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

3.6 Solving Simultaneous Equations Further

A system of $n$ linear equations in $n$ unknowns may be written as $A\mathbf{x} = \mathbf{b}$. The nature of the solution set is determined by $\det(A)$:

Theorem 3.3 — Classification of Solutions

For $2 \times 2$ systems, the geometric interpretation is: two lines in the plane either intersect at a unique point (unique solution), are parallel but distinct (no solution), or are the same line (infinitely many solutions).

For $3 \times 3$ systems, three planes in $\mathbb{R}^3$ may: meet at a unique point; form a triangular prism (no solution — parallel pairwise intersections); share a common line (infinitely many); or coincide on a plane.

Example 3.9 — Unique Solution via Inverse

Solve the system $2x + y = 7$, $x + 3y = 11$ using matrices.

Matrix form: $\begin{pmatrix}2 & 1\\1 & 3\end{pmatrix}\begin{pmatrix}x\\y\end{pmatrix} = \begin{pmatrix}7\\11\end{pmatrix}$.

$\det(A) = 6 - 1 = 5$. $A^{-1} = \dfrac{1}{5}\begin{pmatrix}3 & -1\\-1 & 2\end{pmatrix}$.

$\begin{pmatrix}x\\y\end{pmatrix} = \dfrac{1}{5}\begin{pmatrix}3 & -1\\-1 & 2\end{pmatrix}\begin{pmatrix}7\\11\end{pmatrix} = \dfrac{1}{5}\begin{pmatrix}21-11\\-7+22\end{pmatrix} = \dfrac{1}{5}\begin{pmatrix}10\\15\end{pmatrix} = \begin{pmatrix}2\\3\end{pmatrix}$.

Example 3.10 — Singular Case: Determining Consistency

For what values of $k$ does the system $kx + 2y = 4$, $3x + 6y = 12$ have (a) no solution, (b) infinitely many solutions?

Write as $A\mathbf{x} = \mathbf{b}$ with $A = \begin{pmatrix}k & 2\\3 & 6\end{pmatrix}$.

$\det(A) = 6k - 6$. Setting $\det(A) = 0$: $k = 1$.

When $k = 1$: Row 1 is $x + 2y = 4$; Row 2 is $3x + 6y = 12 = 3(x + 2y) = 12$. These are identical, so the system is consistent and dependent — infinitely many solutions for $k = 1$.

There is no value of $k$ giving no solution here (the second equation is a multiple of the first for all $k = 1$, but the right-hand sides are also proportional). For $k \neq 1$, there is a unique solution.

Example 3.11 — 3×3 System with Unique Solution

Solve $x + 2y + z = 4$, $2x + y + z = 3$, $x + y + 2z = 5$.

$A = \begin{pmatrix}1&2&1\\2&1&1\\1&1&2\end{pmatrix}$, $\mathbf{b} = \begin{pmatrix}4\\3\\5\end{pmatrix}$.

$\det(A) = 1(2-1) - 2(4-1) + 1(2-1) = 1 - 6 + 1 = -4 \neq 0$, so there is a unique solution.

Using row reduction (or the inverse): subtracting Row 1 from Row 3: $0 + (-1)y + z = 1$. Subtracting 2×Row 1 from Row 2: $-3y - z = -5$. Adding these two derived equations: $-4y = -4$, so $y = 1$. Then $z = 1 + y = 2$, and $x = 4 - 2(1) - 2 = 0$.

Solution: $x = 0, y = 1, z = 2$.

Figure 3.1 — A parallelogram with vertices at $(0,0)$, $(2,0)$, $(1,3)$, $(3,3)$ (grey) and its image under the matrix $\begin{pmatrix}2&1\\0&3\end{pmatrix}$ (blue). The image has area $= |\det A| \times \text{original area} = 6 \times 2 = 12$ square units.

3.7 Invariant Points and Lines Further

When a matrix $A$ represents a linear transformation, we ask: which points and lines are left unchanged by the transformation?

Definition 3.5 — Invariant Point and Invariant Line

Every linear transformation fixes the origin (since $A\mathbf{0} = \mathbf{0}$). Non-trivial invariant points exist when $(A - I)$ is singular.

Finding Invariant Points

Solve $(A - I)\mathbf{x} = \mathbf{0}$. If $\det(A - I) = 0$, there are infinitely many invariant points forming a line (an eigenspace for eigenvalue 1).

Example 3.12 — Finding Invariant Points

Find all invariant points under $A = \begin{pmatrix}3 & -2\\2 & -1\end{pmatrix}$.

$A - I = \begin{pmatrix}2 & -2\\2 & -2\end{pmatrix}$. $\det(A - I) = 2(-2) - (-2)(2) = -4 + 4 = 0$, so there is a line of invariant points.

$(A-I)\mathbf{x} = \mathbf{0}$: $2x - 2y = 0$, i.e. $x = y$. The line of invariant points is $y = x$.

Finding Invariant Lines

An invariant line $y = mx + c$ satisfies: if $(x, mx + c)$ is on the line, then $A\begin{pmatrix}x\\mx+c\end{pmatrix}$ also lies on a line of the same gradient $m$ (but possibly different intercept). This gives conditions on $m$ first, then on $c$.

Example 3.13 — Finding Invariant Lines

Find all invariant lines of the form $y = mx$ through the origin for $A = \begin{pmatrix}2 & 1\\1 & 2\end{pmatrix}$.

A line $y = mx$ through the origin is invariant if $A\begin{pmatrix}1\\m\end{pmatrix}$ is proportional to $\begin{pmatrix}1\\m\end{pmatrix}$, i.e. $\begin{pmatrix}2+m\\1+2m\end{pmatrix} = \lambda\begin{pmatrix}1\\m\end{pmatrix}$ for some scalar $\lambda$.

From the first component: $\lambda = 2 + m$. Substituting into the second: $1 + 2m = m(2 + m) = 2m + m^2$, so $m^2 = 1$, giving $m = 1$ or $m = -1$.

Invariant lines through the origin: $y = x$ (with scale factor $\lambda = 3$) and $y = -x$ (with scale factor $\lambda = 1$).

Example 3.14 — Invariant Lines Not Through the Origin

Find all invariant lines of $A = \begin{pmatrix}1 & 2\\0 & 1\end{pmatrix}$ (a shear).

Apply $A$ to a general point $(x, mx + c)$ on $y = mx + c$:

$A\begin{pmatrix}x\\mx+c\end{pmatrix} = \begin{pmatrix}x + 2(mx+c)\\mx+c\end{pmatrix} = \begin{pmatrix}(1+2m)x + 2c\\mx+c\end{pmatrix}$.

For this to lie on $y = mx + c$, we need $mx + c = m\big((1+2m)x + 2c\big) + c$, i.e. $mx + c = m(1+2m)x + 2mc + c$.

Equating coefficients of $x$: $m = m(1 + 2m)$, so $2m^2 = 0$, giving $m = 0$.

Equating constants (with $m = 0$): $c = c$ (always true). So any horizontal line $y = c$ is an invariant line. This reflects the fact that a horizontal shear maps every horizontal line to itself.

Figure 3.2 — Three lines representing a consistent and dependent 3×3 system (all three planes sharing a common line, shown as three coplanar lines in the $xy$-plane), contrasted with an inconsistent arrangement where no common intersection exists.

Exam Tip — Invariant Lines vs. Lines of Invariant Points

These are distinct concepts. A line of invariant points requires every point on the line to be fixed. An invariant line only requires the line itself (as a set) to be preserved — individual points on it may move. Many questions on Further Pure papers ask specifically for invariant lines, so ensure your method correctly identifies which type is requested.

Practice Problems

Problem 1 — Matrix Addition

Given $A = \begin{pmatrix}3 & -1\\2 & 5\end{pmatrix}$ and $B = \begin{pmatrix}-2 & 4\\1 & -3\end{pmatrix}$, find $3A + 2B$.

Show solution

$3A = \begin{pmatrix}9 & -3\\6 & 15\end{pmatrix}$, $2B = \begin{pmatrix}-4 & 8\\2 & -6\end{pmatrix}$.

$3A + 2B = \begin{pmatrix}5 & 5\\8 & 9\end{pmatrix}$.

Problem 2 — Matrix Multiplication

Given $C = \begin{pmatrix}1 & 0 & 2\\3 & 1 & 0\end{pmatrix}$ and $D = \begin{pmatrix}2 & 1\\0 & 3\\1 & -1\end{pmatrix}$, compute $CD$ and state its order.

Show solution

$C$ is $2 \times 3$, $D$ is $3 \times 2$. $CD$ is $2 \times 2$.

$CD = \begin{pmatrix}1(2)+0(0)+2(1) & 1(1)+0(3)+2(-1)\\3(2)+1(0)+0(1) & 3(1)+1(3)+0(-1)\end{pmatrix} = \begin{pmatrix}4 & -1\\6 & 6\end{pmatrix}$.

Problem 3 — Determinant and Singularity

Find the value(s) of $k$ for which the matrix $\begin{pmatrix}k & 3\\2 & k-1\end{pmatrix}$ is singular.

Show solution

$\det = k(k-1) - 6 = k^2 - k - 6 = (k-3)(k+2)$.

Singular when $k = 3$ or $k = -2$.

Problem 4 — 2×2 Inverse

Find the inverse of $M = \begin{pmatrix}7 & 3\\4 & 2\end{pmatrix}$ and verify by computing $MM^{-1}$.

Show solution

$\det(M) = 14 - 12 = 2$. $M^{-1} = \dfrac{1}{2}\begin{pmatrix}2 & -3\\-4 & 7\end{pmatrix} = \begin{pmatrix}1 & -3/2\\-2 & 7/2\end{pmatrix}$.

$MM^{-1} = \begin{pmatrix}7&3\\4&2\end{pmatrix}\begin{pmatrix}1 & -3/2\\-2 & 7/2\end{pmatrix} = \begin{pmatrix}7-6 & -21/2+21/2\\4-4 & -6+7\end{pmatrix} = \begin{pmatrix}1&0\\0&1\end{pmatrix}$ ✓.

Problem 5 — 3×3 Determinant

Compute $\det(B)$ for $B = \begin{pmatrix}2 & -1 & 3\\0 & 4 & -2\\1 & 0 & 5\end{pmatrix}$.

Show solution

Expand along row 1:

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

$= 2(20-0) + 1(0+2) + 3(0-4) = 40 + 2 - 12 = 30$.

Problem 6 — Solving a 2×2 System

Solve the system $3x - y = 5$, $x + 4y = 9$ using matrix methods.

Show solution

$A = \begin{pmatrix}3 & -1\\1 & 4\end{pmatrix}$, $\det(A) = 12 + 1 = 13$.

$A^{-1} = \dfrac{1}{13}\begin{pmatrix}4 & 1\\-1 & 3\end{pmatrix}$.

$\begin{pmatrix}x\\y\end{pmatrix} = \dfrac{1}{13}\begin{pmatrix}4 & 1\\-1 & 3\end{pmatrix}\begin{pmatrix}5\\9\end{pmatrix} = \dfrac{1}{13}\begin{pmatrix}29\\22\end{pmatrix}$. So $x = \dfrac{29}{13}$, $y = \dfrac{22}{13}$.

Problem 7 — Singular 3×3 System

Show that the system $x + y + z = 6$, $2x + y + 3z = 14$, $x + 2y = 4$ is consistent with infinitely many solutions and find the general solution.

Show solution

Matrix $A = \begin{pmatrix}1&1&1\\2&1&3\\1&2&0\end{pmatrix}$.

$\det(A) = 1(0-6) - 1(0-3) + 1(4-1) = -6 + 3 + 3 = 0$. Singular — so the system is either inconsistent or has infinitely many solutions.

Row reduce the augmented matrix $[A|\mathbf{b}]$: $R_2 \leftarrow R_2 - 2R_1$, $R_3 \leftarrow R_3 - R_1$:

$\begin{pmatrix}1&1&1&|&6\\0&-1&1&|&2\\0&1&-1&|&-2\end{pmatrix}$. Row 3 is $-1$ times Row 2, so the system is consistent with one free variable.

Let $z = t$. From $R_2$: $y = t - 2$. From $R_1$: $x = 6 - y - z = 6 - (t-2) - t = 8 - 2t$.

General solution: $x = 8 - 2t$, $y = t - 2$, $z = t$ for $t \in \mathbb{R}$.

Problem 8 — Invariant Points

Find all invariant points of the transformation given by $T = \begin{pmatrix}5 & -4\\3 & -2\end{pmatrix}$.

Show solution

Solve $(T - I)\mathbf{x} = \mathbf{0}$: $\begin{pmatrix}4&-4\\3&-3\end{pmatrix}\begin{pmatrix}x\\y\end{pmatrix} = \mathbf{0}$.

$\det(T-I) = 4(-3) - (-4)(3) = -12 + 12 = 0$, so infinitely many solutions.

Row 1: $4x - 4y = 0 \Rightarrow x = y$. Line of invariant points: $y = x$.

Problem 9 — Invariant Lines

Find all invariant lines of the form $y = mx + c$ under $A = \begin{pmatrix}4 & 1\\2 & 3\end{pmatrix}$.

Show solution

Apply $A$ to the general point $(x, mx+c)$:

$\begin{pmatrix}4x + (mx+c)\\2x + 3(mx+c)\end{pmatrix} = \begin{pmatrix}(4+m)x + c\\(2+3m)x + 3c\end{pmatrix}$.

For this to lie on $y = mx + c$: $y\text{-value} = m \times x\text{-value} + c$, i.e. $(2+3m)x + 3c = m(4+m)x + mc + c$.

Coefficient of $x$: $2 + 3m = 4m + m^2$, so $m^2 + m - 2 = 0$, giving $(m+2)(m-1) = 0$, so $m = 1$ or $m = -2$.

Constants: $3c = mc + c$, i.e. $c(2 - m) = 0$. For $m = 1$: $(2-1)c = 0 \Rightarrow c = 0$. For $m = -2$: $(2+2)c = 0 \Rightarrow c = 0$.

Invariant lines: $y = x$ and $y = -2x$ (both through the origin).

Problem 10 — Area Scaling by Determinant

A triangle with vertices at $(0,0)$, $(3,0)$, $(1,2)$ is transformed by the matrix $M = \begin{pmatrix}2 & 1\\-1 & 3\end{pmatrix}$. Find the area of the image triangle.

Show solution

Original area $= \frac{1}{2}|3 \times 2 - 0 \times 1| = \frac{1}{2}(6) = 3$ square units (using the cross-product formula for the triangle with vertices at the origin).

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

Area of image $= |\det(M)| \times$ original area $= 7 \times 3 = 21$ square units.