Further Pure — Matrices: A-level Further Maths OCR revision notes

4.03

Matrix arithmetic and determinants

Matrices, order and multiplication 4.03

Definitions

Order: an $m \times n$ matrix has $m$ rows and $n$ columns. Matrices of the same order add entry by entry.
Identity matrix $I$ : ones on the leading diagonal, zeros elsewhere; $A I = I A = A$ .
Zero matrix $0$ : every entry zero.

Key results

$A B$ is defined when the number of columns of $A$ equals the number of rows of $B$ ; an $m \times n$ matrix times an $n \times p$ matrix is $m \times p$ .
The entry in row $i$ , column $j$ of $A B$ is row $i$ of $A$ multiplied term by term with column $j$ of $B$ , summed.
Multiplication is associative, $(A B) C = A (BC)$ , and distributive, but not commutative: in general $A B \neq = B A$ .

Notes

Because order matters, $(A + B)^{2} = A^{2} + A B + B A + B^{2}$ , not $A^{2} + 2 A B + B^{2}$ .
$A B = 0$ does not imply $A = 0$ or $B = 0$ : for example, $(1111) (1 - 1 - 1 1) = 0$ . There is no cancelling of matrices without an inverse.

Determinants and what they measure 4.03

Key results

$det (a c b d) = a d - b c$ .
For a $3 \times 3$ matrix, expand along the first row: $det A = a e h f i - b d g f i + c d g e h$ for $A = a d g b e h c f i$ .
$det (A B) = det A det B$ .
$∣ det A ∣$ is the factor by which the transformation $A$ multiplies areas (in 2D) or volumes (in 3D). A negative determinant means orientation is reversed.

Notes

The signs in the expansion follow the pattern $+ - + - + - + - +$ ; expanding along a row or column with zeros saves work.
Swapping two rows changes the sign of the determinant; a row of zeros, or two proportional rows, makes it zero.
A matrix with $det A = 0$ is singular: it squashes the plane onto a line (or space onto a plane or line), so it cannot be undone.

Worked example

Find $det A$ for:

A = 210132 0 - 1 4

A solid of volume $3$ is transformed by $A$ .

Find the volume of its image.

Show worked solution

Expanding along the first row:

det A = 2 32 - 1 4 - 1 10 - 1 4 + 0 = 2 (12 + 2) - 1 (4 - 0) = 24

The image has volume $3 \times 24 = 72$ .

4.03

Matrices as transformations

Transformations of the plane and of space 4.03

Key results

A matrix maps the position vector $x$ to $A x$ . Its columns are the images of the unit vectors: in 2D, $i \mapsto$ column 1 and $j \mapsto$ column 2.
Rotation anticlockwise through $θ$ about O: $(cos θ sin θ - sin θ cos θ)$ .
Reflection in $y = x tan θ$ : $(cos 2 θ sin 2 θ sin 2 θ - cos 2 θ)$ .
Enlargement scale factor $k$ : $k I$ ; stretch parallel to the $x$ -axis: $(k 0 01)$ ; shear with the $x$ -axis fixed: $(10 k 1)$ .
In 3D: rotations about the coordinate axes and reflections in the planes $x = 0$ , $y = 0$ , $z = 0$ , $y = \pm x$ , $z = \pm x$ , $z = \pm y$ .

Method

To find the matrix of a transformation, find the images of $i$ and $j$ (and $k$ ) and write them as the columns.
To identify a matrix, look at what it does to the unit square, and check its determinant: $+ 1$ for a rotation, $- 1$ for a reflection, $1$ for a shear.

Notes

Every linear transformation fixes the origin. A translation is not linear and has no matrix of this kind.
Shears and rotations have determinant 1: they move points but keep area.

Combining transformations 4.03

Key results

The transformation $A$ followed by $B$ is $B A$ : the matrix nearest the vector acts first.
$(A B)^{- 1} = B^{- 1} A^{- 1}$ : to undo $A$ -then- $B$ , undo $B$ first.
Two reflections in lines through O at angle $α$ to each other combine to a rotation through $2 α$ .

Notes

The reversed order is the commonest error in the topic. Read $B A x$ from the right: first $A x$ , then $B$ applied to the result.
Since $det (B A) = det B det A$ , the area scale factor of a combination is the product of the separate factors.

Invariant points and invariant lines 4.03

Definitions

Invariant point: a point with $A x = x$ ; it is not moved.
Invariant line: a line whose every point is mapped to a point on the same line (possibly a different point of it).
Line of invariant points: a line every point of which is itself invariant.

Method

Invariant points: solve $A (x y) = (x y)$ . If the two equations are the same line, that line is a line of invariant points.
Invariant lines $y = m x + c$ : take a general point $(x, m x + c)$ , find its image $(x',y')$ , and require $y'=mx'+c$ for all $x$ . Equate coefficients of $x$ and the constant terms to find $m$ and $c$ .

Notes

Invariant lines through O ( $c = 0$ ) are exactly the eigenvector directions of the matrix (next subtopic).
A line of invariant points is always an invariant line, but an invariant line need not consist of invariant points: under the shear on the sheet, $y = 1$ is invariant, yet each of its points slides along it.

Worked example

Find the invariant lines through the origin of the transformation:

M = (1141)

Show worked solution

A point $(x, m x)$ maps to:

(x + 4 m x, x + m x)

For the line $y = m x$ to be invariant,

x + m x = m (x + 4 m x)

for all $x$ , so $1 + m = m + 4 m^{2}$ , giving:

m^{2} = \frac{1}{4}

and:

m = \pm \frac{1}{2}

The invariant lines are:

y = \frac{1}{2} x

and:

y = - \frac{1}{2} x

4.03

Inverse matrices and systems of equations

Inverse matrices 4.03

Definitions

Inverse: $A^{- 1}$ satisfies $A A^{- 1} = A^{- 1} A = I$ . It exists if and only if $det A \neq = 0$ .

Key results

$(a c b d)^{- 1} = \frac{1}{a d - b c} (d - c - b a)$ .
For $3 \times 3$ : $A^{- 1} = \frac{1}{det A} adj A$ , where the adjugate $adj A$ is the transpose of the matrix of cofactors.
$det (A^{- 1}) = \frac{1}{det A}$ .

Method

Cofactor of an entry: the $2 \times 2$ determinant left after deleting that entry's row and column, with sign from the $+ - +$ pattern.
Form the matrix of cofactors, transpose it, divide by $det A$ . Check by multiplying by $A$ : the result should be $I$ .

Notes

Calculators find $3 \times 3$ inverses, and in many questions that is expected; but a question that says "show" or involves an unknown $k$ needs the working.
With an unknown entry, the matrix is singular for the values of $k$ that make $det A = 0$ ; the inverse exists for every other value.

Simultaneous equations and three planes 4.03

Key results

Three linear equations in $x, y, z$ are $A x = b$ . If $det A \neq = 0$ , there is exactly one solution, $x = A^{- 1} b$ , and the three planes meet at a single point.
If $det A = 0$ , the system has either no solution (inconsistent) or infinitely many (consistent).

Method

When $det A = 0$ , eliminate one variable to obtain two equations in two unknowns. If they agree, the system is consistent: the planes form a sheaf meeting in a line (or two or three coincide). If they contradict each other, it is inconsistent.
Then decide the geometry from the normal vectors (the rows of $A$ ): if no two normals are parallel, an inconsistent system is a prism; if two normals are parallel, two planes are parallel.

Notes

The sheet shows the singular cases edge-on: when $det A = 0$ all three normals lie in one plane, so the three planes share a common direction, and looking along it each plane appears as a line.
To describe a sheaf fully, find the common line: set $z = t$ and solve for $x$ and $y$ in terms of $t$ .

Worked examples

Worked example

Solve $x + y + z = 6$ , $2 x - y + z = 3$ , $x + 2 y - z = 2$ using the inverse matrix.

Show worked solution

A = 121 1 - 1 2 11 - 1

has:

det A = 1 (1 - 2) - 1 (- 2 - 1) + 1 (4 + 1) = 7

The cofactors give:

adj A = - 1 35 3 - 2 - 1 21 - 3

so:

x = \frac{1}{7} - 1 35 3 - 2 - 1 21 - 3 632 = \frac{1}{7} 71421

So $x = 1$ , $y = 2$ , $z = 3$ ; check: $1 + 2 + 3 = 6$ , $2 - 2 + 3 = 3$ , $1 + 4 - 3 = 2$ .

Worked example

The planes $x + y + z = 1$ , $x - y + 2 z = 3$ and $2 x + k z = c$ do not meet at a single point.

Find $k$ , and describe the arrangement of the planes when $c = 4$ and when $c \neq = 4$ .

Show worked solution

The determinant is:

1 (- k - 0) - 1 (k - 4) + 1 (0 + 2) = 6 - 2 k

which is zero when $k = 3$ .

With $k = 3$ , the third left-hand side equals the sum of the first two, so the system is consistent only if $c = 1 + 3 = 4$ .

When $c = 4$ the planes form a sheaf, meeting in a common line.

When $c \neq = 4$ the system is inconsistent, and since no two normals $(1, 1, 1)$ , $(1, - 1, 2)$ , $(2, 0, 3)$ are parallel, the planes form a prism.

Extension: eigenvalues and eigenvectors

Eigenvalues and eigenvectors Beyond 4.03

Definitions

Eigenvector of $A$ : a non-zero vector $v$ with $A v = λ v$ for some scalar $λ$ , its eigenvalue. The transformation only stretches $v$ (by factor $λ$ ), without turning it.

Key results

Eigenvalues are the roots of the characteristic equation $det (A - λ I) = 0$ .
For a $2 \times 2$ matrix: $λ^{2} - (tr A) λ + det A = 0$ , where $tr A = a + d$ . So the eigenvalues sum to the trace and multiply to the determinant.
If $A$ has eigenvectors $v_{1}, v_{2}$ with eigenvalues $λ_{1}, λ_{2}$ , then with $P = (v_{1} v_{2})$ and $D = (λ_{1} 0 0 λ_{2})$ : $A = P D P^{- 1}$ and $A^{n} = P D^{n} P^{- 1}$ .

Method

Solve $det (A - λ I) = 0$ for each $λ$ .
For each eigenvalue, solve $(A - λ I) v = 0$ . The equations are dependent, so they give a direction; any non-zero multiple is an eigenvector.

Notes

Beyond the OCR Pure Core: section 4.03 stops at invariant points and lines. Eigenvectors are included because an invariant line through the origin is exactly the direction of an eigenvector, so this is the natural next step, and it is useful for university mathematics.
OCR's Pure Core does not require eigenvalues; it is included here because it explains invariant lines through the origin, which it does require: each such line is an eigenvector direction.
A rotation through an angle other than $0$ or $π$ has no real eigenvalues: it turns every direction.

Worked example

Find the eigenvalues and eigenvectors of:

A = (2112)

and hence show that:

A^{n} = \frac{1}{2} (3^{n} + 1 3^{n} - 1 3^{n} - 1 3^{n} + 1)

Show worked solution

det (A - λ I) = (2 - λ)^{2} - 1 = 0

gives $λ = 1$ or $3$ .

For $λ = 3$ : $- x + y = 0$ , so:

v = (11)

For $λ = 1$ : $x + y = 0$ , so:

u = (1 - 1)

With:

P = (11 1 - 1)

P^{- 1} = \frac{1}{2} (11 1 - 1)

and:

A^{n} = P (3^{n} 0 01) P^{- 1} = \frac{1}{2} (3^{n} + 1 3^{n} - 1 3^{n} - 1 3^{n} + 1)

Check $n = 1$ :

\frac{1}{2} (4224) = A