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Matrices

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FP3 · Chapter 6

Matrices

A matrix isn't just a grid of numbers — it's a machine that transforms space. Once you can find its determinant, invert it, and see what it does to lines and planes, you can predict exactly how any shape moves, stretches or flattens under that transformation.

Summary

Dimension of a matrix = (rows) × (columns).
Transpose A^T swaps rows and columns; a symmetric matrix satisfies S = S^T.
Identity I leaves everything unchanged; the zero matrix sends everything to zero.
The determinant of a 3×3 matrix is found by cofactor expansion along any row or column.
A matrix is singular if det = 0 (no inverse exists), non-singular if det ≠ 0.
The inverse of a 3×3 matrix is built from the matrix of cofactors, transposed, then divided by the determinant.
Matrices represent linear transformations: functions that preserve addition and scalar multiplication.
The columns of a transformation matrix are exactly where the base vectors i, j, k land.
To find the image of a line, apply the transformation to the position vector and the direction vector separately.
To find the image of a plane, first rewrite it in vector form r = a + λb + μc, then transform each piece.

1. Basic Definitions

Dimension of a matrix

A matrix with r rows and c columns has dimension r × c (rows always come first — "Railway Carriage" is the classic way to remember the order). So a matrix with 3 rows and 2 columns is a "3 × 2" matrix. In FP3 you'll mostly work with 3 × 3 square matrices, since those represent transformations of 3-D space.

Transpose and symmetric matrices

The transpose of a matrix A, written A^T, is found by turning rows into columns (or equivalently, reflecting every entry in the leading diagonal).

Transpose

A =

a	b	c
d	e	f
g	h	i

⇒ A^T =

a	d	g
b	e	h
c	f	i

A really important (and easy to forget) rule is what happens when you transpose a product:

Transpose of a product (AB)^T = B^TA^T — the order flips.

A matrix S is symmetric if it looks the same when reflected in its leading diagonal — formally, if S = S^T. Visually, the entry in row i, column j always matches the entry in row j, column i.

Example — spotting a symmetric matrix

S =

a	b	c
b	d	e
c	e	f

is symmetric — notice the off-diagonal entries mirror each other perfectly: top-right = bottom-left in each pair.

Identity and zero matrices

The identity matrix I is the "do nothing" transformation — multiplying any vector or matrix by I leaves it unchanged, just like multiplying a number by 1. The zero matrix sends every vector to the origin, just like multiplying by 0.

Identity & zero matrix (3×3)

I =

1	0	0
0	1	0
0	0	1

0 =

0	0	0
0	0	0
0	0	0

Practice — Q1 Write down the transpose of B = [[2,-1,4],[0,3,5],[7,1,-2]] (rows given left to right, top to bottom), and state whether B is symmetric.

2. Determinant of a 3 × 3 Matrix

The determinant is a single number attached to a square matrix that tells you how the transformation scales volume (and whether it flips orientation, if it's negative). You'll already have met 2 × 2 determinants — the 3 × 3 version builds directly on that by expanding along a row using smaller 2 × 2 determinants.

Determinant of a 3×3 matrix (expand along top row)

det(A) = Δ =

a	b	c
d	e	f
g	h	i

= a(ei − fh) − b(di − fg) + c(dh − eg)

Notice the alternating + − + signs. Each term is (entry) × (the 2×2 determinant you get by covering up that entry's row and column).

Properties of the determinant

You can expand along any row or column — not just the top row — as long as you use the correct sign pattern:

+ − +
− + −
+ − +

Choosing a row or column with a zero in it saves a lot of work.
Swapping two rows (or columns) flips the sign of the determinant.
Two identical rows (or columns) ⇒ determinant is 0. This makes sense: swapping the identical rows should flip the sign, but the matrix is unchanged, so Δ = −Δ, forcing Δ = 0.
det(AB) = det(A) × det(B) — determinants multiply when matrices multiply.

Worked example — computing a 3×3 determinant Find det(A) where A =

1	2	0
3	-1	4
-2	1	5

Expand along the top row (nice, because of the 0):
det(A) = 1·[(-1)(5) − (4)(1)] − 2·[(3)(5) − (4)(-2)] + 0
= 1·(−5 − 4) − 2·(15 + 8) + 0
= 1·(−9) − 2·(23)
= −9 − 46 = −55

Practice — Q2 Find the determinant of

2	0	1
-1	3	2
4	1	-2

by expanding along the middle column (it has a zero-free but small-looking entry — try it for practice with the sign pattern).

3. Singular and Non-Singular Matrices

This is a short but crucial definition — everything about invertibility hinges on it.

Definitions A matrix A is singular if det(A) = 0 — it has no inverse.
A matrix A is non-singular if det(A) ≠ 0 — it does have an inverse.

Why does det = 0 mean "no inverse"? Geometrically, a singular matrix squashes 3-D space down onto a plane, a line, or even a single point — it collapses volume to zero. Once information is squashed flat like that, there's no way to "undo" it and recover the original 3-D shape, because many different starting points get mapped to the same squashed image. No unique way back = no inverse.

4. Inverse of a 3 × 3 Matrix

Finding a 3×3 inverse is a longer process than the 2×2 case, but it's entirely mechanical — if you follow the steps carefully and don't rush the sign pattern, you won't go wrong.

Cofactors

For a matrix A =

a	b	c
d	e	f
g	h	i

, the cofactor of each entry is the 2×2 determinant left over when you delete that entry's row and column — with the correct sign attached from the +/− grid.

The nine cofactors

A = +|e f; h i| B = −|d f; g i| C = +|d e; g h|
D = −|b c; h i| E = +|a c; g i| F = −|a b; g h|
G = +|b c; e f| H = −|a c; d f| I = +|a b; d e|

(Here |p q; r s| means the 2×2 determinant ps − qr.) These are exactly the same 2×2 determinants used to expand along a row when finding det(A) — you're just calculating all nine of them instead of only three.

Finding the inverse — the four steps

Find det(A). If det(A) = 0, stop — A is singular and has no inverse.
Find the matrix of cofactors, C =
A B C
D E F
G H I
Transpose it: C^T =
A D G
B E H
C F I
(this is called the adjugate matrix).
Divide every entry by det(A): A⁻¹ = ¹⁄_det(A) · C^T

Worked example — finding a 3×3 inverse Find the inverse of A =

1	0	2
-1	3	1
2	1	0

Step 1: determinant (expand top row)
det(A) = 1(3·0 − 1·1) − 0(...) + 2(−1·1 − 3·2)
= 1(−1) − 0 + 2(−1 − 6) = −1 + 2(−7) = −1 − 14 = −15 (non-zero, good — an inverse exists)

Step 2: cofactors

A = +(3·0−1·1) = −1 B = −(−1·0−1·2) = 2 C = +(−1·1−3·2) = −7
D = −(0·0−2·1) = 2 E = +(1·0−2·2) = −4 F = −(1·1−0·2) = −1
G = +(0·1−2·3) = −6 H = −(1·1−2·(−1)) = −3 I = +(1·3−0·(−1)) = 3

So the cofactor matrix is

-1	2	-7
2	-4	-1
-6	-3	3

Step 3: transpose it

C^T =

-1	2	-6
2	-4	-3
-7	-1	3

Step 4: divide by det(A) = −15

A⁻¹ = ¹⁄₋₁₅

-1	2	-6
2	-4	-3
-7	-1	3

Always worth a quick sanity check: multiply A by A⁻¹ mentally on one row/column to confirm you get something close to the identity — catching an arithmetic slip here saves the whole question.

Properties of the inverse

Three key properties 1) A⁻¹A = AA⁻¹ = I
2) (AB)⁻¹ = B⁻¹A⁻¹ — order reverses, just like the transpose rule
3) det(A⁻¹) = ¹⁄_det(A)

Memory hook "Reverse, reverse" — both (AB)^T and (AB)⁻¹ flip the order to B first, A second. If you remember that one rule shape, you get both properties for free.

Practice — Q3 A matrix M has det(M) = 4. What is det(M⁻¹)? And if N = M⁻¹M^T, write down (N)⁻¹ in terms of M and M^T using the reversal rule.

5. Matrices and Linear Transformations

Here's where matrices stop being abstract number grids and start being machines that move space around. A transformation T is called linear if it satisfies two conditions:

Definition of a linear transformation (i) T(x₁ + x₂) = T(x₁) + T(x₂) — transforming a sum = summing the transforms
(ii) T(kx) = kT(x) — scaling before or after T gives the same result

In plain words: a linear transformation doesn't care whether you combine/scale vectors before or after applying it — you get the same answer either way. This is exactly the property that lets every linear transformation be represented by a single matrix, and it's why we can transform a whole line or plane just by transforming a handful of defining vectors (as you'll see in sections 7 and 8).

Worked example — proving a transformation is linear Show that T(x, y, z) = (2x, x+y, −z) is a linear transformation.

Check (i): Let x₁ = (x₁,y₁,z₁), x₂ = (x₂,y₂,z₂).
T(x₁+x₂) = T(x₁+x₂, y₁+y₂, z₁+z₂) = (2(x₁+x₂), x₁+x₂+y₁+y₂, −(z₁+z₂))
= (2x₁, x₁+y₁, −z₁) + (2x₂, x₂+y₂, −z₂) = T(x₁) + T(x₂) ✓

Check (ii): T(kx) = T(kx,ky,kz) = (2kx, kx+ky, −kz) = k(2x, x+y, −z) = kT(x) ✓

Both conditions hold, so T is a linear transformation.

Good news Every matrix automatically represents a linear transformation — you'll rarely be asked to prove linearity from scratch. The real skill tested in this chapter is finding the matrix for a given transformation, which is exactly what base vectors let you do next.

6. Base Vectors i, j, k

This is the single most useful idea in the whole chapter for building transformation matrices quickly. Watch what happens when the standard basis vectors i = (1,0,0), j = (0,1,0), k = (0,0,1) are multiplied by a general matrix:

The columns ARE the images of i, j, k Under the transformation with matrix

a	b	c
d	e	f
g	h	i

:

i → (column 1) j → (column 2) k → (column 3)

This means: if you know where i, j and k each land, you can write down the whole matrix immediately — the image of i becomes the first column, the image of j the second column, and the image of k the third column. No multiplication required.

Worked example — building a matrix from a geometric description Find the matrix for a reflection in the plane y = x.

Think about where each axis vector lands under this reflection:

The z-axis lies in the plane y = x, so k = (0,0,1) is unchanged → column 3 = (0,0,1)
i = (1,0,0) reflects to (0,1,0) → column 1 = (0,1,0)
j = (0,1,0) reflects to (1,0,0) → column 2 = (1,0,0)

Assembling the columns:

Matrix =

0	1	0
1	0	0
0	0	1

Worked example — building a matrix from three mapped points (not just i, j, k) Find the matrix T given that (1,0,0)→(3,4,2), (1,1,0)→(6,1,5), (2,1,−4)→(1,1,−1).

(1,0,0) = i, so column 1 of T is (3,4,2) immediately.

(1,1,0) = i + j, and by linearity T(i+j) = T(i) + T(j), so:
T(j) = (6,1,5) − (3,4,2) = (3,−3,3) → this is column 2.

(2,1,−4) = 2i + j − 4k, so by linearity:
2T(i) + T(j) − 4T(k) = (1,1,−1)
2(3,4,2) + (3,−3,3) − 4T(k) = (1,1,−1)
(9,5,7) − 4T(k) = (1,1,−1) ⇒ 4T(k) = (8,4,8) ⇒ T(k) = (2,1,2) → column 3.

So T =

3	3	2
4	-3	1
2	3	2

This example shows the real power of linearity: you never actually need i, j, k themselves to be given directly — as long as you can express any mapped point as a combination of i, j, k, linearity lets you peel out each column one at a time.

Practice — Q4 Find the matrix representing an enlargement by scale factor 3 in the x-direction only, leaving y and z unchanged.

7. Image of a Line

A line has vector equation r = a + λb — a fixed point a plus a variable multiple of a direction vector b. Because T is linear, transforming the whole line is as easy as transforming a and b separately:

Image of a line under transformation T T(r) = T(a + λb) = T(a) + λT(b)

The image is a new line through the point T(a), parallel to the vector T(b).

Worked example Find the image of the line r = (2,0,−3) + λ(3,−2,1) under T =

3	-2	1
1	3	4
2	-1	1

Transform the point (2,0,−3):

T(2,0,−3) = (3·2 + (−2)·0 + 1·(−3), 1·2 + 3·0 + 4·(−3), 2·2 + (−1)·0 + 1·(−3))
= (6 − 3, 2 − 12, 4 − 3) = (3, −10, 1)

Transform the direction vector (3,−2,1):

T(3,−2,1) = (3·3 + (−2)(−2) + 1·1, 1·3 + 3(−2) + 4·1, 2·3 + (−1)(−2) + 1·1)
= (9+4+1, 3−6+4, 6+2+1) = (14, 1, 9)

So the image line is: r = (3,−10,1) + λ(14,1,9)

Watch out Transform the point and the direction vector completely separately — never add them together first and transform the sum. Keeping them as two clean matrix multiplications avoids arithmetic slip-ups.

Practice — Q5 Find the image of the line r = (1,1,0) + λ(0,1,2) under M = [[1,0,0],[0,2,0],[0,0,-1]].

8. Image of a Plane

A plane in vector form r = a + λb + μc transforms exactly like a line, just with one extra direction vector:

Image of a plane under transformation T T(r) = T(a) + λT(b) + μT(c)

The catch is that planes are usually given as Cartesian equations like ax + by + cz = d, not in vector form — so the real skill here is converting the Cartesian equation into r = a + λb + μc first. The PDF gives two methods for doing this conversion.

Method 1 — spot a point and two parallel vectors

Set two of the variables to convenient values (often 0) and solve for the third to find one point on the plane.
The normal vector n is read directly from the Cartesian equation's coefficients. Any vector perpendicular to n is parallel to the plane — find two such vectors by inspection (check their scalar product with n is zero).
Write r = (point) + λ(vector 1) + μ(vector 2), then apply T to each piece.

Worked example — Method 1 Find the image of the plane 3x − 2y + 4z = 6 under a linear transformation T.

Point on the plane: put x = z = 0 ⇒ y = −3, so (0, −3, 0) lies on the plane.

Vectors parallel to the plane: n = (3, −2, 4). We need vectors ⊥ to n. By inspection, using the top two coordinates, (2,3,0) works since 3(2)+(−2)(3)+4(0) = 0. Using the bottom two coordinates, (0,2,1) works since 3(0)+(−2)(2)+4(1) = 0.

Vector equation: r = (0,−3,0) + λ(2,3,0) + μ(0,2,1)

Image: T(r) = T(0,−3,0) + λT(2,3,0) + μT(0,2,1) — now apply T to each of these three vectors to get the image plane's vector equation.

Method 2 — parametrise directly (as in the textbook)

Method 1 breaks down if the normal vector has a zero component, since then "using the top two coordinates" or "bottom two coordinates" might not give you a valid second vector by inspection. Method 2 sidesteps this by parametrising the equation algebraically instead of hunting for vectors by eye.

Set x = λ and y = μ directly in the Cartesian equation, then solve for z in terms of λ and μ.
Split the resulting vector into a constant part (λ=μ=0) plus a λ-part plus a μ-part.
Multiply through to clear fractions in the direction vectors if it makes the numbers nicer (this doesn't change the plane, just the parametrisation).

Worked example — Method 2 Find the image of the plane 3x − 2y + 4z = 6 under T (same plane as before, different method).

Put x = λ, y = μ: 3λ − 2μ + 4z = 6 ⇒ z = (6 − 3λ + 2μ)/4

(x,y,z) = (λ, μ, (6−3λ+2μ)/4) = (0, 0, 6/4) + λ(1, 0, −3/4) + μ(0, 1, 1/2)

Multiplying the direction vectors by 4 to clear fractions (allowed — it just rescales λ and μ):

r = (0, 0, 3/2) + λ(4, 0, −3) + μ(0, 2, 1)

Now continue exactly as in Method 1: apply T to the point and to each direction vector.

Critical trap — do NOT do this It's tempting to find a normal vector for the image plane by computing M(b) × M(c). This does not work. The cross product isn't a linear operation in the way matrix transformations are, so M(b × c) ≠ M(b) × M(c) in general. Always transform the point and the two direction vectors from the vector-equation form — never try to shortcut through the normal vector.

Practice — Q6 Convert the plane r · (2,−5,0) = 8 into vector form r = a + λb + μc, ready for transforming. (Hint: this is 2x − 5y = 8, with z completely free.)

What to Memorise

Definitions

Dimension = rows × columns. Symmetric: S = S^T. Singular: det(A) = 0 (no inverse). Non-singular: det(A) ≠ 0 (inverse exists).

Determinant (3×3)

det(A) = a(ei−fh) − b(di−fg) + c(dh−eg), sign grid +−+ / −+− / +−+

Inverse (3×3)

A⁻¹ = ¹⁄_det(A) × (matrix of cofactors)^T

Key relationships

(AB)^T = B^TA^T (AB)⁻¹ = B⁻¹A⁻¹ det(A⁻¹) = 1/det(A) det(AB) = det(A)det(B)

Base vectors

Columns of a transformation matrix = images of i, j, k respectively.

Image of a line

r = a + λb ⟶⟶⟶ T(r) = T(a) + λT(b)

Image of a plane

Convert Cartesian → vector form r = a + λb + μc first, then T(r) = T(a) + λT(b) + μT(c). Never transform b × c directly.

Concepts Checklist

I can state the dimension of a matrix and find its transpose.
I can recognise and test whether a matrix is symmetric.
I know the identity and zero matrices and what they do.
I can find the determinant of a 3×3 matrix by expanding along any row or column.
I know all four properties of determinants (sign grid, row swap, identical rows, det(AB)).
I can identify a matrix as singular or non-singular from its determinant.
I can find the matrix of cofactors for a 3×3 matrix.
I can find the full inverse of a 3×3 matrix using cofactors, transpose, and determinant.
I know the properties of the inverse, including the order-reversal rule for (AB)⁻¹.
I can verify whether a transformation is linear using the two defining conditions.
I can build a transformation matrix directly from where i, j and k land.
I can build a transformation matrix from three mapped points using linearity, even without i, j, k directly given.
I can find the image of a line under a linear transformation.
I can convert a Cartesian plane equation into vector form using Method 1 (spotting vectors).
I can convert a Cartesian plane equation into vector form using Method 2 (parametrising) when a coefficient is zero.
I can find the image of a plane under a linear transformation, without falling into the M(b)×M(c) trap.

Exam Tips

Sign errors kill marks The +−+ / −+− / +−+ pattern trips up more students than anything else in this chapter — both in the determinant expansion and in the cofactor matrix. Write the sign grid out in the margin every single time until it's automatic.

Choose your expansion row wisely Always scan for a row or column with a zero before expanding a determinant — it removes an entire term and halves your arithmetic. Examiners often build a convenient zero into the matrix on purpose.

Cofactors ≠ the inverse yet A very common slip is forgetting to transpose the cofactor matrix before dividing by the determinant. The cofactor matrix on its own is not the inverse — always write "C, then C^T, then divide" as three distinct lines of working so you don't skip the transpose under exam pressure.

"Order reverses" — say it out loud Both (AB)^T = B^TA^T and (AB)⁻¹ = B⁻¹A⁻¹ reverse the order. Students frequently write A⁻¹B⁻¹ instead — an easy mark to lose on a "state the property" question.

Linear transformation proofs need BOTH conditions Showing only T(x₁+x₂) = T(x₁)+T(x₂) is not enough — mark schemes expect both T(x₁+x₂) = T(x₁)+T(x₂) and T(kx) = kT(x) checked and stated explicitly, even if they feel like "the same idea."

Image of a plane: transform the pieces, never the cross product If you're tempted to find a new normal vector via M(b) × M(c), stop — this is mathematically invalid for a general linear transformation. Always transform the point and each direction vector separately from the vector-equation form.

Base vectors = free columns If a question tells you directly where i, j, k map to, you can write down the whole matrix in one line — no calculation needed. Don't overcomplicate an easy mark by multiplying things out unnecessarily.

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