Introduction to Matrices and Matrix Operations

A matrix is a rectangular array of numbers. That deceptively simple object is one of the most powerful in all of mathematics: a single matrix can encode a system of simultaneous equations, a geometric transformation of the plane, the structure of a network, or the dynamics of a population model. This opening lesson builds the language — order, entries, special matrices, the transpose — and the three structure-preserving operations (addition, subtraction, scalar multiplication) on which everything later in the course rests. Master the notation here and matrix multiplication, determinants, inverses, transformations and eigentheory will all feel like natural extensions.

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1. Where this sits in AQA 7367

Matrices are compulsory pure content, assessed in Paper 1 and Paper 2 (each 2 hours, 100 marks, weighted 33⅓%). This first lesson is almost entirely AO1 — accurate use of standard techniques and notation — but the equality-of-matrices method (equate corresponding entries to solve for unknowns) is genuine AO2 reasoning and recurs throughout the matrices strand. The full AQA matrices specification runs from these definitions through multiplication, determinants and inverses (2×2 and 3×3), linear transformations in 2D and 3D, solving systems of equations, and on to eigenvalues, eigenvectors, diagonalisation and the Cayley–Hamilton theorem. Everything starts with getting the vocabulary watertight.

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2. Core theory: what a matrix is

A matrix is a rectangular array of numbers arranged in rows and columns. The order (or size) of a matrix with $m$m rows and $n$n columns is written $m \times n$m×n — always rows first, columns second.

$$A = \begin{pmatrix} 1 & 2 & 3 \\ 4 & 5 & 6 \end{pmatrix}$$A=(14​25​36​)

This $A$A is a $2 \times 3$2×3 matrix: $2$2 rows, $3$3 columns. We name the entry in row $i$i, column $j$j as $a_{ij}$aij​ — again row index first. For the matrix above,

$$a_{11} = 1, \quad a_{12} = 2, \quad a_{13} = 3, \quad a_{21} = 4, \quad a_{23} = 6.$$a11​=1,a12​=2,a13​=3,a21​=4,a23​=6.

A compact way to define a whole matrix is by a rule for its general entry. For example, the rule $a_{ij} = i + j$aij​=i+j generates, for a $3\times 3$3×3 matrix,

$$\begin{pmatrix} 1+1 & 1+2 & 1+3 \\ 2+1 & 2+2 & 2+3 \\ 3+1 & 3+2 & 3+3 \end{pmatrix} = \begin{pmatrix} 2 & 3 & 4 \\ 3 & 4 & 5 \\ 4 & 5 & 6 \end{pmatrix},$$​1+12+13+1​1+22+23+2​1+32+33+3​​=​234​345​456​​,

which happens to be symmetric (because $i + j = j + i$i+j=j+i). The main diagonal consists of the entries $a_{ii}$aii​ (top-left to bottom-right); here it is $2, 4, 6$2,4,6. Reading off a single entry from such a rule is a routine AO1 check: $a_{23} = 2 + 3 = 5$a23​=2+3=5, the entry in row $2$2, column $3$3.

Special matrices

Matrix type	Description	Example
Square matrix	Same number of rows and columns ($n \times n$n×n)	$\begin{pmatrix} 1 & 2 \\ 3 & 4 \end{pmatrix}$(13​24​)
Identity matrix $I$I	Square; $1$1s on the main diagonal, $0$0s elsewhere	$\begin{pmatrix} 1 & 0 \\ 0 & 1 \end{pmatrix}$(10​01​)
Zero matrix $O$O	Every entry is $0$0	$\begin{pmatrix} 0 & 0 \\ 0 & 0 \end{pmatrix}$(00​00​)
Diagonal matrix	Square; non-zero entries only on the main diagonal	$\begin{pmatrix} 3 & 0 \\ 0 & 7 \end{pmatrix}$(30​07​)
Symmetric matrix	$A = A^{T}$A=AT, i.e. $a_{ij} = a_{ji}$aij​=aji​	$\begin{pmatrix} 1 & 5 \\ 5 & 2 \end{pmatrix}$(15​52​)
Skew-symmetric	$A^{T} = -A$AT=−A; forces $a_{ii} = 0$aii​=0	$\begin{pmatrix} 0 & 3 \\ -3 & 0 \end{pmatrix}$(0−3​30​)
Column vector	$n \times 1$n×1 matrix	$\begin{pmatrix} 3 \\ -1 \end{pmatrix}$(3−1​)
Row vector	$1 \times n$1×n matrix	$\begin{pmatrix} 2 & 5 \end{pmatrix}$(2​5​)

The identity $I$I and zero matrix $O$O will turn out to behave like the numbers $1$1 and $0$0 under multiplication and addition respectively — the analogy is exact and worth holding onto.

Why matrices matter

It is worth seeing, before the abstraction piles up, what a matrix is for. Three motivating uses, all developed later in this course:

• Simultaneous equations. The system $2x + y = 5$2x+y=5, $3x + 2y = 8$3x+2y=8 is captured by the single matrix equation $\begin{pmatrix} 2 & 1 \\ 3 & 2 \end{pmatrix}\begin{pmatrix} x \\ y \end{pmatrix} = \begin{pmatrix} 5 \\ 8 \end{pmatrix}$(23​12​)(xy​)=(58​) — the coefficient matrix, the unknown vector, and the constant vector. Solving it becomes a single operation $\mathbf{x} = A^{-1}\mathbf{b}$x=A−1b (lesson 5).
• Geometric transformations. Every rotation, reflection, enlargement and shear of the plane is a $2\times2$2×2 matrix acting on position vectors; composing transformations becomes matrix multiplication (lessons 6–7).
• Iterated processes. Population models, Markov chains and recurrence systems take the form $\mathbf{x}_{n+1} = A\mathbf{x}_n$xn+1​=Axn​; the long-term behaviour is read off the eigenvalues of $A$A (lessons 8–9).

Each of these reduces a messy problem to clean matrix algebra — which is exactly why getting the algebra of addition, scaling and multiplication right matters so much.

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3. Matrix addition, subtraction and scalar multiplication

Addition and subtraction

Two matrices can be added or subtracted only if they have the same order. The operation is performed entry-by-entry:

$$\begin{pmatrix} a & b \\ c & d \end{pmatrix} + \begin{pmatrix} e & f \\ g & h \end{pmatrix} = \begin{pmatrix} a+e & b+f \\ c+g & d+h \end{pmatrix}.$$(ac​bd​)+(eg​fh​)=(a+ec+g​b+fd+h​).

A quick concrete subtraction, showing the double-negative care that earns accuracy marks:

$$\begin{pmatrix} 3 & 1 \\ -2 & 4 \end{pmatrix} - \begin{pmatrix} 0 & -1 \\ 5 & 2 \end{pmatrix} = \begin{pmatrix} 3-0 & 1-(-1) \\ -2-5 & 4-2 \end{pmatrix} = \begin{pmatrix} 3 & 2 \\ -7 & 2 \end{pmatrix}.$$(3−2​14​)−(05​−12​)=(3−0−2−5​1−(−1)4−2​)=(3−7​22​).

Addition inherits all the structure of ordinary number addition:

Property	Statement
Commutative	$A + B = B + A$A+B=B+A
Associative	$(A + B) + C = A + (B + C)$(A+B)+C=A+(B+C)
Additive identity	$A + O = A$A+O=A
Additive inverse	$A + (-A) = O$A+(−A)=O

Scalar multiplication

Multiplying a matrix by a scalar $k$k multiplies every entry by $k$k:

$$k\begin{pmatrix} a & b \\ c & d \end{pmatrix} = \begin{pmatrix} ka & kb \\ kc & kd \end{pmatrix}.$$k(ac​bd​)=(kakc​kbkd​).

These two operations make matrices of a fixed order into a vector space — which is exactly why so much vector geometry later carries across to matrices.

Worked Example 1 — combined operations (with mark scheme)

Given $A = \begin{pmatrix} 3 & 1 \\ -2 & 4 \end{pmatrix}$A=(3−2​14​) and $B = \begin{pmatrix} 0 & -1 \\ 5 & 2 \end{pmatrix}$B=(05​−12​), find $2A - 3B$2A−3B.

$$\begin{aligned} 2A &= \begin{pmatrix} 6 & 2 \\ -4 & 8 \end{pmatrix}, \qquad 3B = \begin{pmatrix} 0 & -3 \\ 15 & 6 \end{pmatrix} \\ 2A - 3B &= \begin{pmatrix} 6-0 & 2-(-3) \\ -4-15 & 8-6 \end{pmatrix} = \begin{pmatrix} 6 & 5 \\ -19 & 2 \end{pmatrix}. \end{aligned}$$2A2A−3B​=(6−4​28​),3B=(015​−36​)=(6−0−4−15​2−(−3)8−6​)=(6−19​52​).​

Mark scheme: M1 for correctly forming both scalar multiples $2A$2A and $3B$3B; M1 for subtracting corresponding entries; A1 for the fully correct matrix $\begin{pmatrix} 6 & 5 \\ -19 & 2 \end{pmatrix}$(6−19​52​). A sign slip on $2-(-3)$2−(−3) loses only the A1 if both M marks are clearly earned.

Worked Example 2 — solving a matrix equation for an unknown matrix (with mark scheme)

Because addition and scalar multiplication obey the usual algebraic laws, we can solve a matrix equation for an unknown matrix exactly as we would solve a scalar equation. Find the $2\times2$2×2 matrix $X$X satisfying

$$3X - 2\begin{pmatrix} 1 & 0 \\ -2 & 3 \end{pmatrix} = \begin{pmatrix} 7 & -6 \\ 2 & 9 \end{pmatrix}.$$3X−2(1−2​03​)=(72​−69​).

Rearrange to isolate $3X$3X, then divide by the scalar $3$3:

$$\begin{aligned} 3X &= \begin{pmatrix} 7 & -6 \\ 2 & 9 \end{pmatrix} + 2\begin{pmatrix} 1 & 0 \\ -2 & 3 \end{pmatrix} = \begin{pmatrix} 7 & -6 \\ 2 & 9 \end{pmatrix} + \begin{pmatrix} 2 & 0 \\ -4 & 6 \end{pmatrix} = \begin{pmatrix} 9 & -6 \\ -2 & 15 \end{pmatrix} \\ X &= \tfrac{1}{3}\begin{pmatrix} 9 & -6 \\ -2 & 15 \end{pmatrix} = \begin{pmatrix} 3 & -2 \\ -\tfrac{2}{3} & 5 \end{pmatrix}. \end{aligned}$$3XX​=(72​−69​)+2(1−2​03​)=(72​−69​)+(2−4​06​)=(9−2​−615​)=31​(9−2​−615​)=(3−32​​−25​).​

Mark scheme: M1 for rearranging to $3X = \dots$3X=… (adding $2$2 times the known matrix to both sides); A1 for $3X = \begin{pmatrix} 9 & -6 \\ -2 & 15 \end{pmatrix}$3X=(9−2​−615​); A1 for dividing every entry by $3$3 to give the exact $X$X (the $-\tfrac{2}{3}$−32​ must be kept exact, not rounded). Note we could only "divide by 3" because it is scalar division — there is no division by a matrix here.

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4. The transpose

The transpose of a matrix $A$A, written $A^{T}$AT, is obtained by swapping rows and columns — row $i$i of $A$A becomes column $i$i of $A^{T}$AT. An $m \times n$m×n matrix transposes to an $n \times m$n×m matrix.

$$A = \begin{pmatrix} 1 & 2 & 3 \\ 4 & 5 & 6 \end{pmatrix} \implies A^{T} = \begin{pmatrix} 1 & 4 \\ 2 & 5 \\ 3 & 6 \end{pmatrix}.$$A=(14​25​36​)⟹AT=​123​456​​.

Key properties (the last one is the one students most often get wrong):

$$(A^{T})^{T} = A, \qquad (A + B)^{T} = A^{T} + B^{T}, \qquad (kA)^{T} = kA^{T}, \qquad (AB)^{T} = B^{T}A^{T}.$$(AT)T=A,(A+B)T=AT+BT,(kA)T=kAT,(AB)T=BTAT.

Note the reversal of order in $(AB)^{T} = B^{T}A^{T}$(AB)T=BTAT. The transpose also gives the cleanest definition of symmetry: $A$A is symmetric exactly when $A = A^{T}$A=AT, and skew-symmetric when $A^{T} = -A$AT=−A.

Worked Example 3 — transpose and symmetry (with mark scheme)

The matrix $A = \begin{pmatrix} 2 & p & 5 \\ 3 & 4 & q \\ r & -1 & 0 \end{pmatrix}$A=​23r​p4−1​5q0​​ is symmetric. Find $p$p, $q$q and $r$r, and write down $A^{T}$AT.

A symmetric matrix satisfies $a_{ij} = a_{ji}$aij​=aji​, so we match each off-diagonal entry with its mirror image across the main diagonal:

$$\begin{aligned} a_{12} = a_{21} &\implies p = 3, \\ a_{13} = a_{31} &\implies 5 = r, \\ a_{23} = a_{32} &\implies q = -1. \end{aligned}$$a12​=a21​a13​=a31​a23​=a32​​⟹p=3,⟹5=r,⟹q=−1.​

Hence $A = \begin{pmatrix} 2 & 3 & 5 \\ 3 & 4 & -1 \\ 5 & -1 & 0 \end{pmatrix}$A=​235​34−1​5−10​​, and because $A$A is symmetric, $A^{T} = A$AT=A — that is the whole point of symmetry.

Mark scheme: M1 for using $a_{ij} = a_{ji}$aij​=aji​ on at least two pairs; A1 for all three values $p = 3,\ q = -1,\ r = 5$p=3, q=−1, r=5; B1 for stating $A^{T} = A$AT=A with the reason (symmetric $\Rightarrow$⇒ equals its transpose). A common loss is matching $a_{13}$a13​ with $a_{23}$a23​ rather than with its true mirror $a_{31}$a31​.

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5. Equality of matrices — the workhorse method

Two matrices $A$A and $B$B are equal if and only if (i) they have the same order, and (ii) corresponding entries are equal: $a_{ij} = b_{ij}$aij​=bij​ for all $i, j$i,j. A single matrix equation of order $m \times n$m×n is therefore shorthand for $mn$mn simultaneous scalar equations — exactly as a complex-number equation splits into "real parts equal, imaginary parts equal".

Worked Example 2 — solving by equating entries (with mark scheme)

Find $x$x, $y$y and $z$z given

$$\begin{pmatrix} 2x & 3 \\ y+1 & 2z-1 \end{pmatrix} = \begin{pmatrix} 6 & 3 \\ 4 & z+4 \end{pmatrix}.$$(2xy+1​32z−1​)=(64​3z+4​).

Equate corresponding entries:

$$\begin{aligned} 2x = 6 &\implies x = 3, \\ y + 1 = 4 &\implies y = 3, \\ 2z - 1 = z + 4 &\implies z = 5. \end{aligned}$$2x=6y+1=42z−1=z+4​⟹x=3,⟹y=3,⟹z=5.​

Mark scheme: M1 for equating at least two pairs of corresponding entries; A1 for $x = 3$x=3 and $y = 3$y=3; A1 for solving the linear equation $2z - 1 = z + 4$2z−1=z+4 to get $z = 5$z=5. The $(1,2)$(1,2) entry $3 = 3$3=3 is automatically consistent and carries no unknown — recognising that it imposes no extra condition is part of the reasoning.

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6. Specimen-style exam question

(Specimen-style — not a real past paper.) The matrices $P = \begin{pmatrix} 3 & a \\ -1 & 2 \end{pmatrix}$P=(3−1​a2​) and $Q = \begin{pmatrix} b & 4 \\ 5 & -2 \end{pmatrix}$Q=(b5​4−2​) satisfy $P + Q = \begin{pmatrix} 1 & 1 \\ 4 & 0 \end{pmatrix}$P+Q=(14​10​). Find $a$a and $b$b, and hence write down $P - Q$P−Q. (4 marks)

Model solution. Adding entry-by-entry, $P + Q = \begin{pmatrix} 3+b & a+4 \\ 4 & 0 \end{pmatrix}$P+Q=(3+b4​a+40​). Comparing with $\begin{pmatrix} 1 & 1 \\ 4 & 0 \end{pmatrix}$(14​10​):

$$3 + b = 1 \implies b = -2, \qquad a + 4 = 1 \implies a = -3.$$3+b=1⟹b=−2,a+4=1⟹a=−3.

The $(2,1)$(2,1) and $(2,2)$(2,2) entries ($4=4$4=4, $0=0$0=0) are automatically consistent. So $P = \begin{pmatrix} 3 & -3 \\ -1 & 2 \end{pmatrix}$P=(3−1​−32​), $Q = \begin{pmatrix} -2 & 4 \\ 5 & -2 \end{pmatrix}$Q=(−25​4−2​), giving

$$P - Q = \begin{pmatrix} 3-(-2) & -3-4 \\ -1-5 & 2-(-2) \end{pmatrix} = \begin{pmatrix} 5 & -7 \\ -6 & 4 \end{pmatrix}.$$P−Q=(3−(−2)−1−5​−3−42−(−2)​)=(5−6​−74​).

(M1 equate entries; A1 $b=-2$b=−2; A1 $a=-3$a=−3; A1 correct $P-Q$P−Q.)

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7. Synoptic links

• Complex numbers (compulsory pure): equating corresponding matrix entries is the same logical move as equating real and imaginary parts of $z = x + iy$z=x+iy. The complex number $a + bi$a+bi can itself be represented by the matrix $\begin{pmatrix} a & -b \\ b & a \end{pmatrix}$(ab​−ba​), with $i \leftrightarrow \begin{pmatrix} 0 & -1 \\ 1 & 0 \end{pmatrix}$i↔(01​−10​) — a link you will appreciate after the eigenvalue lessons.
• Vectors (A-Level Maths): a column vector is an $n\times 1$n×1 matrix; the dot product is a row vector times a column vector. This bridges directly into the transformation lessons.
• Later in matrices: the transpose underlies symmetric matrices and the adjugate route to inverses; the trace introduced in the stretch section returns as the sum of eigenvalues.
• Mechanics and statistics options (Paper 3): matrices encode systems of forces resolved in two directions and the transition matrices of probabilistic models; the algebra is identical, only the interpretation changes.

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8. Mark-scheme literacy

Examiners reward method that is visible and correct entry-by-entry. On these foundational questions:

• "Find" expects a final exact matrix or values — a decimal where an exact value is available can be penalised.
• For combined operations, the scalar multiples are method marks: write $2A$2A and $3B$3B out explicitly before combining, so a single arithmetic slip costs one accuracy mark, not the whole answer.
• For equating entries, line up the equations clearly; the command word is usually "hence find", which signals that you must use the matrix-equality argument rather than guessing.
• Follow-through (ft) marks are common: if you misread one entry but apply the correct method, later marks can still be earned. Showing structure protects you.
• When a question gives a constraint such as "$A$A is symmetric" or "$B$B is singular", that phrase is doing mathematical work — quote it explicitly and use it, because a mark is usually allocated for recognising and applying the property rather than ignoring it.
• For a "write down" instruction, no working is expected and none is credited — but a "find" or "determine" expects justified working, and an unsupported correct answer can score zero method marks even when the final value is right.

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9. Grade-band model answers

Question. The matrix $M = \begin{pmatrix} a & 2 \\ b & a \end{pmatrix}$M=(ab​2a​) is symmetric and satisfies $M + M^{T} = \begin{pmatrix} 4 & 4 \\ 4 & 4 \end{pmatrix}$M+MT=(44​44​). Find $a$a and $b$b, showing your reasoning. (4 marks)

Mid-band response. "$M^{T} = \begin{pmatrix} a & b \\ 2 & a \end{pmatrix}$MT=(a2​ba​). Adding: $\begin{pmatrix} 2a & 2+b \\ 2+b & 2a \end{pmatrix} = \begin{pmatrix} 4 & 4 \\ 4 & 4 \end{pmatrix}$(2a2+b​2+b2a​)=(44​44​). So $2a = 4$2a=4, $a = 2$a=2 and $2 + b = 4$2+b=4, $b = 2$b=2."

Examiner-style commentary: correct values and the key transpose step is shown, so this scores well. It does not, however, use the stated fact that $M$M is symmetric, nor check consistency — a marker looking for AO2 communication would want to see that loose end addressed.

Stronger response. "Since $M$M is symmetric, $M = M^{T}$M=MT, so the off-diagonal entries must agree: $2 = b$2=b. Then $M + M^{T} = 2M = \begin{pmatrix} 2a & 4 \\ 4 & 2a \end{pmatrix}$M+MT=2M=(2a4​42a​). Comparing with $\begin{pmatrix} 4 & 4 \\ 4 & 4 \end{pmatrix}$(44​44​) gives $2a = 4$2a=4, so $a = 2$a=2."

Examiner-style commentary: this uses symmetry up front to get $b = 2$b=2 directly, then exploits $M + M^{T} = 2M$M+MT=2M for symmetric $M$M — economical and fully justified. Every claim is warranted.

Top-band response. "Symmetry forces $b = 2$b=2 (off-diagonal entries equal), so $M = \begin{pmatrix} a & 2 \\ 2 & a \end{pmatrix}$M=(a2​2a​), which already equals its transpose. Hence $M + M^{T} = 2M$M+MT=2M, and the equation reduces to $2\begin{pmatrix} a & 2 \\ 2 & a \end{pmatrix} = \begin{pmatrix} 4 & 4 \\ 4 & 4 \end{pmatrix}$2(a2​2a​)=(44​44​). The off-diagonal entries are automatically $4$4 (consistent), and the diagonal gives $2a = 4 \Rightarrow a = 2$2a=4⇒a=2. Thus $a = b = 2$a=b=2, and $M = \begin{pmatrix} 2 & 2 \\ 2 & 2 \end{pmatrix}$M=(22​22​), which indeed satisfies the condition."

Examiner-style commentary: full marks. The candidate states the logical force of each hypothesis ("symmetry forces…", "automatically consistent…"), reduces the problem before computing, and verifies the final matrix against the original condition. This is precisely the AO2 communication top answers display.

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10. Common misconceptions

1. "Order is columns × rows." No — it is rows × columns, and $a_{ij}$aij​ is row $i$i, column $j$j. Reversing this corrupts every later calculation.
2. "You can add any two matrices." Addition needs identical orders. A $2\times 3$2×3 and a $3\times 2$3×2 cannot be added even though both have six entries.
3. "The transpose of a product is $A^{T}B^{T}$ATBT." It is $(AB)^{T} = B^{T}A^{T}$(AB)T=BTAT — the order reverses.
4. "A diagonal matrix must be the identity." A diagonal matrix only requires zeros off the diagonal; the diagonal entries can be any numbers, e.g. $\operatorname{diag}(3,7)$diag(3,7).
5. "Symmetric means the diagonal is symmetric." Symmetry is about $a_{ij} = a_{ji}$aij​=aji​ (mirror across the diagonal); the diagonal entries themselves are unconstrained.
6. "Scalar multiplication multiplies just the first entry / the diagonal." It multiplies every entry by the scalar.
7. "$O$O and $I$I are interchangeable." $O$O is the additive identity ($A + O = A$A+O=A); $I$I is the multiplicative identity ($AI = A$AI=A, next lesson). They play opposite roles.

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11. Common errors

• Writing the order columns-first (calling a $2\times3$2×3 matrix "$3\times2$3×2").
• Sign slips when subtracting, especially $b - (-3)$b−(−3) type entries — write the double negative explicitly.
• Forgetting to multiply every entry in a scalar multiple (commonly leaving a row untouched).
• Transposing only the off-diagonal entries and forgetting that the shape changes for non-square matrices.
• In equality problems, stopping after one equation instead of solving the full set, or missing that one entry is automatically consistent.
• Confusing the zero matrix $O$O with the scalar $0$0 when a question mixes the two.

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12. Going further (stretch)

The set $M_{n}(\mathbb{R})$Mn​(R) of $n\times n$n×n real matrices, with addition and scalar multiplication, is a vector space of dimension $n^{2}$n2 — a natural basis is the matrices $E_{ij}$Eij​ with a single $1$1 in position $(i,j)$(i,j) and $0$0s elsewhere. A neat STEP/MAT-flavoured fact: every square matrix splits uniquely as the sum of a symmetric and a skew-symmetric part,

$$A = \underbrace{\tfrac{1}{2}(A + A^{T})}_{\text{symmetric}} + \underbrace{\tfrac{1}{2}(A - A^{T})}_{\text{skew-symmetric}}.$$A=symmetric21​(A+AT)​​+skew-symmetric21​(A−AT)​​.

Check the symmetric part: $\left(\tfrac12(A+A^{T})\right)^{T} = \tfrac12(A^{T} + (A^{T})^{T}) = \tfrac12(A^{T} + A)$(21​(A+AT))T=21​(AT+(AT)T)=21​(AT+A), which is the matrix itself — so it is symmetric. Likewise $\left(\tfrac12(A-A^{T})\right)^{T} = \tfrac12(A^{T} - A) = -\tfrac12(A - A^{T})$(21​(A−AT))T=21​(AT−A)=−21​(A−AT), so the second part is skew-symmetric. Adding the two brackets gives $\tfrac12(A + A^{T}) + \tfrac12(A - A^{T}) = A$21​(A+AT)+21​(A−AT)=A. The decomposition is also unique: if $A = S + K$A=S+K with $S$S symmetric and $K$K skew, then transposing gives $A^{T} = S - K$AT=S−K, and adding/subtracting recovers $S = \tfrac12(A+A^{T})$S=21​(A+AT), $K = \tfrac12(A-A^{T})$K=21​(A−AT). This decomposition reappears in mechanics (strain vs rotation) and in pure analysis, and is a favourite "show that" springboard. A further observation: $\operatorname{tr}(A) = \sum_i a_{ii}$tr(A)=∑i​aii​, the trace, satisfies $\operatorname{tr}(A+B) = \operatorname{tr}(A) + \operatorname{tr}(B)$tr(A+B)=tr(A)+tr(B), $\operatorname{tr}(kA) = k\operatorname{tr}(A)$tr(kA)=ktr(A) and $\operatorname{tr}(A^{T}) = \operatorname{tr}(A)$tr(AT)=tr(A) — it will return as the sum of eigenvalues.

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13. Additional A* practice (with worked answers)

P1. Given $A = \begin{pmatrix} 1 & -2 \\ 3 & 0 \end{pmatrix}$A=(13​−20​) and $B = \begin{pmatrix} 4 & 1 \\ -1 & 5 \end{pmatrix}$B=(4−1​15​), find $3A - 2B$3A−2B.

Answer: $3A = \begin{pmatrix} 3 & -6 \\ 9 & 0 \end{pmatrix}$3A=(39​−60​), $2B = \begin{pmatrix} 8 & 2 \\ -2 & 10 \end{pmatrix}$2B=(8−2​210​), so $3A - 2B = \begin{pmatrix} -5 & -8 \\ 11 & -10 \end{pmatrix}$3A−2B=(−511​−8−10​).

P2. Find $a$a and $b$b if $\begin{pmatrix} a + 1 & 2b \\ 3 & a - b \end{pmatrix} = \begin{pmatrix} 4 & 6 \\ 3 & 0 \end{pmatrix}$(a+13​2ba−b​)=(43​60​).

Answer: $a + 1 = 4 \Rightarrow a = 3$a+1=4⇒a=3; $2b = 6 \Rightarrow b = 3$2b=6⇒b=3; check the $(2,2)$(2,2) entry: $a - b = 0$a−b=0 ✓ (consistent).

P3. Write $A = \begin{pmatrix} 2 & 5 \\ -1 & 4 \end{pmatrix}$A=(2−1​54​) as the sum of a symmetric and a skew-symmetric matrix.

Answer: $\tfrac12(A + A^{T}) = \begin{pmatrix} 2 & 2 \\ 2 & 4 \end{pmatrix}$21​(A+AT)=(22​24​) (symmetric) and $\tfrac12(A - A^{T}) = \begin{pmatrix} 0 & 3 \\ -3 & 0 \end{pmatrix}$21​(A−AT)=(0−3​30​) (skew-symmetric); their sum is $A$A ✓.

P4. A $3\times3$3×3 matrix has entries $a_{ij} = i - j$aij​=i−j. Write it out and state whether it is symmetric, skew-symmetric or neither.

Answer: $\begin{pmatrix} 0 & -1 & -2 \\ 1 & 0 & -1 \\ 2 & 1 & 0 \end{pmatrix}$​012​−101​−2−10​​. Since $a_{ji} = j - i = -(i-j) = -a_{ij}$aji​=j−i=−(i−j)=−aij​, it is skew-symmetric (note the zero diagonal).

P5. Find the $2\times2$2×2 matrix $X$X such that $2X^{T} = \begin{pmatrix} 4 & 6 \\ -2 & 8 \end{pmatrix}$2XT=(4−2​68​).

Answer: First $X^{T} = \begin{pmatrix} 2 & 3 \\ -1 & 4 \end{pmatrix}$XT=(2−1​34​). Transposing both sides (and using $(X^{T})^{T} = X$(XT)T=X) gives $X = \begin{pmatrix} 2 & -1 \\ 3 & 4 \end{pmatrix}$X=(23​−14​). (Check: transpose $X$X back and double it — you recover the original.)

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Board alignment

This content is aligned to AQA A-Level Further Mathematics (7367), compulsory pure (Papers 1 & 2). The definitions, transpose properties and equality method are identical in Edexcel (9FM0) and OCR (H245 / MEI H645) — only question styling differs, so this lesson serves all three boards.

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Visual summary

$$\boxed{\;\text{order } m\times n \;\;\big|\;\; a_{ij}=\text{row }i,\text{ col }j \;\;\big|\;\; (kA)_{ij}=ka_{ij} \;\;\big|\;\; (A^{T})_{ij}=a_{ji} \;\;\big|\;\; A=A^{T}\Leftrightarrow\text{symmetric}\;}$$order m×n​aij​=row i, col j​(kA)ij​=kaij​​(AT)ij​=aji​​A=AT⇔symmetric​

Operation	Condition	Result order
$A + B$A+B	same order $m\times n$m×n	$m\times n$m×n
$kA$kA	any $A$A	same as $A$A
$A^{T}$AT	any $A$A ($m\times n$m×n)	$n\times m$n×m
equality $A=B$A=B	same order	$mn$mn scalar equations

Exam takeaway: state orders explicitly, double-negate carefully when subtracting, remember $(AB)^{T}=B^{T}A^{T}$(AB)T=BTAT, and treat a matrix equation as a tidy bundle of scalar equations.