Matrices — Order, Types, Equality and the Linear Operations

A matrix is an ordered rectangular array; with $m$ rows and $n$ columns its order is $m \times n$ (rows first, always) and it has $mn$ entries, the entry in row $i$, column $j$ written $a_{ij}$. A matrix with $k$ entries can take one order for each ordered factor pair — $18$ entries allow $6$ orders: $1 \times 18, 2 \times 9, 3 \times 6, 6 \times 3, 9 \times 2, 18 \times 1$.

The type gallery: row matrix ($1 \times n$), column matrix ($m \times 1$), square ($m = n$), diagonal (square, $a_{ij} = 0$ off the diagonal), scalar (diagonal with equal diagonal entries $k$ — i.e. $kI$), identity $I$ (scalar with $k = 1$), and zero matrix $O$. Equality $A = B$ needs the same order and every corresponding entry equal — matching some entries is worth nothing; this is what turns matrix equations into systems of ordinary equations.

Addition and scalar multiplication

Defined entrywise, addition only for matrices of the same order. The full rulebook: $A + B = B + A$, $(A + B) + C = A + (B + C)$, $A + O = A$, $A + (-A) = O$, $k(A + B) = kA + kB$, $(k + l)A = kA + lA$. Matrix equations like $2A + 3X = 5B$ solve exactly like linear equations: $X = \frac{1}{3}(5B - 2A)$ — isolate, then compute entrywise.

Multiplication — the Row-by-Column Rule and Its Surprises

$AB$ exists only when (columns of $A$) $=$ (rows of $B$): an $m \times n$ times an $n \times p$ gives an $m \times p$, with $$c_{ik} = \sum_{j=1}^{n} a_{ij}b_{jk}$$ — row $i$ of $A$ against column $k$ of $B$, multiply pairwise, add.

What survives from ordinary algebra: associativity $A(BC) = (AB)C$, distributivity over addition, and $AI = IA = A$. What breaks — the two facts every board and MCQ paper tests:

1. $AB \neq BA$ in general; indeed $BA$ may not even exist, or exist with a different order.
2. $AB = O$ does not force $A = O$ or $B = O$ — nonzero zero divisors exist, so a matrix factor can never be cancelled: $AB = AC$ does not give $B = C$.

Transpose, Symmetric and Skew-Symmetric

$A'$ (or $A^T$) swaps rows and columns. The four laws: $$(A')' = A, \qquad (kA)' = kA', \qquad (A + B)' = A' + B', \qquad (AB)' = B'A'$$ The reversal in the product law is the one to memorise consciously.

Symmetric: $A' = A$. Skew-symmetric: $A' = -A$, which forces every diagonal entry to satisfy $a_{ii} = -a_{ii}$, i.e. all diagonal entries are $0$. For any square $A$: $A + A'$ is symmetric, $A - A'$ is skew-symmetric, and hence the chapter's flagship theorem — $$\boxed{\;A = \underbrace{\tfrac{1}{2}(A + A')}_{\text{symmetric}} + \underbrace{\tfrac{1}{2}(A - A')}_{\text{skew-symmetric}}\;}$$ — every square matrix splits uniquely into a symmetric plus a skew-symmetric part. In the exam, compute $A'$, form both halves, verify each half's property, and check the halves add back to $A$.

Invertible Matrices

For square $A$, $B$ of the same order: $B$ is the inverse of $A$ if $AB = BA = I$; then $B = A^{-1}$. Two theorems close the chapter:

1. Uniqueness: if $B$ and $C$ both invert $A$, then $B = BI = B(AC) = (BA)C = IC = C$ — three lines, pure associativity, and a favourite 2-mark proof.
2. Reversal: $(AB)^{-1} = B^{-1}A^{-1}$ — same order-flip as the transpose law.

Worked Examples — Basics and Linear Operations

Example 1 — Counting possible orders

A matrix has 18 entries. What orders can it have?

Step 1 — ordered factor pairs: $mn = 18$ gives $(1,18), (2,9), (3,6), (6,3), (9,2), (18,1)$.

Answer: $6$ possible orders — ordered pairs, so $3 \times 6$ and $6 \times 3$ both count.

Example 2 — Building a matrix from its rule

Construct the $2 \times 2$ matrix $A = [a_{ij}]$ with $a_{ij} = \dfrac{(i + 2j)^2}{2}$.

Step 1 — evaluate each position: $a_{11} = \frac{9}{2}$, $a_{12} = \frac{25}{2}$, $a_{21} = \frac{16}{2} = 8$, $a_{22} = \frac{36}{2} = 18$.

Answer: $A = \begin{pmatrix} \frac{9}{2} & \frac{25}{2} \\ 8 & 18 \end{pmatrix}$ — remember $i$ is the row index, $j$ the column.

Example 3 — Equality as a system of equations

Find $x$ and $y$ if $\begin{pmatrix} x + 3 & 4 \\ y - 2 & 7 \end{pmatrix} = \begin{pmatrix} 5 & 4 \\ 3 & 7 \end{pmatrix}$.

Step 1 — match entry by entry: $x + 3 = 5$ and $y - 2 = 3$.

Answer: $x = 2$, $y = 5$ — the other two positions already agree, and all four had to be checked.

Example 4 — Solving a matrix equation

Find $X$ if $X + \begin{pmatrix} 1 & 2 \\ 3 & 4 \end{pmatrix} = \begin{pmatrix} 5 & 5 \\ 5 & 5 \end{pmatrix}$.

Step 1 — isolate: $X = \begin{pmatrix} 5 & 5 \\ 5 & 5 \end{pmatrix} - \begin{pmatrix} 1 & 2 \\ 3 & 4 \end{pmatrix}$.

Answer: $X = \begin{pmatrix} 4 & 3 \\ 2 & 1 \end{pmatrix}$ — matrix equations in one unknown move exactly like ordinary linear equations.

Worked Examples — Multiplication

Example 5 — $AB$ and $BA$, side by side

For $A = \begin{pmatrix} 1 & 2 \\ 3 & 4 \end{pmatrix}$ and $B = \begin{pmatrix} 2 & 0 \\ 1 & 3 \end{pmatrix}$, compute $AB$ and $BA$.

Step 1 — row-by-column for $AB$: $(1,1)$: $1(2) + 2(1) = 4$; $(1,2)$: $0 + 6 = 6$; $(2,1)$: $6 + 4 = 10$; $(2,2)$: $0 + 12 = 12$.

Step 2 — and for $BA$: $(1,1)$: $2(1) + 0(3) = 2$; $(1,2)$: $4$; $(2,1)$: $1 + 9 = 10$; $(2,2)$: $2 + 12 = 14$.

Answer: $AB = \begin{pmatrix} 4 & 6 \\ 10 & 12 \end{pmatrix} \neq \begin{pmatrix} 2 & 4 \\ 10 & 14 \end{pmatrix} = BA$ — one computed pair is a complete proof that multiplication is not commutative.

Example 6 — Zero divisors

Show that $AB = O$ is possible with $A \neq O$ and $B \neq O$.

Step 1 — the standard pair: $$\begin{pmatrix} 0 & -1 \\ 0 & 2 \end{pmatrix}\begin{pmatrix} 3 & 5 \\ 0 & 0 \end{pmatrix} = \begin{pmatrix} 0 & 0 \\ 0 & 0 \end{pmatrix}$$ (each row of $A$ meets each column of $B$ where the nonzero entries miss each other).

Answer: both factors nonzero, product zero — hence cancellation of matrix factors is illegal in general.

Example 7 — A matrix satisfying a quadratic

Show that $A = \begin{pmatrix} 2 & 0 \\ 0 & 3 \end{pmatrix}$ satisfies $A^2 - 5A + 6I = O$.

Step 1 — compute the pieces: $A^2 = \begin{pmatrix} 4 & 0 \\ 0 & 9 \end{pmatrix}$, $5A = \begin{pmatrix} 10 & 0 \\ 0 & 15 \end{pmatrix}$, $6I = \begin{pmatrix} 6 & 0 \\ 0 & 6 \end{pmatrix}$.

Step 2 — combine entrywise: $(4 - 10 + 6, \; 9 - 15 + 6)$ on the diagonal — both zero.

Answer: $A^2 - 5A + 6I = O$. ✓ (The diagonal entries $2, 3$ are the roots of $k^2 - 5k + 6 = 0$ — no coincidence.)

Example 8 — Conformability

$A$ is $2 \times 3$ and $B$ is $3 \times 4$. Which of $AB$, $BA$ exist, and with what order?

Step 1 — $AB$: inner dimensions $3 = 3$ match: $AB$ exists with order $2 \times 4$.

Step 2 — $BA$: would need $4 = 2$ — does not exist.

Answer: only $AB$, of order $2 \times 4$ — existence of $AB$ says nothing about $BA$.

Worked Examples — Transpose, Symmetry, Inverse

Example 9 — Verifying the reversal law

For $A = \begin{pmatrix} 1 & 2 \\ 0 & 3 \end{pmatrix}$ and $B = \begin{pmatrix} 2 & 1 \\ 4 & 0 \end{pmatrix}$, verify $(AB)' = B'A'$.

Step 1 — left side: $AB = \begin{pmatrix} 10 & 1 \\ 12 & 0 \end{pmatrix}$, so $(AB)' = \begin{pmatrix} 10 & 12 \\ 1 & 0 \end{pmatrix}$.

Step 2 — right side: $B'A' = \begin{pmatrix} 2 & 4 \\ 1 & 0 \end{pmatrix}\begin{pmatrix} 1 & 0 \\ 2 & 3 \end{pmatrix} = \begin{pmatrix} 10 & 12 \\ 1 & 0 \end{pmatrix}$.

Answer: equal — and note $A'B'$ would give a different matrix: the reversal is essential, not cosmetic.

Example 10 — The symmetric plus skew split

Express $A = \begin{pmatrix} 3 & 5 \\ 1 & -1 \end{pmatrix}$ as the sum of a symmetric and a skew-symmetric matrix.

Step 1 — transpose: $A' = \begin{pmatrix} 3 & 1 \\ 5 & -1 \end{pmatrix}$.

Step 2 — the two halves: $$P = \tfrac{1}{2}(A + A') = \begin{pmatrix} 3 & 3 \\ 3 & -1 \end{pmatrix}, \qquad Q = \tfrac{1}{2}(A - A') = \begin{pmatrix} 0 & 2 \\ -2 & 0 \end{pmatrix}$$

Step 3 — verify all three claims: $P' = P$ ✓, $Q' = -Q$ ✓, and $P + Q = A$ ✓.

Answer: $A = P + Q$ as displayed — the three verification lines are where the marks sit.

Example 11 — Why the halves work in general

Show that for any square matrix $A$, $A + A'$ is symmetric and $A - A'$ is skew-symmetric.

Step 1 — transpose the sum: $(A + A')' = A' + (A')' = A' + A = A + A'$ — equals itself: symmetric. ✓

Step 2 — transpose the difference: $(A - A')' = A' - A = -(A - A')$ — equals its own negative: skew-symmetric. ✓

Answer: proved in two lines from the transpose laws — this is the standard 2-mark theory question behind Example 10.

Example 12 — Verifying an inverse

Show that $B = \begin{pmatrix} 2 & -3 \\ -1 & 2 \end{pmatrix}$ is the inverse of $A = \begin{pmatrix} 2 & 3 \\ 1 & 2 \end{pmatrix}$.

Step 1 — $AB$: $(1,1)$: $4 - 3 = 1$; $(1,2)$: $-6 + 6 = 0$; $(2,1)$: $2 - 2 = 0$; $(2,2)$: $-3 + 4 = 1$ — so $AB = I$.

Step 2 — $BA$: the same arithmetic gives $BA = I$.

Answer: $AB = BA = I$, so $B = A^{-1}$ — the definition asks for both products, and the board answer should show both.

Example 13 — Uniqueness of the inverse

Prove that if a square matrix has an inverse, it has only one.

Step 1 — suppose two: let $B$ and $C$ both satisfy $AB = BA = I$ and $AC = CA = I$.

Step 2 — the associativity chain: $$B = BI = B(AC) = (BA)C = IC = C$$

Answer: $B = C$ — uniqueness proved. ∎ Learn the chain as one sentence; it reappears in Determinants when $A^{-1}$ gets its formula.