Matrix, Order and the $a_{ij}$ Addressing Scheme

What a matrix is

A matrix is a rectangular array of numbers (or functions), arranged in rows and columns and enclosed in brackets. The entries are its elements. Matrices compress bookkeeping: the numbers of men and women in three families, the marks of a class in four subjects, the coefficients of a system of equations — each becomes a single object that can be computed with.

Order and addressing

A matrix with $m$ rows and $n$ columns has order $m \times n$ (read "$m$ by $n$" — rows first, always) and $mn$ elements. We write $$A = [a_{ij}]_{m \times n}$$ where $a_{ij}$ is the entry sitting in row $i$, column $j$. So $a_{23}$ lives in the second row, third column — the first subscript never means column.

Counting possible orders. A matrix with 8 elements can have any order $m \times n$ with $mn = 8$: namely $1 \times 8, \ 2 \times 4, \ 4 \times 2, \ 8 \times 1$ — one order per ordered factor pair. A matrix with a prime number of elements (say 13) has only the two extreme orders $1 \times 13$ and $13 \times 1$.

Building a matrix from a rule

Exam staple: construct the matrix whose entries obey a formula.

Worked construction. Build the $3 \times 2$ matrix with $a_{ij} = \frac{1}{2}\vert i - 3j \vert$.

Step 1 — list the addresses: rows $i = 1, 2, 3$ and columns $j = 1, 2$.

Step 2 — evaluate each: $$a_{11} = \tfrac{1}{2}\vert 1 - 3\vert = 1, \quad a_{12} = \tfrac{1}{2}\vert 1 - 6\vert = \tfrac{5}{2}, \quad a_{21} = \tfrac{1}{2}\vert 2 - 3 \vert = \tfrac{1}{2}$$ $$a_{22} = \tfrac{1}{2}\vert 2 - 6\vert = 2, \quad a_{31} = \tfrac{1}{2}\vert 3 - 3 \vert = 0, \quad a_{32} = \tfrac{1}{2}\vert 3 - 6\vert = \tfrac{3}{2}$$

Step 3 — assemble (rows across, columns down): first row $1, \frac{5}{2}$; second row $\frac{1}{2}, 2$; third row $0, \frac{3}{2}$.

The only real danger is swapping $i$ and $j$ — always evaluate with the address order (row, column).

Types of Matrices and Equality

The type gallery

Column matrix — one column, order $m \times 1$. Row matrix — one row, order $1 \times n$.

Square matrix — rows $=$ columns, called "of order $n$". Its entries $a_{11}, a_{22}, \ldots, a_{nn}$ form the diagonal.

Diagonal matrix — square with every off-diagonal entry zero: $b_{ij} = 0$ for $i \neq j$. (Diagonal entries may be anything, including zero.)

Scalar matrix — a diagonal matrix whose diagonal entries are all equal to some constant $k$.

Identity matrix $I_n$ — the scalar matrix with $k = 1$: ones on the diagonal, zeros elsewhere. Every identity matrix is scalar; a scalar matrix is an identity matrix only when $k = 1$.

Zero (null) matrix $O$ — all entries zero, of any order (not necessarily square).

The chain worth remembering: $$\text{identity} \subset \text{scalar} \subset \text{diagonal} \subset \text{square}$$

Equality of matrices

Definition. $A = [a_{ij}]$ and $B = [b_{ij}]$ are equal exactly when (i) they have the same order, and (ii) $a_{ij} = b_{ij}$ for every address $(i, j)$.

Equality is entry-by-entry, so a single matrix equation unpacks into a system of scalar equations — the chapter's standard "find the values" problem.

Worked unpacking. Find $a, b, c, d$ if $$\begin{pmatrix} a - b & 2a + c \\ 2a - b & 3c + d \end{pmatrix} = \begin{pmatrix} -1 & 5 \\ 0 & 13 \end{pmatrix}$$

Step 1 — equate corresponding entries: $a - b = -1$, $\ 2a + c = 5$, $\ 2a - b = 0$, $\ 3c + d = 13$.

Step 2 — solve in a smart order: subtracting the first from the third: $a = 1$; then $b = 2$, $c = 5 - 2 = 3$, $d = 13 - 9 = 4$.

Answer: $a = 1, \ b = 2, \ c = 3, \ d = 4$.

Consistency matters. If the system unpacked from a matrix equation is inconsistent, the matrices simply cannot be equal for any values — "not possible to find" is then the correct answer, and one such question is famous enough to appear in the exercises.

Common mistakes to avoid

Mistake 1 — reading $a_{ij}$ as (column, row). The first index is the row. $a_{12}$ and $a_{21}$ are different entries.

Mistake 2 — order written backwards. A matrix with 3 rows and 4 columns is $3 \times 4$, never $4 \times 3$.

Mistake 3 — comparing matrices of different orders. A $2 \times 3$ matrix is never equal to a $3 \times 2$ matrix, however similar the entries look.

Mistake 4 — 'diagonal' in a non-square matrix. The diagonal, and hence diagonal/scalar/identity types, exist only for square matrices; the zero matrix is the one special type of arbitrary shape.

Mistake 5 — solving only some of the unpacked equations. Every entry gives an equation; a candidate solution must satisfy all of them (else the correct verdict may be "not possible").

Solved Examples

Example 1 — Reading off order and entries

For $A = \begin{pmatrix} 2 & 5 & 19 & -7 \\ 35 & -2 & \frac{5}{2} & 12 \\ \sqrt 3 & 1 & -5 & 17 \end{pmatrix}$, find (i) the order, (ii) the number of elements, (iii) the entries $a_{13}, a_{21}, a_{33}, a_{24}, a_{23}$.

Step 1 — order: 3 rows, 4 columns: $3 \times 4$.

Step 2 — count: $3 \times 4 = 12$ elements.

Step 3 — address each entry (row first): $$a_{13} = 19, \quad a_{21} = 35, \quad a_{33} = -5, \quad a_{24} = 12, \quad a_{23} = \frac{5}{2}$$

Answer: order $3 \times 4$; 12 elements; values as above.

Example 2 — Counting possible orders

If a matrix has 8 elements, what are its possible orders? What if it has 5 elements?

Step 1 — factor pairs of 8: $(1, 8), (2, 4), (4, 2), (8, 1)$ — four possible orders.

Step 2 — 5 is prime: only $(1, 5)$ and $(5, 1)$ — two orders.

Answer: 8 elements allow $1 \times 8, 2 \times 4, 4 \times 2, 8 \times 1$; 5 elements allow only $1 \times 5$ and $5 \times 1$. (Ordered pairs: $2 \times 4$ and $4 \times 2$ are different orders.)

Example 3 — Construction from a rule

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

Step 1 — evaluate all four addresses: $$a_{11} = \frac{4}{2} = 2, \qquad a_{12} = \frac{9}{2}, \qquad a_{21} = \frac{9}{2}, \qquad a_{22} = \frac{16}{2} = 8$$

Step 2 — assemble: $$A = \begin{pmatrix} 2 & \frac{9}{2} \\ \frac{9}{2} & 8 \end{pmatrix}$$

Answer: as above — note the symmetry, inevitable since the rule depends only on $i + j$.

Example 4 — A data matrix

A shop stocks 15 red pens, 20 blue pens in its first branch and 12 red, 28 blue in its second. Record the stock as a matrix and state what $a_{21}$ represents.

Step 1 — choose the convention: rows = branches, columns = colours (red, blue): $$A = \begin{pmatrix} 15 & 20 \\ 12 & 28 \end{pmatrix}$$

Step 2 — interpret: $a_{21} = 12$ is the entry in row 2 (second branch), column 1 (red pens).

Answer: $a_{21}$ = red pens in the second branch. A matrix is only meaningful together with its row/column convention — state it.

Example 5 — Classify the types

Classify: $A = [7]$, $\ B = \begin{pmatrix} 2 & 0 \\ 0 & 3 \end{pmatrix}$, $\ C = \begin{pmatrix} 4 & 0 \\ 0 & 4 \end{pmatrix}$, $\ D = \begin{pmatrix} 1 & 0 \\ 0 & 1 \end{pmatrix}$, $\ E = \begin{pmatrix} 0 & 0 \\ 0 & 0 \end{pmatrix}$.

Step 1 — work down the chain: all five are square. $B$: off-diagonal zeros, unequal diagonal → diagonal (not scalar). $C$: diagonal with equal entries 4 → scalar (not identity). $D$: scalar with $k = 1$ → identity. $E$: all zeros → zero matrix (also diagonal and scalar with $k = 0$). $A$: a $1 \times 1$ matrix is simultaneously diagonal and scalar.

Answer: $B$ diagonal, $C$ scalar, $D$ identity, $E$ zero — the gallery in action, with the inclusions identity $\subset$ scalar $\subset$ diagonal visible.

Example 6 — Equality: solving the system

Find $x, y, z$ if $\begin{pmatrix} x + y & 2 \\ 5 + z & xy \end{pmatrix} = \begin{pmatrix} 6 & 2 \\ 5 & 8 \end{pmatrix}$.

Step 1 — unpack: $x + y = 6$, $\ 5 + z = 5$, $\ xy = 8$ (and $2 = 2$ is automatic).

Step 2 — solve: $z = 0$; and $x + y = 6$, $xy = 8$ make $x, y$ the roots of $t^2 - 6t + 8 = 0$: $\ t = 2, 4$.

Answer: $z = 0$ and $\{x, y\} = \{2, 4\}$ (either assignment) — matrix equality routinely produces small nonlinear systems like this.

Example 7 — Equality: the inconsistent case

Can values of $x$ and $y$ make $\begin{pmatrix} 3x + 7 & 5 \\ y + 1 & 2 - 3x \end{pmatrix}$ and $\begin{pmatrix} 0 & y - 2 \\ 8 & 4 \end{pmatrix}$ equal?

Step 1 — unpack all four equations: $3x + 7 = 0$, $\ 5 = y - 2$, $\ y + 1 = 8$, $\ 2 - 3x = 4$.

Step 2 — test consistency: the first gives $x = -\frac{7}{3}$; the fourth gives $x = -\frac{2}{3}$ — contradiction (the two $y$-equations agree on $y = 7$, but $x$ cannot satisfy both of its equations).

Answer: not possible — no values of $x$ and $y$ make the matrices equal. Checking every equation is the entire point of this classic.