How a Shopkeeper's Stock Register Is Already a Matrix

A shopkeeper's register showing how many bags of rice, dal and sugar are stored in two godowns is a rectangular arrangement of numbers — a matrix.

This part covers what a matrix is and when two are equal, types of matrices, addition and scalar multiplication, and solving matrix equations.

**A matrix is a rectangular array of numbers in rows and columns, its order $m\times n$ gives the number of rows and columns, and two matrices are equal only when they have the same order and every pair of corresponding entries is equal.

Notation.** A matrix is written $A=[a_{ij}{]}_{m\times n}$, where $a_{ij}$ is the entry in the **$i$th row and $j$th column**. An $m\times n$ matrix has $mn$ entries.

Worked example 1 — reading entries.

$$A=\left[\begin{array}{ccc}2& 5& -1\\ 0& 3& 7\end{array}\right]$$

The order is $2\times 3$, so there are $6$ entries; $a_{12}=5$ and $a_{23}=7$.

Possible orders. A matrix with $12$ entries can have order $1\times 12$, $2\times 6$, $3\times 4$, $4\times 3$, $6\times 2$ or $12\times 1$ — six orders in all.

An everyday example. A school timetable with 6 days as rows and 8 periods as columns is a $6\times 8$ matrix of subjects.

The substance. Matrices of different orders are never equal, even if they contain the same numbers.

A row matrix has one row, a column matrix has one column, a square matrix has as many rows as columns, a diagonal matrix is square with zeros off the main diagonal, a scalar matrix is diagonal with equal diagonal entries, an identity matrix is scalar with diagonal entries 1, and a zero matrix has every entry 0.

Types with examples:

- Row matrix — order $1\times n$: $\left[\begin{array}{ccc}3& -1& 4\end{array}\right]$
- Column matrix — order $m\times 1$: $\left[\begin{array}{c}2\\ 5\end{array}\right]$
- Square matrix — $m=n$; its main diagonal is $a_{11},a_{22},\dots ,a_{nn}$
- Diagonal matrix — square, with $a_{ij}=0$ whenever $i\ne j$: $\left[\begin{array}{cc}4& 0\\ 0& -3\end{array}\right]$
- Scalar matrix — diagonal, with every diagonal entry equal to some $k$: $\left[\begin{array}{cc}5& 0\\ 0& 5\end{array}\right]$
- Identity matrix — scalar with $k=1$: $I_2=\left[\begin{array}{cc}1& 0\\ 0& 1\end{array}\right]$
- Zero matrix $O$ — every entry $0$, of any order

An everyday example. One column listing every student's marks in a single test is a column matrix; one student's marks across five subjects is a row matrix.

The substance. A diagonal matrix may have zeros on its diagonal too — the only condition is that off-diagonal entries are zero.

**Matrices of the same order are added or subtracted entry by entry, a matrix is multiplied by a scalar by multiplying every entry, and matrix addition is commutative and associative, with the zero matrix as additive identity and $-A$ as additive inverse.

Operations:

- Addition** — $A+B=[a_{ij}+b_{ij}]$, only when $A$ and $B$ have the same order
- Scalar multiplication — $kA=[k{a}_{ij}]$
- Negative and subtraction — $-A=(-1)A$ and $A-B=A+(-1)B$

Worked example. Let $A=\left[\begin{array}{cc}1& 3\\ -2& 4\end{array}\right]$ and $B=\left[\begin{array}{cc}5& -1\\ 0& 2\end{array}\right]$.

$$A+B=\left[\begin{array}{cc}6& 2\\ -2& 6\end{array}\right]2A-B=\left[\begin{array}{cc}-3& 7\\ -4& 6\end{array}\right]$$

Properties of addition for matrices of the same order:

- Commutative — $A+B=B+A$
- Associative — $(A+B)+C=A+(B+C)$
- Additive identity — $A+O=A$
- Additive inverse — $A+(-A)=O$

An everyday example. Two stationery shops' monthly sales matrices of pens and notebooks add entry by entry to give combined sales, and multiplying by $2$ shows the sales for two such months.

The substance. **Adding a $2\times 3$ matrix to a $3\times 2$ matrix is undefined**, even though both have six entries.

To solve a matrix equation, simplify each side into a single matrix using addition and scalar multiplication, then equate corresponding entries and solve the resulting ordinary equations.

Worked example 1. Find $x$ and $y$ if

$$2\left[\begin{array}{cc}1& 3\\ 0& x\end{array}\right]+\left[\begin{array}{cc}y& 0\\ 1& 2\end{array}\right]=\left[\begin{array}{cc}5& 6\\ 1& 8\end{array}\right]$$

The left side is $\left[\begin{array}{cc}2+y& 6\\ 1& 2x+2\end{array}\right]$. Equating entries:

$$2+y=5\Rightarrow y=32x+2=8\Rightarrow x=3$$

Worked example 2 — two unknown matrices. Solve

$$X+Y=\left[\begin{array}{cc}7& 0\\ 2& 5\end{array}\right]X-Y=\left[\begin{array}{cc}3& 0\\ 0& 3\end{array}\right]$$

Adding the equations gives $2X=\left[\begin{array}{cc}10& 0\\ 2& 8\end{array}\right]$, so $X=\left[\begin{array}{cc}5& 0\\ 1& 4\end{array}\right]$. Subtracting gives $2Y=\left[\begin{array}{cc}4& 0\\ 2& 2\end{array}\right]$, so $Y=\left[\begin{array}{cc}2& 0\\ 1& 1\end{array}\right]$.

An everyday example. Knowing a shop's closing stock matrix and its sales matrix for a month gives the opening stock matrix by the same entry-by-entry reasoning.

The substance. Substitute your answers back to check, since one wrong entry makes the matrices unequal.