Multiplication of Matrices

The rule: rows meet columns

Definition. The product $AB$ is defined only when the number of columns of $A$ equals the number of rows of $B$. If $A = [a_{ik}]_{m \times n}$ and $B = [b_{kj}]_{n \times p}$, then $AB = [c_{ij}]_{m \times p}$ with $$c_{ij} = \sum_{k=1}^{n} a_{ik}\,b_{kj} = a_{i1}b_{1j} + a_{i2}b_{2j} + \cdots + a_{in}b_{nj}$$ — the entry at $(i, j)$ is row $i$ of $A$ times column $j$ of $B$: multiply across the row and down the column, then add.

Order bookkeeping: $(m \times n)(n \times p) \to m \times p$ — the inner numbers must match and cancel; the outer numbers survive.

Worked product. $$AB = \begin{pmatrix} 1 & 2 \\ 3 & 4 \end{pmatrix}\begin{pmatrix} 2 & 0 \\ 1 & 3 \end{pmatrix} = \begin{pmatrix} 1\cdot 2 + 2 \cdot 1 & 1 \cdot 0 + 2 \cdot 3 \\ 3 \cdot 2 + 4 \cdot 1 & 3 \cdot 0 + 4 \cdot 3 \end{pmatrix} = \begin{pmatrix} 4 & 6 \\ 10 & 12 \end{pmatrix}$$

Surprise 1: $AB \neq BA$

With the same matrices in the other order: $$BA = \begin{pmatrix} 2 & 0 \\ 1 & 3 \end{pmatrix}\begin{pmatrix} 1 & 2 \\ 3 & 4 \end{pmatrix} = \begin{pmatrix} 2 & 4 \\ 10 & 14 \end{pmatrix} \neq AB$$ Matrix multiplication is not commutative in general. Worse, $BA$ may not even exist: if $A$ is $2 \times 3$ and $B$ is $3 \times 4$, then $AB$ is $2 \times 4$ but $BA$ pairs $(3 \times 4)(2 \times 3)$ — inner numbers 4 and 2 clash. Both products exist and have the same order only when $A$ and $B$ are square of the same order — and even then they usually differ.

Surprise 2: zero divisors

For numbers, $ab = 0$ forces $a = 0$ or $b = 0$. For matrices: $$\begin{pmatrix} 0 & -1 \\ 0 & 2 \end{pmatrix}\begin{pmatrix} 3 & 5 \\ 0 & 0 \end{pmatrix} = \begin{pmatrix} 0 & 0 \\ 0 & 0 \end{pmatrix}$$ — the product of two nonzero matrices can be the zero matrix. Consequently there is no cancellation law: $AB = AC$ does not imply $B = C$.

Properties That Do Survive

The good news

For matrices with compatible orders:

Associativity is what makes powers unambiguous: $A^3 = AAA$ needs no brackets. And because multiplication does not commute, left and right distributivity are separate facts — expand $(A + B)^2$ carefully: $$(A + B)^2 = A^2 + AB + BA + B^2 \qquad (\text{NOT } A^2 + 2AB + B^2 \text{ unless } AB = BA)$$ The same caution kills the familiar identities: $(A+B)(A-B) = A^2 - AB + BA - B^2 \neq A^2 - B^2$ in general.

Row and column vectors as products

A $1 \times n$ row times an $n \times 1$ column is a $1 \times 1$ matrix — essentially a number: $$\begin{pmatrix} 1 & -2 & 3 \end{pmatrix}\begin{pmatrix} 2 \\ 3 \\ 1 \end{pmatrix} = [\,1 \cdot 2 - 2 \cdot 3 + 3 \cdot 1\,] = [-1]$$ In the other order, $(n \times 1)(1 \times n)$ blows up into an $n \times n$ matrix — a dramatic illustration of how much the order matters.

Multiplication as aggregation (the word problems)

A bookshop holds 120 chemistry, 96 physics and 120 economics books, selling at 80, 60 and 40 rupees. The total revenue is the product $$\begin{pmatrix} 120 & 96 & 120 \end{pmatrix}\begin{pmatrix} 80 \\ 60 \\ 40 \end{pmatrix} = [\,9600 + 5760 + 4800\,] = [20160]$$ i.e. ₹ 20160 — quantities-row times prices-column. Every cost/revenue/audience word problem in the exercises is this pattern, sometimes with several rows (several shops) or several columns (several price lists) at once.

Common mistakes to avoid

Mistake 1 — multiplying entrywise. $(AB)_{ij}$ is a sum of products along row $i$ and column $j$, never $a_{ij}b_{ij}$.

Mistake 2 — assuming $AB = BA$. Never assume it; when a problem says the matrices commute, that is a special hypothesis to be used.

Mistake 3 — expanding $(A+B)^2$ with the binomial formula. Keep $AB$ and $BA$ separate unless told they are equal.

Mistake 4 — cancelling matrices. From $AB = AC$ you may NOT conclude $B = C$ (no division; zero divisors exist).

Mistake 5 — order bookkeeping. Before computing anything, check conformability and write down the product's order — half the MCQs are settled by the orders alone.

Solved Examples

Example 1 — A full $2 \times 2$ product, both orders

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

Step 1 — $AB$ row-by-column: $$AB = \begin{pmatrix} 4 & 6 \\ 10 & 12 \end{pmatrix}$$

Step 2 — $BA$: $$BA = \begin{pmatrix} 2 \cdot 1 + 0 \cdot 3 & 2\cdot 2 + 0 \cdot 4 \\ 1 \cdot 1 + 3 \cdot 3 & 1 \cdot 2 + 3 \cdot 4 \end{pmatrix} = \begin{pmatrix} 2 & 4 \\ 10 & 14 \end{pmatrix}$$

Answer: $AB \neq BA$ — already the $(1,1)$ entries (4 vs 2) disagree. One disagreement suffices.

Example 2 — Row times column, column times row

For $A = \begin{pmatrix} 1 & -2 & 3 \end{pmatrix}$ and $B = \begin{pmatrix} 2 \\ 3 \\ 1 \end{pmatrix}$, find $AB$ and the order of $BA$.

Step 1 — $AB$: $(1 \times 3)(3 \times 1) \to 1 \times 1$: $$AB = [\,2 - 6 + 3\,] = [-1]$$

Step 2 — $BA$: $(3 \times 1)(1 \times 3) \to 3 \times 3$ — a full $3 \times 3$ matrix (each entry $b_i a_j$).

Answer: $AB = [-1]$ while $BA$ is $3 \times 3$ — the two orders do not even have the same shape.

Example 3 — Nonzero matrices with zero product

Show that $A = \begin{pmatrix} 0 & -1 \\ 0 & 2 \end{pmatrix}$ and $B = \begin{pmatrix} 3 & 5 \\ 0 & 0 \end{pmatrix}$ satisfy $AB = O$ although $A \neq O$ and $B \neq O$.

Step 1 — compute: $$AB = \begin{pmatrix} 0 \cdot 3 + (-1) \cdot 0 & 0 \cdot 5 + (-1) \cdot 0 \\ 0 \cdot 3 + 2 \cdot 0 & 0 \cdot 5 + 2 \cdot 0 \end{pmatrix} = \begin{pmatrix} 0 & 0 \\ 0 & 0 \end{pmatrix}$$

Answer: $AB = O$ with both factors nonzero — matrices admit zero divisors, so "$AB = O \Rightarrow A = O$ or $B = O$" is FALSE, and cancellation fails with it.

Example 4 — Conformability bookkeeping

$A$ is $2 \times 3$, $B$ is $3 \times 4$, $C$ is $4 \times 2$. Which of $AB$, $BA$, $BC$, $CA$, $(AB)C$, $C(AB)$ are defined, and with what orders?

Step 1 — test inner numbers pair by pair: $AB$: $(2 \times 3)(3 \times 4) \to 2 \times 4$ ✓. $\ BA$: $(3 \times 4)(2 \times 3)$ — inner 4 vs 2 ✗. $\ BC$: $(3 \times 4)(4 \times 2) \to 3 \times 2$ ✓. $\ CA$: $(4 \times 2)(2 \times 3) \to 4 \times 3$ ✓.

Step 2 — chain: $(AB)C$: $(2 \times 4)(4 \times 2) \to 2 \times 2$ ✓; $\ C(AB)$: $(4 \times 2)(2 \times 4) \to 4 \times 4$ ✓.

Answer: all but $BA$ are defined, with the orders listed — order bookkeeping first, arithmetic second.

Example 5 — Verifying associativity on numbers

For $A = \begin{pmatrix} 1 & 1 \\ 0 & 2 \end{pmatrix}$, $B = \begin{pmatrix} 2 & 0 \\ 1 & 1 \end{pmatrix}$, $C = \begin{pmatrix} 1 & 0 \\ 1 & 1 \end{pmatrix}$, verify $(AB)C = A(BC)$.

Step 1 — left grouping: $AB = \begin{pmatrix} 3 & 1 \\ 2 & 2 \end{pmatrix}$, then $(AB)C = \begin{pmatrix} 3 + 1 & 1 \\ 2 + 2 & 2 \end{pmatrix} = \begin{pmatrix} 4 & 1 \\ 4 & 2 \end{pmatrix}$.

Step 2 — right grouping: $BC = \begin{pmatrix} 2 & 0 \\ 2 & 1 \end{pmatrix}$, then $A(BC) = \begin{pmatrix} 2 + 2 & 1 \\ 4 & 2 \end{pmatrix} = \begin{pmatrix} 4 & 1 \\ 4 & 2 \end{pmatrix}$.

Answer: both groupings agree — associativity in action (it holds always; a numerical verification is how boards ask it).

Example 6 — The identity in action

Show that $AI = IA = A$ for $A = \begin{pmatrix} a & b \\ c & d \end{pmatrix}$ and $I = \begin{pmatrix} 1 & 0 \\ 0 & 1 \end{pmatrix}$.

Step 1 — $AI$: row-by-column gives $\begin{pmatrix} a & b \\ c & d \end{pmatrix}$ back (each row of $A$ picks out the matching column of $I$).

Step 2 — $IA$: likewise returns $A$.

Answer: $I$ is the multiplicative identity — the one matrix that commutes with everything of its order.

Example 7 — A revenue product

A bookshop has 120 chemistry, 96 physics and 120 economics books, priced at 80, 60 and 40 rupees respectively. Use a matrix product to find the total revenue if all are sold.

Step 1 — quantities row times prices column: $$\begin{pmatrix} 120 & 96 & 120 \end{pmatrix}\begin{pmatrix} 80 \\ 60 \\ 40 \end{pmatrix} = [\,9600 + 5760 + 4800\,] = [20160]$$

Answer: ₹ 20160. The row-times-column sum is the natural "quantity times price, added up" — which is why matrix multiplication is defined the way it is.