Matrices of the same order are added entry by entry and scaled by multiplying every entry. The product $AB$ is built row by column — and, unlike numbers, $AB$ need not equal $BA$.
The stationery chain of Section 3.1 now has two weekly sales matrices. To get the fortnight's sales it adds them position by position; to plan for a festival week in which every branch expects to sell one and a half times as much, it multiplies every entry by $1.5$. These two operations — addition and scalar multiplication — behave exactly as you would hope, because they are just ordinary arithmetic done in every cell at once.
The third operation is different. To turn “how many of each item” into “how much money”, each quantity must be multiplied by its own price and the results added up: a row of quantities meets a column of prices. Repeating this for every row and every column gives the product of matrices. It is the most useful operation in the subject — it also composes geometric transformations and solves systems of equations — but it has two habits that numbers do not: the order of the factors matters, and a product can be zero when neither factor is.
Addition and Scalar Multiplication
📘 Definition 3.6 — Sum of two matrices
If $A=[a_{ij}]$ and $B=[b_{ij}]$ have the same order $m\times n$, their sum is the $m\times n$ matrix
$$A+B=[a_{ij}+b_{ij}]_{m\times n}$$
obtained by adding corresponding entries. If $A$ and $B$ have different orders, $A+B$ is not defined.
📘 Definition 3.7 — Scalar multiple, negative and difference
For a real number $k$ (a scalar) and $A=[a_{ij}]_{m\times n}$, the scalar multiple $kA$ is the $m\times n$ matrix $kA=[k\,a_{ij}]$: every entry is multiplied by $k$. The negative of $A$ is $-A=(-1)A=[-a_{ij}]$, and for $A,B$ of the same order the difference is $A-B=A+(-B)=[a_{ij}-b_{ij}]$.
✏️ Example 1 — Adding and scaling
Let
$$A=\begin{bmatrix}2&-1&3\\0&4&-5\end{bmatrix}$$
$$B=\begin{bmatrix}1&3&-2\\-4&0&6\end{bmatrix}$$
Find $A+B$ and $2A-3B$. Is $A+C$ defined when $C=\begin{bmatrix}1&2\\3&4\end{bmatrix}$?
💡 Show solution▶
A + B Both are $2\times3$, so add position by position:
$$A+B=\begin{bmatrix}3&2&1\\-4&4&1\end{bmatrix}$$
2A − 3B $2A=\begin{bmatrix}4&-2&6\\0&8&-10\end{bmatrix}$ and $3B=\begin{bmatrix}3&9&-6\\-12&0&18\end{bmatrix}$, so
$$2A-3B=\begin{bmatrix}1&-11&12\\12&8&-28\end{bmatrix}$$
A + C Not defined: $A$ is $2\times3$ and $C$ is $2\times2$. There is no entry of $C$ to add to $a_{13}$.
Check One entry from scratch: position $(2,3)$ of $2A-3B$ is $2(-5)-3(6)=-28$. ✓
Because each entry of $A+B$ or $kA$ is an ordinary sum or product of real numbers, the familiar laws of arithmetic carry over position by position.
📐 Theorem 3.2 — Properties of matrix addition
Let $A$, $B$, $C$ be matrices of the same order $m\times n$, and let $O$ be the $m\times n$ zero matrix. Then
(commutative) $A+B=B+A$;
(associative) $(A+B)+C=A+(B+C)$;
(additive identity) $A+O=O+A=A$;
(additive inverse) $A+(-A)=(-A)+A=O$.
📐 Theorem 3.3 — Properties of scalar multiplication
Let $A$, $B$ be matrices of the same order and $k$, $l$ real numbers. Then $k(A+B)=kA+kB$, $\ (k+l)A=kA+lA$, $\ k(lA)=(kl)A$, $\ 1\cdot A=A$ and $0\cdot A=O$.
Why. Look at one position $(i,j)$. For commutativity, the $(i,j)$ entry of $A+B$ is $a_{ij}+b_{ij}$ and that of $B+A$ is $b_{ij}+a_{ij}$; these are equal real numbers. Since this holds in every position and both matrices have the same order, $A+B=B+A$ by Definition 3.5. For $k(A+B)=kA+kB$ the entries are $k(a_{ij}+b_{ij})$ and $ka_{ij}+kb_{ij}$, equal by the distributive law for numbers. The other laws are proved the same way.
These laws let you solve matrix equations involving sums exactly as you solve equations with numbers — as long as you never divide by a matrix.
✏️ Example 2 — Solving for unknown matrices
(a) Find matrices $X$ and $Y$ with
$$X+Y=\begin{bmatrix}5&2\\0&9\end{bmatrix}$$
$$X-Y=\begin{bmatrix}3&6\\0&-1\end{bmatrix}$$
(b) Find $X$ if $3X-2A=B$, where $A=\begin{bmatrix}1&-2\\3&0\end{bmatrix}$ and $B=\begin{bmatrix}4&1\\-1&3\end{bmatrix}$.
💡 Show solution▶
(a) add Adding the equations, $2X=\begin{bmatrix}8&8\\0&8\end{bmatrix}$, so $X=\begin{bmatrix}4&4\\0&4\end{bmatrix}$.
(a) subtract Subtracting, $2Y=\begin{bmatrix}2&-4\\0&10\end{bmatrix}$, so $Y=\begin{bmatrix}1&-2\\0&5\end{bmatrix}$. Check: $X+Y=\begin{bmatrix}5&2\\0&9\end{bmatrix}$ ✓.
(b) Add $2A$ to both sides: $3X=B+2A=\begin{bmatrix}6&-3\\5&3\end{bmatrix}$. Multiply by $\tfrac13$:
$$X=\begin{bmatrix}2&-1\\\tfrac53&1\end{bmatrix}$$
Check Position $(2,1)$: $3\cdot\tfrac53-2\cdot3=5-6=-1=b_{21}$ ✓.
Multiplying Two Matrices
A customer buys $3$ pens, $2$ notebooks and $5$ geometry boxes priced at ₹$10$, ₹$40$ and ₹$75$. The bill is $3\cdot10+2\cdot40+5\cdot75=485$ rupees: multiply matching quantities and prices, then add. Written with matrices, a row of quantities times a column of prices gives a $1\times1$ matrix:
$$\begin{bmatrix}3&2&5\end{bmatrix}\begin{bmatrix}10\\40\\75\end{bmatrix}=\begin{bmatrix}485\end{bmatrix}$$
This only works because the row and the column have the same length — one price for each item. The product of two matrices does this for every row of the first and every column of the second.
📘 Definition 3.8 — Product of two matrices
Let $A=[a_{ij}]$ be an $m\times n$ matrix and $B=[b_{jk}]$ an $n\times p$ matrix — the number of columns of $A$ equals the number of rows of $B$. The product $AB$ is the $m\times p$ matrix $C=[c_{ik}]$ with
$$c_{ik}=a_{i1}b_{1k}+a_{i2}b_{2k}+\cdots+a_{in}b_{nk}=\sum_{j=1}^{n}a_{ij}b_{jk}$$
In words: the entry in row $i$ and column $k$ of $AB$ is “row $i$ of $A$ times column $k$ of $B$”. If the number of columns of $A$ is not the number of rows of $B$, then $AB$ is not defined.
order check$(m\times\underline{n})\cdot(\underline{n}\times p)=m\times p$ — the inner numbers must match, the outer numbers give the order of $AB$
one entry$c_{ik}$: walk along row $i$ of $A$ and down column $k$ of $B$, multiply pairs, add
✏️ Example 3 — A product and its reverse
Let
$$A=\begin{bmatrix}1&2&-1\\3&0&4\end{bmatrix}$$
$$B=\begin{bmatrix}2&1\\0&-3\\5&2\end{bmatrix}$$
Find $AB$ and $BA$, if they are defined.
💡 Show solution▶
Orders $A$ is $2\times\underline3$ and $B$ is $\underline3\times2$: inner numbers match, so $AB$ is defined and is $2\times2$. Also $B$ is $3\times\underline2$ and $A$ is $\underline2\times3$, so $BA$ is defined and is $3\times3$.
AB $c_{11}=1\cdot2+2\cdot0+(-1)\cdot5=-3$; $c_{12}=1\cdot1+2(-3)+(-1)\cdot2=-7$; $c_{21}=3\cdot2+0+4\cdot5=26$; $c_{22}=3\cdot1+0+4\cdot2=11$:
$$AB=\begin{bmatrix}-3&-7\\26&11\end{bmatrix}$$
BA Row 1 of $B$ is $(2,1)$: $(2+3,\ 4+0,\ -2+4)=(5,4,2)$. Row 2 is $(0,-3)$: $(-9,0,-12)$. Row 3 is $(5,2)$: $(5+6,\ 10+0,\ -5+8)=(11,10,3)$:
$$BA=\begin{bmatrix}5&4&2\\-9&0&-12\\11&10&3\end{bmatrix}$$
Lesson $AB$ is $2\times2$ and $BA$ is $3\times3$ — they cannot possibly be equal. Step through both products in Interactive 1.
First the order check: the inner numbers of $(m\times n)(n\times p)$ must match. Then Play builds $AB$ one entry at a time: row $i$ of $A$ meets column $k$ of $B$, the pairs are multiplied and added, and the result drops into position $(i,k)$. Press Swap to compute the product the other way round.
orders—
product—
current entry—
✏️ Example 4 — Takings and costs of two branches
In one week the Delhi and Pune branches sold the quantities in $Q$ (columns: pens, notebooks, geometry boxes). Matrix $P$ gives each item's selling price and cost price in rupees (rows: the same three items).
$$Q=\begin{bmatrix}30&20&10\\25&40&15\end{bmatrix}$$
$$P=\begin{bmatrix}10&7\\40&30\\75&60\end{bmatrix}$$
Find $QP$, explain its entries and find each branch's profit.
💡 Show solution▶
Orders $(2\times3)(3\times2)$: defined, and $QP$ is $2\times2$ (rows: branches; columns: selling, cost).
Compute Delhi: $30\cdot10+20\cdot40+10\cdot75=1850$ and $30\cdot7+20\cdot30+10\cdot60=1410$. Pune: $25\cdot10+40\cdot40+15\cdot75=2975$ and $25\cdot7+40\cdot30+15\cdot60=2275$:
$$QP=\begin{bmatrix}1850&1410\\2975&2275\end{bmatrix}$$
Meaning Column 1 is each branch's takings, column 2 the cost of the goods sold, in rupees.
📐 Theorem 3.4 — Properties of matrix multiplication
Whenever the products and sums below are defined (that is, the orders fit),
(associative) $(AB)C=A(BC)$;
(distributive) $A(B+C)=AB+AC$ and $(A+B)C=AC+BC$;
(identity) if $A$ is $m\times n$, then $I_mA=A=AI_n$;
(scalars) $k(AB)=(kA)B=A(kB)$ for every real $k$.
Why (distributive law). The $(i,k)$ entry of $A(B+C)$ is $\sum_j a_{ij}(b_{jk}+c_{jk})=\sum_j a_{ij}b_{jk}+\sum_j a_{ij}c_{jk}$, which is the $(i,k)$ entry of $AB+AC$. For the identity, the $(i,k)$ entry of $AI_n$ is $\sum_j a_{ij}\delta_{jk}$, where $\delta_{jk}$ is $1$ if $j=k$ and $0$ otherwise; only the term $j=k$ survives, leaving $a_{ik}$. The associative law is proved by writing out the $(i,l)$ entry of both sides as the same double sum $\sum_j\sum_k a_{ij}b_{jk}c_{kl}$.
What is missing from the list is just as important. For numbers, $ab=ba$ always, and $ab=0$ forces $a=0$ or $b=0$. Neither is true for matrices. For $BA$ to equal $AB$, three things must go right, and any of them can fail: $BA$ might not be defined at all; it might be defined but of a different order (Example 3); or it might have the same order but different entries.
✏️ Example 5 — Same order, different products
Let $A=\begin{bmatrix}1&2\\0&1\end{bmatrix}$ and $B=\begin{bmatrix}1&0\\3&1\end{bmatrix}$. (a) Show that $AB\ne BA$. (b) Is $(A+B)^2=A^2+2AB+B^2$?
💡 Show solution▶
(a) $AB=\begin{bmatrix}1+6&0+2\\0+3&0+1\end{bmatrix}=\begin{bmatrix}7&2\\3&1\end{bmatrix}$ but $BA=\begin{bmatrix}1&2\\3&6+1\end{bmatrix}=\begin{bmatrix}1&2\\3&7\end{bmatrix}$. The $(1,1)$ entries differ, so $AB\ne BA$.
(b) left $A+B=\begin{bmatrix}2&2\\3&2\end{bmatrix}$, so $(A+B)^2=\begin{bmatrix}4+6&4+4\\6+6&6+4\end{bmatrix}=\begin{bmatrix}10&8\\12&10\end{bmatrix}$.
(b) right $A^2=\begin{bmatrix}1&4\\0&1\end{bmatrix}$, $B^2=\begin{bmatrix}1&0\\6&1\end{bmatrix}$, $2AB=\begin{bmatrix}14&4\\6&2\end{bmatrix}$, so $A^2+2AB+B^2=\begin{bmatrix}16&8\\12&4\end{bmatrix}$. Not equal.
Why Expanding with the distributive law, $(A+B)^2=(A+B)(A+B)=A^2+AB+BA+B^2$. This equals $A^2+2AB+B^2$ only when $BA=AB$. Here the difference is $AB-BA=\begin{bmatrix}6&0\\0&-6\end{bmatrix}$ — exactly the gap between the two answers.
✏️ Example 6 — A zero product of non-zero matrices
Let $A=\begin{bmatrix}1&2\\2&4\end{bmatrix}$ and $B=\begin{bmatrix}2&-6\\-1&3\end{bmatrix}$. Find $AB$ and $BA$. What does this say about “cancelling” in matrix equations?
💡 Show solution▶
AB $\begin{bmatrix}2-2&-6+6\\4-4&-12+12\end{bmatrix}=\begin{bmatrix}0&0\\0&0\end{bmatrix}=O$, although $A\ne O$ and $B\ne O$.
BA $\begin{bmatrix}2-12&4-24\\-1+6&-2+12\end{bmatrix}=\begin{bmatrix}-10&-20\\5&10\end{bmatrix}\ne O$. So even $AB=O$ does not give $BA=O$.
Cancelling Since $AB=O=AO$, we have $AB=AC$ with $C=O$, yet $B\ne C$. So from $AB=AC$ you may not conclude $B=C$ — unless $A$ has an inverse (Section 3.4).
Why it happens Each row of $A$ is a multiple of $(1,2)$, and each column of $B$ is a multiple of $(2,-1)$; the “row times column” of these is $1\cdot2+2(-1)=0$ every time.
Powers. For a square matrix $A$ we write $A^2=AA$, $A^3=A^2A$, and so on (and $A^0=I$). Powers of a non-square matrix are not defined, since $AA$ needs the columns of $A$ to match its rows. By the associative law $A^2A=AA^2$, so powers of the same matrix always commute with each other. A polynomial expression such as $A^2-5A+2I$ makes sense for square $A$: the constant term becomes a multiple of $I$.
Interactive 2 — Why AB ≠ BA: Doing Two Moves in Two Orders
A $2\times2$ matrix moves the plane (see the chapter teaser). The product $AB$ means “first $B$, then $A$”, because $(AB)X=A(BX)$ by the associative law. Play applies the two moves to the unit square and its flag in both orders: left panel $AB$ (first $B$, then $A$), right panel $BA$ (first $A$, then $B$). Find pairs that end in the same place — they are exactly the pairs with $AB=BA$.
AB—
BA—
AB = BA ?—
⚠️ Common mistakes when studying alone
1. Multiplying entry by entry. $\begin{bmatrix}1&2\\3&4\end{bmatrix}\begin{bmatrix}5&6\\7&8\end{bmatrix}$ is not $\begin{bmatrix}5&12\\21&32\end{bmatrix}$. It is $\begin{bmatrix}19&22\\43&50\end{bmatrix}$: always row times column.
2. Skipping the order check. Write the orders side by side, $(m\times n)(n\times p)$, before multiplying. If the inner numbers differ, the answer is “not defined” — not a matrix of zeros.
3. Swapping factors. $AB$ and $BA$ are different products. In a proof, never replace $AB$ by $BA$ unless you have been told the matrices commute. The same goes for $(A+B)^2$ and $(A-B)(A+B)=A^2-B^2$, which need $AB=BA$.
4. Cancelling a matrix. From $AB=AC$ or $AB=O$ nothing follows about $B$ in general (Example 6). Numbers can be cancelled because non-zero numbers have reciprocals; most matrices do not.
5. Forgetting $I$ in polynomials. In $A^2-5A+6$ the “$6$” must be $6I$; a matrix minus a number is not defined.
🚀 Beyond the board syllabus (JEE Main · JEE Advanced)
In JEE Main and JEE Advanced 2026; not examined in the CBSE board exam. JEE often asks for a high power $A^n$ or uses a polynomial relation satisfied by $A$.
Powers by spotting a pattern. For $N=\begin{bmatrix}1&1\\0&1\end{bmatrix}$, $N^2=\begin{bmatrix}1&2\\0&1\end{bmatrix}$ and $N^3=\begin{bmatrix}1&3\\0&1\end{bmatrix}$; if $N^k=\begin{bmatrix}1&k\\0&1\end{bmatrix}$ then $N^{k+1}=N^kN=\begin{bmatrix}1&k+1\\0&1\end{bmatrix}$, so by induction $N^n=\begin{bmatrix}1&n\\0&1\end{bmatrix}$ — for example $N^7=\begin{bmatrix}1&7\\0&1\end{bmatrix}$.
Relations $A^2-kA+lI=O$. Every $2\times2$ matrix $A=\begin{bmatrix}a&b\\c&d\end{bmatrix}$ satisfies $A^2-(a+d)A+(ad-bc)I=O$ (multiply it out to check). Such a relation turns every power into a combination of $A$ and $I$.
Worked example. $A=\begin{bmatrix}2&3\\-1&1\end{bmatrix}$ has $a+d=3$, $ad-bc=2+3=5$; indeed $A^2=\begin{bmatrix}1&9\\-3&-2\end{bmatrix}$ and $A^2-3A+5I=O$. So $A^2=3A-5I$ and $A^3=A\cdot A^2=3A^2-5A=3(3A-5I)-5A=4A-15I=\begin{bmatrix}-7&12\\-4&-11\end{bmatrix}$, without ever multiplying three matrices.
Exercises
I
Addition and scalar multiples. Let $A=\begin{bmatrix}3&-1\\2&4\end{bmatrix}$, $B=\begin{bmatrix}-2&5\\1&0\end{bmatrix}$, $C=\begin{bmatrix}0&2\\-3&1\end{bmatrix}$.
1 Find $A+B$.
2 Find $A-C$.
3 Find $3A-2B$.
4 Find $A+B-C$.
5 Verify that $(A+B)+C=A+(B+C)$.
6 Find the matrix $X$ with $A+X=O$.
7 Find $X$ if $2X-A=B$.
8 Is $\begin{bmatrix}1&2&3\end{bmatrix}+\begin{bmatrix}1\\2\\3\end{bmatrix}$ defined? Explain.
9 Find $X$ and $Y$ if $X+Y=\begin{bmatrix}7&0\\2&5\end{bmatrix}$ and $X-Y=\begin{bmatrix}3&0\\0&3\end{bmatrix}$.
10 Find $x$ and $y$ if $2\begin{bmatrix}1&3\\0&x\end{bmatrix}+\begin{bmatrix}y&0\\1&2\end{bmatrix}=\begin{bmatrix}5&6\\1&8\end{bmatrix}$.
6 $A$ is $3\times4$, $B$ is $4\times2$ and $C$ is $2\times3$. Which of $AB$, $BA$, $AC$, $CA$, $BC$, $CB$ are defined, and of what order? What is the order of $ABC$?
7 For $A=\begin{bmatrix}1&2\\3&4\end{bmatrix}$, find $A^2$ and show that $A^2-5A-2I=O$.
8 Show that $A=\begin{bmatrix}0&1\\0&0\end{bmatrix}$ satisfies $A^2=O$ although $A\ne O$.
9 Find all $x$ with $\begin{bmatrix}x&1\end{bmatrix}\begin{bmatrix}1&2\\3&0\end{bmatrix}\begin{bmatrix}x\\1\end{bmatrix}=\begin{bmatrix}0\end{bmatrix}$.
4 $(2\times3)(3\times1)$: $\begin{bmatrix}2+0-2\\-2+3-1\end{bmatrix}=\begin{bmatrix}0\\0\end{bmatrix}$, a zero column although neither factor is zero.
5 $\begin{bmatrix}a^2+b^2&-ab+ab\\-ab+ab&b^2+a^2\end{bmatrix}=(a^2+b^2)I$, a scalar matrix.
6 $AB$: $(3\times4)(4\times2)$, defined, $3\times2$. $BC$: $(4\times2)(2\times3)$, defined, $4\times3$. $CA$: $(2\times3)(3\times4)$, defined, $2\times4$. $BA$ ($2\ne3$), $AC$ ($4\ne2$), $CB$ ($3\ne4$) are not defined. $ABC=(AB)C$ is $(3\times2)(2\times3)=3\times3$.
7 $A^2=\begin{bmatrix}1+6&2+8\\3+12&6+16\end{bmatrix}=\begin{bmatrix}7&10\\15&22\end{bmatrix}$. Then $A^2-5A-2I=\begin{bmatrix}7-5-2&10-10\\15-15&22-20-2\end{bmatrix}=O$ ✓.
8 $A^2=\begin{bmatrix}0\cdot0+1\cdot0&0\cdot1+1\cdot0\\0&0\end{bmatrix}=\begin{bmatrix}0&0\\0&0\end{bmatrix}$. A non-zero matrix can have square $O$.
9 $\begin{bmatrix}x&1\end{bmatrix}\begin{bmatrix}1&2\\3&0\end{bmatrix}=\begin{bmatrix}x+3&2x\end{bmatrix}$; times $\begin{bmatrix}x\\1\end{bmatrix}$ gives $\begin{bmatrix}x^2+3x+2x\end{bmatrix}=\begin{bmatrix}x^2+5x\end{bmatrix}$. So $x(x+5)=0$: $x=0$ or $x=-5$.
III
Properties, commuting and cancelling. In parts 1, 4 and 5, $A=\begin{bmatrix}1&2\\3&0\end{bmatrix}$ and $B=\begin{bmatrix}0&1\\1&1\end{bmatrix}$.
1 Find $AB$ and $BA$. Do $A$ and $B$ commute?
2 Verify $P(Q+R)=PQ+PR$ for $P=\begin{bmatrix}1&-1\\2&0\end{bmatrix}$, $Q=\begin{bmatrix}3&1\\0&2\end{bmatrix}$, $R=\begin{bmatrix}-1&0\\4&1\end{bmatrix}$.
3 Give two non-zero $2\times2$ matrices whose product is $O$.
4 Show that $(A+B)^2\ne A^2+2AB+B^2$, and identify the difference.
5 Find $(A-B)(A+B)$ and $A^2-B^2$. Are they equal?
6 Let $P=\begin{bmatrix}1&1\\1&1\end{bmatrix}$, $Q=\begin{bmatrix}1&2\\3&0\end{bmatrix}$, $R=\begin{bmatrix}2&1\\2&1\end{bmatrix}$. Show that $PQ=PR$ although $Q\ne R$.
7 For $A=\begin{bmatrix}1&0\\0&-1\end{bmatrix}$ find $A^2$ and $A^3$.
8 For which real $k$ does $\begin{bmatrix}k&0\\0&k\end{bmatrix}^2=4I$?
9 Prove: if $A$ and $B$ are square matrices of the same order with $AB=BA$, then $(A+B)^2=A^2+2AB+B^2$.
💡 Show solutions▶
1 $AB=\begin{bmatrix}2&3\\0&3\end{bmatrix}$, $BA=\begin{bmatrix}3&0\\4&2\end{bmatrix}$. Different, so they do not commute.
2 $Q+R=\begin{bmatrix}2&1\\4&3\end{bmatrix}$ and $P(Q+R)=\begin{bmatrix}-2&-2\\4&2\end{bmatrix}$. $PQ=\begin{bmatrix}3&-1\\6&2\end{bmatrix}$, $PR=\begin{bmatrix}-5&-1\\-2&0\end{bmatrix}$, sum $\begin{bmatrix}-2&-2\\4&2\end{bmatrix}$ ✓.
3 For example $\begin{bmatrix}1&0\\0&0\end{bmatrix}\begin{bmatrix}0&0\\0&1\end{bmatrix}=O$ (or the pair in Example 6). Many answers are possible — check yours by multiplying.
4 $(A+B)^2=\begin{bmatrix}1&3\\4&1\end{bmatrix}^2=\begin{bmatrix}13&6\\8&13\end{bmatrix}$. $A^2=\begin{bmatrix}7&2\\3&6\end{bmatrix}$, $B^2=\begin{bmatrix}1&1\\1&2\end{bmatrix}$, $2AB=\begin{bmatrix}4&6\\0&6\end{bmatrix}$, so $A^2+2AB+B^2=\begin{bmatrix}12&9\\4&14\end{bmatrix}$. The difference $(A+B)^2-(A^2+2AB+B^2)=BA-AB=\begin{bmatrix}1&-3\\4&-1\end{bmatrix}$.
5 $(A-B)(A+B)=A^2+AB-BA-B^2$. Here $A-B=\begin{bmatrix}1&1\\2&-1\end{bmatrix}$ and $(A-B)(A+B)=\begin{bmatrix}5&4\\-2&5\end{bmatrix}$, while $A^2-B^2=\begin{bmatrix}6&1\\2&4\end{bmatrix}$. Not equal (they differ by $AB-BA$).
6 $PQ=\begin{bmatrix}4&2\\4&2\end{bmatrix}$ and $PR=\begin{bmatrix}4&2\\4&2\end{bmatrix}$: equal, though $Q\ne R$. Cancelling $P$ is not allowed.
7 $A^2=\begin{bmatrix}1&0\\0&1\end{bmatrix}=I$ and $A^3=A^2A=IA=A$.
8 $(kI)^2=k^2I$, so $k^2=4$: $k=2$ or $k=-2$.
9 By the distributive law, $(A+B)^2=(A+B)(A+B)=A(A+B)+B(A+B)=A^2+AB+BA+B^2$. Using $BA=AB$, this is $A^2+2AB+B^2$. ∎
IV
Modelling with matrix products.
1Market. A family buys $12$ kg of apples, $6$ kg of bananas and $8$ kg of oranges at ₹$120$, ₹$40$ and ₹$60$ per kg. Write the bill as a product of a row matrix and a column matrix and evaluate it.
2Factory. Each chair needs $3$ hours of labour and $2$ kg of material; each table needs $4$ hours and $5$ kg. An order is for $50$ chairs and $30$ tables. Using $R=\begin{bmatrix}3&2\\4&5\end{bmatrix}$, find the total labour and material as a product. If labour costs ₹$200$ per hour and material ₹$50$ per kg, find the total cost with one more product.
3Investment. A cooperative society invests ₹$50\,000$ in two schemes paying $6\%$ and $8\%$ simple interest per year. Writing the amounts as $\begin{bmatrix}x&50000-x\end{bmatrix}$, use a matrix product to find how much should go into each scheme to earn ₹$3600$ a year.
4Cricket league. Three teams have the win–draw–loss records Mumbai $(5,2,3)$, Chennai $(6,1,3)$ and Kolkata $(4,4,2)$. A win scores $2$ points, a draw $1$ and a loss $0$. Find the points table as a product of a $3\times3$ and a $3\times1$ matrix.
5Monthly sales. Two shops sold the quantities $J=\begin{bmatrix}30&20\\15&25\end{bmatrix}$ in January and $K=\begin{bmatrix}35&18\\20&30\end{bmatrix}$ in February (rows: shops; columns: two products). Find the total sales $J+K$ and the change $K-J$, and interpret the negative entry.
💡 Show solutions▶
1 $\begin{bmatrix}12&6&8\end{bmatrix}\begin{bmatrix}120\\40\\60\end{bmatrix}=\begin{bmatrix}1440+240+480\end{bmatrix}=\begin{bmatrix}2160\end{bmatrix}$: the bill is ₹$2160$.
2 $\begin{bmatrix}50&30\end{bmatrix}R=\begin{bmatrix}150+120&100+150\end{bmatrix}=\begin{bmatrix}270&250\end{bmatrix}$: $270$ hours and $250$ kg. Cost: $\begin{bmatrix}270&250\end{bmatrix}\begin{bmatrix}200\\50\end{bmatrix}=\begin{bmatrix}66500\end{bmatrix}$, i.e. ₹$66\,500$.
3 $\begin{bmatrix}x&50000-x\end{bmatrix}\begin{bmatrix}0.06\\0.08\end{bmatrix}=\begin{bmatrix}0.06x+4000-0.08x\end{bmatrix}=\begin{bmatrix}3600\end{bmatrix}$. So $0.02x=400$, $x=20\,000$: ₹$20\,000$ at $6\%$ and ₹$30\,000$ at $8\%$. Check: $1200+2400=3600$ ✓.
4 $\begin{bmatrix}5&2&3\\6&1&3\\4&4&2\end{bmatrix}\begin{bmatrix}2\\1\\0\end{bmatrix}=\begin{bmatrix}12\\13\\12\end{bmatrix}$: Mumbai $12$, Chennai $13$, Kolkata $12$ points.
5 $J+K=\begin{bmatrix}65&38\\35&55\end{bmatrix}$ and $K-J=\begin{bmatrix}5&-2\\5&5\end{bmatrix}$. The $-2$ means shop 1 sold $2$ fewer units of the second product in February than in January.
How this is examined: the Board asks for the order of a product, a product or polynomial in a $2\times2$ or $3\times3$ matrix, “find $k$” from a matrix relation, and a word problem set up as a product. JEE tests the failures of commutativity and cancellation, and high powers of a matrix.
E1.Board · 1 mark If $A$ is a $3\times4$ matrix and $B$ is a $4\times2$ matrix, the order of $AB$ is (A) $3\times2$ (B) $4\times4$ (C) $2\times3$ (D) $AB$ is not defined
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Idea $(3\times\underline4)(\underline4\times2)$: the inner numbers match, and the outer numbers give $3\times2$.
Answer (A). (C) reverses the outer numbers; (D) would be right for $BA$, where the inner numbers are $2\ne3$.
E2.Board · 1 markAssertion–Reason. Assertion (A): for $A=\begin{bmatrix}0&1\\0&0\end{bmatrix}$, $A^2=O$. Reason (R): if $A$ and $B$ are square matrices with $AB=O$, then $A=O$ or $B=O$. (A) Both A and R are true and R is the correct explanation of A (B) Both A and R are true but R is not the correct explanation of A (C) A is true but R is false (D) A is false but R is true
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Idea A: $A^2=\begin{bmatrix}0&0\\0&0\end{bmatrix}$ — true (Exercise II.8). R is false, and A itself is the counterexample: $A\cdot A=O$ with $A\ne O$.
Answer (C).
E3.Board · 3 marks If $A=\begin{bmatrix}4&-3\\3&-2\end{bmatrix}$, find $k$ such that $A^2=kA-I$.
Answer Every equation gives $k=2$. (½ mark — and say that all four agree; if they did not, there would be no such $k$.)
E4.Board · Case study For a charity fair in Chennai, three sections of Class 11 sold handmade cards, candles and cloth bags. The quantities sold are
$$Q=\begin{bmatrix}20&15&5\\25&10&8\\18&20&6\end{bmatrix}$$
(rows: sections A, B, C; columns: cards, candles, bags), and the prices are ₹$30$, ₹$50$ and ₹$120$. (i) Write the price matrix $P$ so that $QP$ gives each section's collection, and find $QP$. [2 marks] (ii) Which section collected the most? [1 mark] (iii) Find the total collected. [1 mark]
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(i) $P$ must be $3\times1$: $P=\begin{bmatrix}30\\50\\120\end{bmatrix}$. Then $QP=\begin{bmatrix}600+750+600\\750+500+960\\540+1000+720\end{bmatrix}=\begin{bmatrix}1950\\2210\\2260\end{bmatrix}$.
(ii) Section C, with ₹$2260$.
(iii) $1950+2210+2260=6420$: ₹$6420$. (As a product: $\begin{bmatrix}1&1&1\end{bmatrix}QP$.)
E5.JEE Main If $A=\begin{bmatrix}1&2\\0&1\end{bmatrix}$, then $A^{10}$ equals (A) $\begin{bmatrix}1&20\\0&1\end{bmatrix}$ (B) $\begin{bmatrix}1&1024\\0&1\end{bmatrix}$ (C) $\begin{bmatrix}10&20\\0&10\end{bmatrix}$ (D) $\begin{bmatrix}1&10\\0&1\end{bmatrix}$
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Pattern $A^2=\begin{bmatrix}1&4\\0&1\end{bmatrix}$, $A^3=\begin{bmatrix}1&6\\0&1\end{bmatrix}$. If $A^n=\begin{bmatrix}1&2n\\0&1\end{bmatrix}$ then $A^{n+1}=A^nA=\begin{bmatrix}1&2+2n\\0&1\end{bmatrix}$, so the pattern continues (🚀 box above).
Answer (A). Trap (B) squares the entry $2$ repeatedly, as if powers of a matrix were powers of its entries; (C) multiplies the matrix by $10$.
⏱ Fast route $A=I+N$ with $N=\begin{bmatrix}0&2\\0&0\end{bmatrix}$, $N^2=O$, and $I$ commutes with $N$: $A^{10}=I+10N$.
E6.JEE Advanced(One or more options correct.) For $2\times2$ real matrices $A$ and $B$, which statements are always true? (A) If $AB=O$ then $BA=O$ (B) If $A^2=O$ then $A=O$ (C) If $AB=BA$ then $(A+B)^2=A^2+2AB+B^2$ (D) If $A$ is a scalar matrix then $AB=BA$ for every $B$
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(A) False $A=\begin{bmatrix}0&1\\0&0\end{bmatrix}$, $B=\begin{bmatrix}1&0\\0&0\end{bmatrix}$: $AB=O$ but $BA=\begin{bmatrix}0&1\\0&0\end{bmatrix}$. (Example 6 is another counterexample.)
(B) False The same $A$ has $A^2=O$.
(C) True Exercise III.9.
(D) True $A=kI$ gives $AB=k(IB)=kB$ and $BA=k(BI)=kB$.
Answer (C), (D). Only scalar matrices commute with every matrix of their order — try $2I$ against each move in Interactive 2.
🔬 Research–Teaching NexusMini project · about 60 minutes
Research question. In a city where commuters choose each month between the metro and buses, what share will each mode have in a year's time — and does the answer depend on where we start?
Why it matters. Transport planners, telecom companies (which network do customers switch to?) and epidemiologists all model “who moves where next step” with a matrix of switching rates. Repeated matrix multiplication predicts the future, and the long-run shares guide decisions such as how many trains to buy.
What to do.
Suppose each month $90\%$ of metro users stay on the metro and $10\%$ switch to buses, while $20\%$ of bus users switch to the metro and $80\%$ stay. Write the switching matrix $T=\begin{bmatrix}0.9&0.2\\0.1&0.8\end{bmatrix}$ (column = where you are now, row = where you go).
Start with shares $X_0=\begin{bmatrix}0.3\\0.7\end{bmatrix}$ (metro, bus). In a spreadsheet use MMULT to compute $X_1=TX_0$, $X_2=TX_1$, … up to $X_{12}$.
Repeat from $X_0=\begin{bmatrix}0.9\\0.1\end{bmatrix}$. What do you notice after a year?
Find shares $X$ with $TX=X$ by solving two linear equations (with $x_1+x_2=1$). Compare with your table.
Tools: Spreadsheet (MMULT), or pen and calculator for the first three months.
Hand in: Your two tables of $X_0,\dots,X_{12}$, a line graph of the metro share, the steady state $X$, and two sentences on what a planner should conclude.
Going further. Change the $10\%$ to $5\%$ (a better metro service). How much does the long-run metro share rise? Why does every column of $T$ add up to $1$?
Instructor note: $X_1=\begin{bmatrix}0.41\\0.59\end{bmatrix}$ from $X_0=(0.3,0.7)$; both starts approach the steady state $\left(\tfrac23,\tfrac13\right)$, which solves $0.1x_1=0.2x_2$. With $5\%$ switching the steady state is $(0.8,0.2)$. The columns add to $1$ because everyone ends up somewhere.
🧪 Self-check quiz
0/5 correct
Q1. $A$ is $2\times3$ and $B$ is $3\times4$. Then $AB$ is
Q2. If $A=\begin{bmatrix}1&2\\3&4\end{bmatrix}$, then $2A-I$ is
Q3. Which statement is true for all square matrices $A$, $B$ of the same order?