Matrices: Class 12 Maths · Chapter 3

Mathematics Part 1 · Companion notes, simulations and practice · Official NCERT chapter PDF ↗

Published 11 Oct 2026 · Updated 11 Oct 2026

120 min
1 simulation
61 practice questions
31 MCQs

What you will learn in Matrices

Read the order of a matrix, locate any entry a_ij, and build a matrix from data or from a rule
Recognise row, column, square, diagonal, scalar, identity and zero matrices, and use equality of matrices to find unknowns
Add, subtract and scale matrices, and use the laws of these operations to solve matrix equations
Multiply matrices by the row-by-column rule, decide when a product is defined, and explain why AB and BA can differ
Find transposes and apply the rules for the transpose of a sum, a multiple and a product
Identify symmetric and skew symmetric matrices and split any square matrix into the two kinds
Check whether two matrices are inverses, and use the uniqueness of the inverse and the rule for the inverse of a product

Original companion text aligned to NCERT Mathematics Class 12, Chapter 3, pp. 34–74. Not affiliated with or endorsed by NCERT.

3.1 Introduction

A cricket scoresheet, a railway reservation chart and a household budget all keep numbers in rows and columns. Once everyone agrees what each row and each column stands for, the labels can be set aside and the block of numbers can itself be added, scaled and multiplied. Such a block is a matrix (आव्यूह), and the rules for computing with it make up matrix algebra.

Matrices began as a tidy way to write a system of linear equations and to solve it by a fixed routine. The same idea now runs computer graphics (each time a photo on your phone is rotated or zoomed, a matrix does the work), the ranking of web pages, weather models, the codes that protect bank messages and the input–output tables that economists use to plan production. This chapter builds the vocabulary and the arithmetic; Chapter 4 puts them to work with determinants.

Fig. 3.1 shows the first step of that routine. The unknowns and the plus signs carry no information once their positions are agreed, so a system of equations shrinks to two blocks of numbers, and the whole system becomes a single matrix equation.

3/3The system becomes a single matrix equation (solved in Chapter 4).

Fig. 3.1A system of linear equations becomes one matrix equation once the positions of the numbers are agreed.NCERT idea of 3.1 · p. 34

Step 1 shows the equations 2x + 3y = 13 and 4x minus y = 5, written in aligned columns. Step 2 fades the letters and signs, leaving the coefficient block 2, 3 over 4, minus 1 in brackets and the constants 13 over 5 in a column. Step 3 writes the whole system as one matrix equation: the coefficient matrix times the column x, y equals the column 13, 5, with the solution x = 2, y = 3 noted underneath.

Quick check

1 of 2

Which matrix holds the coefficients of the system

5 x - 2 y = 7

x + 4 y = 3

[\begin{array}{cc} 5 & - 2 \\ 1 & 4 \end{array}]

[\begin{array}{cc} 5 & 2 \\ 1 & 4 \end{array}]

[\begin{array}{cc} 5 & 1 \\ - 2 & 4 \end{array}]

[\begin{array}{c} 7 \\ 3 \end{array}]

Show answer

Answer (A) $[\begin{array}{cc} 5 & - 2 \\ 1 & 4 \end{array}]$

Each equation gives one row, in the order

x

then

y

, and a minus sign belongs to its coefficient:

- 2

. Writing the equations down the columns gives the third option, a different matrix.

3.2 Matrix

Here is a record from a school canteen. On Monday it sold 42 samosas and 30 plates of idli, on Tuesday 35 samosas and 48 plates of idli, and on Wednesday 50 samosas and 26 plates of idli. A table makes the record easy to read. If we agree that the rows are the days and the columns are the dishes, we can drop the labels and keep only the rectangular block of numbers, enclosed in brackets (Fig. 3.2).

5/5Row 2 meets column 1 at the entry $a_{21} = 35$ .

Fig. 3.2A table of numbers with its labels set aside is a matrix; each entry is located by its row and its column.NCERT idea of the tables on p. 35 · p. 35

Step 1 shows a table with row labels Mon, Tue and Wed and column labels Samosa and Idli, holding 42 and 30, 35 and 48, 50 and 26. Step 2 fades the labels and grid and puts square brackets around the six numbers, marked as a 3 by 2 matrix. Step 3 highlights the second row, 35 and 48, as row 2. Step 4 highlights the first column, 42, 35 and 50, as column 1. Step 5 rings the number 35 where row 2 and column 1 cross and labels it a sub 2 1 equals 35.

Nothing forces this arrangement. We could just as well put the dishes in rows and the days in columns, which gives a block with 2 rows and 3 columns. Both blocks carry the same information; what matters is that the convention is fixed before the numbers are used. In Section 3.5 we will see that the two arrangements are transposes of each other.

Definition

Matrix

A rectangular arrangement of numbers or functions in rows and columns, in a definite order. Each item in it is an element (or entry) of the matrix.

Example $[\begin{array}{cc} 42 & 30 \\ 35 & 48 \\ 50 & 26 \end{array}]$ , the canteen record.

Entries side by side across the page form a row; entries stacked down the page form a column. Rows are numbered from the top, columns from the left, and matrices themselves get capital letters. Entries need not be whole numbers, and they may even be functions of a variable, as in these three matrices:

\begin{array}{c} P = [\begin{array}{cc} - 1 & \frac{3}{4} \\ 0 & \sqrt{5} \end{array}] \\ Q = [\begin{array}{ccc} 2 & - 7 & 0.5 \end{array}] \\ R = [\begin{array}{ccc} \cos t & - \sin t & t \\ \sin t & \cos t & t^{2} \end{array}] \end{array}

$P$ has 2 rows and 2 columns, $Q$ has a single row of 3 elements (अवयव), and $R$ has 2 rows and 3 columns whose entries change with $t$ .

3.2.1 Order of a matrix

A matrix with $m$ rows and $n$ columns has order $m \times n$ (कोटि), read “ $m$ by $n$ ”. Rows always come first. The canteen matrix has order $3 \times 2$ , and $P$ , $Q$ , $R$ above have orders $2 \times 2$ , $1 \times 3$ and $2 \times 3$ . An $m \times n$ matrix has $m n$ entries, one for each choice of a row and a column, so $P$ has 4 entries and $R$ has 6.

A general m × n matrix

\begin{aligned} A & = [a_{i j}]_{m \times n} \\ = [\begin{array}{cccc} a_{11} & a_{12} & \dots & a_{1 n} \\ a_{21} & a_{22} & \dots & a_{2 n} \\ ⋮ & ⋮ & ⋱ & ⋮ \\ a_{m 1} & a_{m 2} & \dots & a_{m n} \end{array}] \end{aligned}

where

$a_{i j}$: the entry in row $i$ and column $j$ , called the $(i, j)$ th element
$i, j$: natural numbers with $1 \leq i \leq m$ and $1 \leq j \leq n$

The first subscript names the row and the second names the column. Row $i$ therefore reads $a_{i 1}, a_{i 2}, \dots, a_{i n}$ and column $j$ reads $a_{1 j}, a_{2 j}, \dots, a_{m j}$ . In the canteen matrix $a_{21} = 35$ , because row 2 is Tuesday and column 1 is samosas, while $a_{12} = 30$ is Monday's idli. In this chapter every entry is a real number or a real-valued function, and $A = [a_{i j}]_{m \times n}$ is our shorthand for “ $A$ is an $m \times n$ matrix with entries $a_{i j}$ ”.

A point of the plane is itself a small matrix: $(x, y)$ can be written as the column $[\begin{array}{c} x \\ y \end{array}]$ or as the row $[x y]$ . Stand the columns of several points side by side and one matrix stores a whole figure. For the quadrilateral $K L M N$ in Fig. 3.3, with $K (- 2, 1)$ , $L (3, 0)$ , $M (4, 4)$ and $N (0, 5)$ , the vertex matrix has one column per vertex, in the order $K, L, M, N$ , so it is $2 \times 4$ :

V = [\begin{array}{cccc} - 2 & 3 & 4 & 0 \\ 1 & 0 & 4 & 5 \end{array}]

SimulationVertices as columns of a matrix

A coordinate grid from minus 5 to 7 in both directions, showing the quadrilateral KLMN with vertices K at (minus 2, 1), L at (3, 0), M at (4, 4) and N at (0, 5), joined by sides with their lengths. You can drag any vertex to a grid point and read its new coordinates. A mirror control starts on the x-axis, showing the reflected quadrilateral K prime L prime M prime N prime, and can switch to the y-axis, the origin or off.

Fig. 3.3A reflection changes every column of the vertex matrix

V

by one rule; dragging a vertex changes just one column.

Try this

The sim opens with the mirror image $K^{'} L^{'} M^{'} N^{'}$ in the $x$ -axis. Write its vertex matrix: what has happened to the first row of $V$ , and to the second?
Set the mirror to the origin. By which single number could you multiply $V$ to get the new vertex matrix?
Turn the mirror off and drag $M$ to $(5, 2)$ . Which column of $V$ changes, and what does it become?

Example 1NCERT Example 1 · p. 37

Seats on three trains

A booking clerk notes the berths still free on three evening trains. Train P has 64 sleeper berths and 12 AC-3 berths free, train Q has 9 sleeper and 27 AC-3, and train R has 0 sleeper and 41 AC-3.

(a) Write this information as a $2 \times 3$ matrix $B$ with the classes as rows. (b) What does $b_{23}$ stand for, and what is its value? (c) What would the order be if the trains were the rows?

Try it yourself first. Then reveal the solution one step at a time.

Example 2NCERT Example 2 · p. 38

Every shape for twelve entries

A matrix has 12 elements. List all the orders it could have.

Try it yourself first. Then reveal the solution one step at a time.

3/6Three rows of four: order $3 \times 4$ .

Fig. 3.4Twelve entries fit six rectangular shapes, one for each ordered factor pair of 12.NCERT idea of Example 2 · p. 38

The same twelve dots are rearranged in each step. Step 1 is a single row of 12, order 1 by 12. Step 2 is 2 rows of 6. Step 3 is 3 rows of 4. Step 4 is 4 rows of 3. Step 5 is 6 rows of 2. Step 6 is one column of 12 dots, order 12 by 1. Each step labels the order and the number of rows and columns.

When the number of elements is a prime $p$ , only two shapes are possible, $1 \times p$ and $p \times 1$ . When it is a perfect square such as 16, one of the shapes, $4 \times 4$ , is the same either way round, so the count of orders is odd.

Example 3NCERT Example 3 · p. 38

A matrix from a rule

Construct the $3 \times 3$ matrix $A = [a_{i j}]$ with $a_{i j} = i^{j}$ , and name the position of the entry 9.

Try it yourself first. Then reveal the solution one step at a time.

3.3 Types of Matrices

Some shapes and patterns of entries turn up so often that they have names. The first three depend only on the order.

Column matrix: a single column, order $m \times 1$ . For example $[\begin{array}{c} 4 \\ - 1 \\ \frac{2}{3} \end{array}]$ is a column matrix of order $3 \times 1$ .
Row matrix: a single row, order $1 \times n$ . For example $[0 \sqrt{3} - 6 1]$ is a row matrix of order $1 \times 4$ .
Square matrix: as many rows as columns. An $n \times n$ matrix is called a square matrix of order $n$ ; $[\begin{array}{ccc} 7 & 0 & - 2 \\ 1 & - 5 & 4 \\ 9 & 3 & 6 \end{array}]$ is square of order 3.

In a square matrix $A = [a_{i j}]_{n \times n}$ the entries $a_{11}, a_{22}, \dots, a_{n n}$ , running from the top-left corner to the bottom-right corner, form the main diagonal (or simply the diagonal). In the order-3 example above the diagonal entries are $7$ , $- 5$ and $6$ . Three important kinds of square matrix are defined by what happens on and off this diagonal, and each kind narrows the previous one (Fig. 3.5).

Diagonal matrix: every entry off the diagonal is 0, that is $a_{i j} = 0$ whenever $i \neq j$ . The diagonal entries may be anything, even 0, so $[\begin{array}{cc} 3 & 0 \\ 0 & 0 \end{array}]$ is diagonal.
Scalar matrix: a diagonal matrix whose diagonal entries all equal one number $k$ , such as $[\begin{array}{ccc} - 2 & 0 & 0 \\ 0 & - 2 & 0 \\ 0 & 0 & - 2 \end{array}]$ . Multiplying by it simply scales by $k$ , which explains the name (Section 3.4.5).
Identity matrix: the scalar matrix with $k = 1$ . The identity of order $n$ is written $I_{n}$ , or just $I$ when the order is clear: $I_{1} = [1]$ , $I_{2} = [\begin{array}{cc} 1 & 0 \\ 0 & 1 \end{array}]$ , and so on. In multiplication it plays the part of the number 1.

3/4Equal diagonal entries ( $k = 4$ ): a scalar matrix.

Fig. 3.5Each special type is a square matrix with one more condition: every identity matrix is scalar, every scalar matrix is diagonal, every diagonal matrix is square.NCERT types (iv)–(vi) · p. 40

On the left is a 3 by 3 matrix with its main diagonal shaded; on the right are four nested boxes labelled square, diagonal, scalar and identity. Step 1 shows a general square matrix with entries 5, minus 2, 7, 1, 3, 0, 4, 8, 6 and the outer box lit. Step 2 turns every off-diagonal entry into 0, leaving 5, 3, 6 on the diagonal, and lights the diagonal box. Step 3 makes all diagonal entries 4 and lights the scalar box. Step 4 makes them all 1 and lights the innermost identity box.

So every identity matrix is scalar, every scalar matrix is diagonal and every diagonal matrix is square, but none of these statements can be turned round. A $1 \times 1$ matrix $[c]$ is square, diagonal and scalar all at once, and it is the identity exactly when $c = 1$ .

Finally, a zero matrix (शून्य आव्यूह), written O, has every entry equal to 0. It may have any order, such as $[0 0 0]$ or $[\begin{array}{cc} 0 & 0 \\ 0 & 0 \\ 0 & 0 \end{array}]$ ; the order is read from the context. A square zero matrix also counts as a scalar matrix, with $k = 0$ .

3.3.1 Equality of matrices

Two matrices can be compared entry by entry only if their entries line up, so equality asks for two things at once.

Definition

Equal matrices

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

Example $[\begin{array}{cc} 1 & 2 \\ 3 & 4 \end{array}] \neq [\begin{array}{cc} 1 & 3 \\ 2 & 4 \end{array}]$ : the same order and the same four numbers, but in different places.

Neither condition can be dropped. The row $[0 0]$ and the column $[\begin{array}{c} 0 \\ 0 \end{array}]$ hold the same numbers yet are not equal, because their orders differ. In the other direction, one matrix equation packs in one ordinary equation for each position. If $[\begin{array}{cc} x & 4 \\ - 1 & y + 2 \end{array}] = [\begin{array}{cc} 3 & 4 \\ - 1 & 0 \end{array}]$ , then $x = 3$ and $y = - 2$ , while the other two positions simply agree.

Example 4NCERT Example 4 · p. 41

Six positions, five unknowns

\begin{array}{c} [\begin{array}{ccc} x - 2 & 3 y + 1 & 4 \\ a + 5 & 2 b - 1 & c / 2 \end{array}] \\ = [\begin{array}{ccc} 5 & 2 y + 4 & 4 \\ - 1 & b + 6 & 3 \end{array}] \end{array}

Find $a$ , $b$ , $c$ , $x$ and $y$ .

Try it yourself first. Then reveal the solution one step at a time.

Example 5NCERT Example 5 · p. 41

Equality gives a system

\begin{array}{c} [\begin{array}{cc} p + q & 2 p - q \\ 3 r + s & r - 2 s \end{array}] \\ = [\begin{array}{cc} 6 & 9 \\ 5 & - 3 \end{array}] \end{array}

Find $p$ , $q$ , $r$ and $s$ .

Try it yourself first. Then reveal the solution one step at a time.

3.4 Operations on Matrices

Numbers can be added, subtracted and multiplied, and so can matrices, but every operation comes with a condition on the orders. We follow one small business through all of them. A juice seller runs two kiosks, one at a railway station and one at a college gate, and sells three juices: mango, sugarcane and mosambi (sweet lime). In every matrix below, row 1 is the station, row 2 is the college, and the columns are the three juices in that order.

3.4.1 Addition of matrices

On one day the morning shift and the evening shift sold these numbers of glasses:

\begin{array}{c} M = [\begin{array}{ccc} 40 & 55 & 30 \\ 25 & 20 & 35 \end{array}] \\ E = [\begin{array}{ccc} 35 & 60 & 20 \\ 30 & 15 & 25 \end{array}] \end{array}

The day's mango sales at the station are $40 + 35 = 75$ glasses, found by adding the two (1, 1) entries. Sugarcane at the station comes from the two (1, 2) entries, and so on for all six positions (Fig. 3.6). The result is again a $2 \times 3$ matrix, the sum of $M$ and $E$ :

\begin{aligned} D & = M + E \\ = [\begin{array}{ccc} 40 + 35 & 55 + 60 & 30 + 20 \\ 25 + 30 & 20 + 15 & 35 + 25 \end{array}] \\ = [\begin{array}{ccc} 75 & 115 & 50 \\ 55 & 35 & 60 \end{array}] \end{aligned}

3/3A week at the same rate: every entry of $D$ is multiplied by 7.

Fig. 3.6Matrices of the same order are added position by position; a scalar multiplies every entry.NCERT idea of 3.4.1–3.4.2 · p. 43

Step 1 shows the morning matrix 40, 55, 30 over 25, 20, 35 and the evening matrix 35, 60, 20 over 30, 15, 25, with an empty bracket for their sum. Step 2 fills in the sum 75, 115, 50 over 55, 35, 60 and rings the matching entries 55, 60 and 115 in the same position. Step 3 adds a second line where 7 times the day matrix gives the week matrix 525, 805, 350 over 385, 245, 420, ringing 115 and 805.

This is the rule in general. If $A = [a_{i j}]$ and $B = [b_{i j}]$ have the same order $m \times n$ , their sum is the $m \times n$ matrix whose entries are the sums of corresponding entries:

Sum of two matrices

A + B = [a_{i j} + b_{i j}]_{m \times n}

Addition is not defined for matrices of different orders. There is no sensible way to add $[\begin{array}{cc} 1 & 0 \\ 2 & 5 \end{array}]$ and $[3 4]$ , because most positions in one have no partner in the other. On the set of all $m \times n$ matrices, however, any two can be added and the answer stays in the set: addition is a binary operation on that set.

Example 6NCERT Example 6 · p. 44

Adding surds and fractions

Find $A + B$ if $A = [\begin{array}{ccc} \sqrt{2} & - 1 & \frac{3}{2} \\ 0 & 4 & - 3 \end{array}]$ and $B = [\begin{array}{ccc} 2 \sqrt{2} & 5 & \frac{1}{2} \\ - 2 & - 4 & 1 \end{array}]$ .

Try it yourself first. Then reveal the solution one step at a time.

3.4.2 Multiplication of a matrix by a scalar

Suppose the kiosks sell exactly the same amounts every day for a week. Then each week's figure is seven times the day's figure, in every position:

\begin{aligned} 7 D & = 7 [\begin{array}{ccc} 75 & 115 & 50 \\ 55 & 35 & 60 \end{array}] \\ = [\begin{array}{ccc} 525 & 805 & 350 \\ 385 & 245 & 420 \end{array}] \end{aligned}

In general, for a matrix $A = [a_{i j}]_{m \times n}$ and a number $k$ (called a scalar, to set it apart from a matrix), the scalar multiple $k A$ is the $m \times n$ matrix obtained by multiplying every entry of $A$ by $k$ . For instance, $- \frac{1}{2} [\begin{array}{cc} 4 & - 6 \\ 1 & 0 \end{array}] = [\begin{array}{cc} - 2 & 3 \\ - \frac{1}{2} & 0 \end{array}]$ .

Scalar multiple

k A = k [a_{i j}]_{m \times n} = [k a_{i j}]_{m \times n}

Two special cases have names. The negative of $A$ is $- A = (- 1) A$ : change the sign of every entry. The difference of two matrices of the same order is $A - B = A + (- 1) B$ , so it is found by subtracting corresponding entries, $A - B = [a_{i j} - b_{i j}]$ . For the kiosks,

M - E = [\begin{array}{ccc} 5 & - 5 & 10 \\ - 5 & 5 & 10 \end{array}]

A positive entry means the morning shift sold more of that juice at that kiosk, and a negative entry means the evening shift did. Negative entries in a matrix carry meaning, just as negative numbers do.

Example 7NCERT Example 7 · p. 46

A combination of two matrices

If $A = [\begin{array}{cc} 2 & - 1 \\ 0 & 3 \\ 1 & 4 \end{array}]$ and $B = [\begin{array}{cc} 1 & 2 \\ - 3 & 5 \\ 4 & - 2 \end{array}]$ , find $3 A - 2 B$ .

Try it yourself first. Then reveal the solution one step at a time.

3.4.3 Properties of matrix addition

Matrix addition is ordinary addition carried out in every position at once, so it inherits the familiar laws of numbers. Let $A$ , $B$ and $C$ be matrices of the same order $m \times n$ .

PropertyLaws of matrix addition

(i) Commutative: $A + B = B + A$ . (ii) Associative: $(A + B) + C = A + (B + C)$ . (iii) Additive identity: the $m \times n$ zero matrix O satisfies $A + O = O + A = A$ . (iv) Additive inverse: $- A = [- a_{i j}]$ satisfies $A + (- A) = (- A) + A = O$ .

3.4.4 Properties of scalar multiplication of a matrix

PropertyLaws of scalar multiplication

For matrices $A$ and $B$ of the same order and scalars $k$ and $l$ : (i) $k (A + B) = k A + k B$ ; (ii) $(k + l) A = k A + l A$ ; (iii) $k (l A) = (k l) A$ ; (iv) $1 \cdot A = A$ .

These laws let us solve simple matrix equations exactly as we solve $3 a - 2 x = b$ for a number $x$ : move terms across, then divide by a non-zero number. The next examples name the law behind each move.

Example 8NCERT Example 8 · p. 47

Solving for an unknown matrix

Given $A = [\begin{array}{cc} 2 & 0 \\ - 1 & 3 \end{array}]$ and $B = [\begin{array}{cc} 4 & - 2 \\ 1 & 5 \end{array}]$ , find the matrix $X$ with $3 A - 2 X = B$ .

Try it yourself first. Then reveal the solution one step at a time.

Example 9NCERT Example 9 · p. 48

Two equations, two unknown matrices

Find $X$ and $Y$ if $2 X + Y = [\begin{array}{cc} 7 & 2 \\ - 1 & 8 \end{array}]$ and $X - Y = [\begin{array}{cc} 2 & 4 \\ - 2 & 1 \end{array}]$ .

Try it yourself first. Then reveal the solution one step at a time.

Example 10NCERT Example 10 · p. 48

Unknown entries inside a combination

Find $x$ and $y$ if $3 [\begin{array}{cc} x & - 1 \\ 2 & y \end{array}] - [\begin{array}{cc} 4 & 2 \\ 6 & 1 \end{array}] = [\begin{array}{cc} 11 & - 5 \\ 0 & 11 \end{array}]$ .

Try it yourself first. Then reveal the solution one step at a time.

Example 11NCERT Example 11 · p. 49

Takings at a crafts fair

At a crafts fair, a pottery stall (row 1) and a bamboo stall (row 2) sell small, medium and large items (columns, in that order). Their takings in rupees were $A$ on Saturday and $B$ on Sunday:

A = [\begin{array}{ccc} 4000 & 2500 & 1800 \\ 3200 & 4500 & 2000 \end{array}]

B = [\begin{array}{ccc} 5200 & 3100 & 2400 \\ 3000 & 6000 & 3500 \end{array}]

(a) Find each stall's weekend takings for each size. (b) Find the change in takings from Saturday to Sunday. (c) The organisers keep a 5 % fee on Sunday's takings. Find the fee for each stall and each size.

Try it yourself first. Then reveal the solution one step at a time.

3.4.5 Multiplication of matrices

Adding and scaling work position by position. Multiplication is different, and the kiosks show why. Suppose a glass of mango juice costs ₹40, sugarcane ₹20 and mosambi ₹30. The station's takings for the day pair each number of glasses with its price, multiply, and add: $75 \times 40 + 115 \times 20 + 50 \times 30 = 3000 + 2300 + 1500 = 6800$ , that is ₹6800. The college kiosk takes $55 \times 40 + 35 \times 20 + 60 \times 30 = 4700$ , that is ₹4700.

Each total pairs a row of the sales matrix $D$ with the column of prices. Now add a second column of summer prices (₹50, ₹25 and ₹35). Every pairing of a kiosk row with a price column produces one entry of the answer:

\begin{aligned} D P & = \underset{2 \times 3}{\underset{⏟}{[\begin{array}{ccc} 75 & 115 & 50 \\ 55 & 35 & 60 \end{array}]}} \underset{3 \times 2}{\underset{⏟}{[\begin{array}{cc} 40 & 50 \\ 20 & 25 \\ 30 & 35 \end{array}]}} \\ = [\begin{array}{cc} 6800 & 8375 \\ 4700 & 5725 \end{array}] \end{aligned}

Read the answer like a table: row 1 is the station and row 2 the college; column 1 uses the usual prices and column 2 the summer prices. So at summer prices the station would take ₹8375 a day. Notice the shapes: a row of $D$ has 3 entries and a column of $P$ has 3 entries, so they pair off exactly.

Definition

Product of matrices

If $A = [a_{i j}]$ is $m \times n$ and $B = [b_{j k}]$ is $n \times p$ , their product $A B$ is the $m \times p$ matrix whose $(i, k)$ entry is found by multiplying row $i$ of $A$ with column $k$ of $B$ entry by entry and adding. If the number of columns of $A$ is not the number of rows of $B$ , then $A B$ is not defined.

Example $[\begin{array}{cc} 1 & 2 \end{array}] [\begin{array}{c} 3 \\ 4 \end{array}] = [1 \cdot 3 + 2 \cdot 4] = [11]$

Entry (i, k) of the product AB

\begin{aligned} c_{i k} & = a_{i 1} b_{1 k} + a_{i 2} b_{2 k} \\ + \dots + a_{i n} b_{n k} \\ = \sum_{j = 1}^{n} a_{i j} b_{j k} \end{aligned}

where

$j$: the running index: it moves along row $i$ of $A$ and down column $k$ of $B$ together
$C = [c_{i k}]_{m \times p}$: the product $A B$

A quick test for whether a product exists is to write the orders side by side, $(m \times n) (n \times p)$ . The two inner numbers must match, and the two outer numbers give the order of the product, $m \times p$ . Fig. 3.7 works through a small product one entry at a time, and the lab after it lets you try your own matrices.

3/5Row 1 × column 2: $2 \cdot 2 + 1 \cdot 0 + 3 \cdot 5 = 19$ .

Fig. 3.7Entry

(i, k)

A B

is row

i

A

times column

k

B

: multiply in pairs, then add.NCERT idea of the product CD on p. 51 · p. 51

The matrix with rows 2, 1, 3 and 0, 4, 1 is multiplied by the matrix with rows 1, 2 and 3, 0 and 1, 5. Step 1 shows the orders 2 by 3 and 3 by 2 with the matching inner 3s and the product order 2 by 2, with an empty answer. In step 2 row 1 and column 1 are highlighted and the working 2 times 1 plus 1 times 3 plus 3 times 1 equals 8 gives the first entry. Step 3 uses row 1 and column 2 to get 19. Step 4 uses row 2 and column 1 to get 13. Step 5 uses row 2 and column 2 to get 5, completing the product 8, 19 over 13, 5.

InteractiveRow-by-column lab

An interactive multiplication lab. A menu offers ready-made pairs: the kiosk sales and prices, two square matrices, a column times a row, a row times a column and a pair whose orders do not fit; buttons change the number of rows and columns. Every entry of A and B can be edited. The panel states both orders and whether the inner numbers match. Tapping a cell of the product highlights the matching row of A and column of B and writes out the sum of products; a Next entry button fills the product one cell at a time, Fill all completes it, and a Swap button shows B times A for comparison.

Fig. 3.8Pick a pair of matrices (or type your own entries), then tap any cell of the product to see the row and column that make it.

Example 12NCERT Example 12 · p. 52

A square matrix times a wide one

Find $A B$ for $A = [\begin{array}{cc} 3 & - 2 \\ 1 & 4 \end{array}]$ and $B = [\begin{array}{ccc} 2 & 0 & - 1 \\ 5 & 1 & 3 \end{array}]$ . Is $B A$ defined?

Try it yourself first. Then reveal the solution one step at a time.

So $A B$ can exist while $B A$ does not. If $A$ is $m \times n$ and $B$ is $k \times l$ , then $A B$ needs $n = k$ and $B A$ needs $l = m$ , and both products exist only when both conditions hold. In particular, for two square matrices of the same order both products always exist.

Order matters: AB and BA

For numbers, $3 \times 5 = 5 \times 3$ . For matrices, even when $A B$ and $B A$ both exist they are usually different, and they may not even have the same order.

Example 13NCERT Example 13 · p. 53

Products of different orders

For $A = [\begin{array}{ccc} 1 & 0 & 2 \\ - 1 & 3 & 1 \end{array}]$ and $B = [\begin{array}{cc} 2 & 1 \\ 0 & - 1 \\ 1 & 4 \end{array}]$ , find $A B$ and $B A$ and show that $A B \neq B A$ .

Try it yourself first. Then reveal the solution one step at a time.

Equal orders do not rescue the rule. Geometry shows why. A $2 \times 2$ matrix $A$ can act on the plane by sending each point $[\begin{array}{c} x \\ y \end{array}]$ to $A [\begin{array}{c} x \\ y \end{array}]$ , so the product $R S$ means “apply $S$ first, then $R$ ”. In Fig. 3.9, $S$ reflects a flag in the $y$ -axis and $R$ turns it through $90 °$ anticlockwise. Done in the two possible orders, the moves leave the flag in two different places.

3/3Then the other move. The flags end in different places, so $R S \neq S R$ .

Fig. 3.9Reflect then turn is not the same as turn then reflect:

R S \neq S R

.NCERT idea of Example 14 · p. 53

Two rows of three small coordinate panels. Top row: S first, then R. Bottom row: R first, then S. Step 1 shows the same starting flag in both rows: a pole standing just right of the y-axis with a pennant pointing right. Step 2 shows the first move: in the top row the flag is reflected in the y-axis, so the pennant points left; in the bottom row it is turned a quarter turn anticlockwise, so the pole lies just above the negative x-axis. Step 3 shows the second move: the top row ends with the pole below the negative x-axis and the pennant hanging down, the bottom row ends with the pole above the positive x-axis and the pennant pointing up. Beside the results are the matrices RS, with rows 0, minus 1 and minus 1, 0, and SR, with rows 0, 1 and 1, 0, which differ.

Example 14NCERT Example 14 · p. 53

Same order, different products

Let $A = [\begin{array}{cc} 2 & 1 \\ 0 & 3 \end{array}]$ and $B = [\begin{array}{cc} 1 & 0 \\ 4 & 1 \end{array}]$ . Show that $A B \neq B A$ .

Try it yourself first. Then reveal the solution one step at a time.

This does not mean that $A B$ and $B A$ always differ. Diagonal matrices of the same order always commute, because each product just multiplies the diagonal entries in place: $[\begin{array}{cc} 2 & 0 \\ 0 & - 1 \end{array}] [\begin{array}{cc} 5 & 0 \\ 0 & 3 \end{array}] = [\begin{array}{cc} 10 & 0 \\ 0 & - 3 \end{array}]$ in either order. Every square matrix also commutes with $I$ , with O and with itself.

A zero product from non-zero matrices

For real numbers, $a b = 0$ forces $a = 0$ or $b = 0$ . Matrices break this rule as well.

Example 15NCERT Example 15 · p. 54

Two non-zero factors, product zero

Find $A B$ and $B A$ for $A = [\begin{array}{cc} 1 & 2 \\ 2 & 4 \end{array}]$ and $B = [\begin{array}{cc} 2 & - 4 \\ - 1 & 2 \end{array}]$ .

Try it yourself first. Then reveal the solution one step at a time.

So from $A B = O$ we cannot conclude that $A = O$ or $B = O$ . The same example shows that $A B = O$ does not even force $B A = O$ .

3.4.6 Properties of multiplication of matrices

Although commutativity fails, the other laws we rely on for numbers survive. We state them, sketch why they hold, and then test them on examples.

PropertyLaws of matrix multiplication

Whenever the products involved are defined: (i) Associative: $(A B) C = A (B C)$ . (ii) Distributive: $A (B + C) = A B + A C$ and $(A + B) C = A C + B C$ . (iii) Multiplicative identity: for a square matrix $A$ of order $n$ , $I_{n} A = A I_{n} = A$ .

(iii) The $(i, k)$ entry of $A I$ is $\sum_{j} a_{i j} e_{j k}$ , where $e_{j k}$ is 1 when $j = k$ and 0 otherwise. Only the term with $j = k$ survives, leaving $a_{i k}$ . The same argument works for $I A$ .

(ii) The $(i, k)$ entry of $A (B + C)$ is $\sum_{j} a_{i j} (b_{j k} + c_{j k}) = \sum_{j} a_{i j} b_{j k} + \sum_{j} a_{i j} c_{j k}$ , which is the $(i, k)$ entry of $A B + A C$ . The second law is similar.

(i) Both $(A B) C$ and $A (B C)$ have $(i, l)$ entry $\sum_{j} \sum_{k} a_{i j} b_{j k} c_{k l}$ ; the two sides only group the same products differently.

Associativity means a product such as $A B C$ needs no brackets, and it makes powers of a square matrix meaningful: $A^{2} = A A$ , $A^{3} = A A A$ , and so on. Polynomials in a matrix, such as $A^{2} - 3 A + 2 I$ , follow, with $I$ standing in for the constant term.

Example 16NCERT Example 16 · p. 54

Grouping a product two ways

Let $A = [\begin{array}{cc} 1 & - 1 \\ 2 & 3 \end{array}]$ , $B = [\begin{array}{ccc} 2 & 0 & 1 \\ - 1 & 4 & 2 \end{array}]$ and $C = [\begin{array}{c} 3 \\ 1 \\ - 2 \end{array}]$ . Find $(A B) C$ and $A (B C)$ , and check that they are equal.

Try it yourself first. Then reveal the solution one step at a time.

Example 17NCERT Example 17 · p. 56

Checking the distributive law

For $A = [\begin{array}{cc} 1 & 2 \\ 0 & - 1 \end{array}]$ , $B = [\begin{array}{cc} 3 & - 1 \\ 2 & 2 \end{array}]$ and $C = [\begin{array}{cc} 2 & 1 \\ - 1 & 4 \end{array}]$ , verify that $A (B + C) = A B + A C$ .

Try it yourself first. Then reveal the solution one step at a time.

Example 18NCERT Example 18 · p. 56

A matrix that satisfies a quadratic

Let $A = [\begin{array}{cc} 2 & 1 \\ 1 & 3 \end{array}]$ . Show that $A^{2} - 5 A + 5 I = O$ , and use this to write $A^{3}$ as $p A + q I$ .

Try it yourself first. Then reveal the solution one step at a time.

Example 19NCERT Example 19 · p. 57

Costing an awareness campaign

A health NGO runs an awareness campaign in two districts using posters (₹12 each), radio spots (₹1500 each) and street plays (₹4000 each). District P gets 2000 posters, 30 radio spots and 12 street plays; district Q gets 3500 posters, 20 radio spots and 18 street plays. Use matrix multiplication to find the cost in each district.

Try it yourself first. Then reveal the solution one step at a time.

Practice · 22 questions

Practice 3.2: questions with answers

Practice for Section 3.4 (matches NCERT Exercise 3.2).

1
Short answerEasyNCERT Exercise 3.2, Q1
Let $A = [\begin{array}{cc} 3 & - 1 \\ 2 & 4 \end{array}]$ , $B = [\begin{array}{cc} 1 & 5 \\ - 2 & 0 \end{array}]$ and $C = [\begin{array}{cc} - 3 & 2 \\ 1 & 6 \end{array}]$ . Find:
1. $A + B$
  Show solution
  Answer $[\begin{array}{cc} 4 & 4 \\ 0 & 4 \end{array}]$
  $A + B = [\begin{array}{cc} 4 & 4 \\ 0 & 4 \end{array}]$ .
2. $A - B$
  Show solution
  Answer $[\begin{array}{cc} 2 & - 6 \\ 4 & 4 \end{array}]$
  $A - B = [\begin{array}{cc} 2 & - 6 \\ 4 & 4 \end{array}]$ .
3. $2 A - 3 C$
  Show solution
  Answer $[\begin{array}{cc} 15 & - 8 \\ 1 & - 10 \end{array}]$
  $2 A - 3 C = [\begin{array}{cc} 6 & - 2 \\ 4 & 8 \end{array}] - [\begin{array}{cc} - 9 & 6 \\ 3 & 18 \end{array}] = [\begin{array}{cc} 15 & - 8 \\ 1 & - 10 \end{array}]$ .
4. $A B$
  Show solution
  Answer $[\begin{array}{cc} 5 & 15 \\ - 6 & 10 \end{array}]$
  $A B = [\begin{array}{cc} 3 + 2 & 15 + 0 \\ 2 - 8 & 10 + 0 \end{array}] = [\begin{array}{cc} 5 & 15 \\ - 6 & 10 \end{array}]$ .
5. $B A$
  Show solution
  Answer $[\begin{array}{cc} 13 & 19 \\ - 6 & 2 \end{array}]$
  $B A = [\begin{array}{cc} 3 + 10 & - 1 + 20 \\ - 6 + 0 & 2 + 0 \end{array}] = [\begin{array}{cc} 13 & 19 \\ - 6 & 2 \end{array}]$ , which differs from $A B$ .
2
Short answerMediumNCERT Exercise 3.2, Q2
Simplify each sum.
1. $[\begin{array}{cc} p & - q \\ q & p \end{array}] + [\begin{array}{cc} p & q \\ - q & p \end{array}]$
  Show solution
  Answer $[\begin{array}{cc} 2 p & 0 \\ 0 & 2 p \end{array}]$ , a scalar matrix.
  The off-diagonal entries cancel: $- q + q = 0$ and $q - q = 0$ .
2. $[\begin{array}{cc} m^{2} + n^{2} & m - n \\ 2 m n & 1 \end{array}] + [\begin{array}{cc} - 2 m n & m + n \\ m^{2} + n^{2} & 3 \end{array}]$
  Show solution
  Answer $[\begin{array}{cc} (m - n)^{2} & 2 m \\ (m + n)^{2} & 4 \end{array}]$
  $m^{2} + n^{2} - 2 m n = (m - n)^{2}$ , $(m - n) + (m + n) = 2 m$ , $2 m n + m^{2} + n^{2} = (m + n)^{2}$ and $1 + 3 = 4$ .
3. $[\begin{array}{ccc} 2 & - 5 & 7 \\ 0 & 4 & - 1 \\ 3 & 6 & 8 \end{array}] + [\begin{array}{ccc} - 2 & 3 & 1 \\ 5 & - 4 & 9 \\ 1 & 0 & - 8 \end{array}]$
  Show solution
  Answer $[\begin{array}{ccc} 0 & - 2 & 8 \\ 5 & 0 & 8 \\ 4 & 6 & 0 \end{array}]$
  Adding position by position gives $[\begin{array}{ccc} 0 & - 2 & 8 \\ 5 & 0 & 8 \\ 4 & 6 & 0 \end{array}]$ ; notice the zero diagonal.
4. $[\begin{array}{cc} \sec^{2} x & 1 \\ \tan^{2} x & {cosec}^{2} x \end{array}] + [\begin{array}{cc} - \tan^{2} x & - \sec^{2} x \\ 1 & - \cot^{2} x \end{array}]$
  Show solution
  Answer $[\begin{array}{cc} 1 & - \tan^{2} x \\ \sec^{2} x & 1 \end{array}]$
  Use $\sec^{2} x - \tan^{2} x = 1$ , $1 - \sec^{2} x = - \tan^{2} x$ , $\tan^{2} x + 1 = \sec^{2} x$ and ${cosec}^{2} x - \cot^{2} x = 1$ .
3
Short answerMediumNCERT Exercise 3.2, Q3
Compute each product.
1. $[\begin{array}{cc} p & q \\ q & p \end{array}] [\begin{array}{cc} p & - q \\ - q & p \end{array}]$
  Show solution
  Answer $[\begin{array}{cc} p^{2} - q^{2} & 0 \\ 0 & p^{2} - q^{2} \end{array}] = (p^{2} - q^{2}) I$
  Entry (1, 1): $p^{2} - q^{2}$ ; entry (1, 2): $- p q + q p = 0$ ; entry (2, 1): $q p - p q = 0$ ; entry (2, 2): $- q^{2} + p^{2}$ .
2. $[\begin{array}{c} 2 \\ - 1 \\ 3 \end{array}] [\begin{array}{ccc} 1 & 4 & - 2 \end{array}]$
  Show solution
  Answer $[\begin{array}{ccc} 2 & 8 & - 4 \\ - 1 & - 4 & 2 \\ 3 & 12 & - 6 \end{array}]$
  A $(3 \times 1) (1 \times 3)$ product is $3 \times 3$ ; each entry is a single product, row number times column number.
  $[\begin{array}{ccc} 2 & 8 & - 4 \\ - 1 & - 4 & 2 \\ 3 & 12 & - 6 \end{array}]$
3. $[\begin{array}{cc} 2 & - 1 \\ 1 & 3 \end{array}] [\begin{array}{ccc} 1 & 0 & 2 \\ - 2 & 1 & 3 \end{array}]$
  Show solution
  Answer $[\begin{array}{ccc} 4 & - 1 & 1 \\ - 5 & 3 & 11 \end{array}]$
  $[\begin{array}{ccc} 2 + 2 & 0 - 1 & 4 - 3 \\ 1 - 6 & 0 + 3 & 2 + 9 \end{array}] = [\begin{array}{ccc} 4 & - 1 & 1 \\ - 5 & 3 & 11 \end{array}]$ .
4. $[\begin{array}{ccc} 1 & 2 & 0 \\ 0 & - 1 & 3 \\ 2 & 1 & 1 \end{array}] [\begin{array}{ccc} 2 & - 1 & 1 \\ 1 & 0 & 3 \\ 0 & 2 & - 1 \end{array}]$
  Show solution
  Answer $[\begin{array}{ccc} 4 & - 1 & 7 \\ - 1 & 6 & - 6 \\ 5 & 0 & 4 \end{array}]$
  Row 1: $(2 + 2 + 0, - 1 + 0 + 0, 1 + 6 + 0)$ . Row 2: $(0 - 1 + 0, 0 + 0 + 6, 0 - 3 - 3)$ . Row 3: $(4 + 1 + 0, - 2 + 0 + 2, 2 + 3 - 1)$ .
  $[\begin{array}{ccc} 4 & - 1 & 7 \\ - 1 & 6 & - 6 \\ 5 & 0 & 4 \end{array}]$
5. $[\begin{array}{cc} 1 & 3 \\ 0 & - 2 \\ 4 & 1 \end{array}] [\begin{array}{ccc} 2 & 0 & - 1 \\ 1 & 5 & 2 \end{array}]$
  Show solution
  Answer $[\begin{array}{ccc} 5 & 15 & 5 \\ - 2 & - 10 & - 4 \\ 9 & 5 & - 2 \end{array}]$
  $(3 \times 2) (2 \times 3)$ gives $3 \times 3$ : row 1 $(2 + 3, 0 + 15, - 1 + 6)$ , row 2 $(0 - 2, 0 - 10, 0 - 4)$ , row 3 $(8 + 1, 0 + 5, - 4 + 2)$ .
  $[\begin{array}{ccc} 5 & 15 & 5 \\ - 2 & - 10 & - 4 \\ 9 & 5 & - 2 \end{array}]$
6. $[\begin{array}{ccc} 3 & 0 & - 1 \\ 1 & 2 & 4 \end{array}] [\begin{array}{cc} 1 & 2 \\ 0 & - 3 \\ 5 & 1 \end{array}]$
  Show solution
  Answer $[\begin{array}{cc} - 2 & 5 \\ 21 & 0 \end{array}]$
  $[\begin{array}{cc} 3 + 0 - 5 & 6 + 0 - 1 \\ 1 + 0 + 20 & 2 - 6 + 4 \end{array}] = [\begin{array}{cc} - 2 & 5 \\ 21 & 0 \end{array}]$ .
Short answerMediumNCERT Exercise 3.2, Q4
Let $A = [\begin{array}{ccc} 2 & - 1 & 0 \\ 3 & 4 & - 2 \\ 1 & 0 & 5 \end{array}]$ , $B = [\begin{array}{ccc} 1 & 3 & - 1 \\ 0 & - 2 & 4 \\ 2 & 1 & 1 \end{array}]$ and $C = [\begin{array}{ccc} - 1 & 2 & 3 \\ 4 & 0 & 1 \\ 2 & - 3 & 0 \end{array}]$ . Find $A + B$ and $B - C$ , and verify that $A + (B - C) = (A + B) - C$ .
Show solution
AnswerBoth sides equal $[\begin{array}{ccc} 4 & 0 & - 4 \\ - 1 & 2 & 1 \\ 1 & 4 & 6 \end{array}]$ .
$\begin{array}{c} A + B = [\begin{array}{ccc} 3 & 2 & - 1 \\ 3 & 2 & 2 \\ 3 & 1 & 6 \end{array}] \\ B - C = [\begin{array}{ccc} 2 & 1 & - 4 \\ - 4 & - 2 & 3 \\ 0 & 4 & 1 \end{array}] \end{array}$
$\begin{array}{c} A + (B - C) = [\begin{array}{ccc} 4 & 0 & - 4 \\ - 1 & 2 & 1 \\ 1 & 4 & 6 \end{array}] \\ (A + B) - C = [\begin{array}{ccc} 4 & 0 & - 4 \\ - 1 & 2 & 1 \\ 1 & 4 & 6 \end{array}] \end{array}$
The results agree, as the associative law (with $B - C = B + (- C)$ ) predicts.
Short answerMediumNCERT Exercise 3.2, Q5
If $A = [\begin{array}{ccc} \frac{1}{2} & \frac{3}{4} & - 1 \\ \frac{5}{4} & 0 & \frac{3}{2} \end{array}]$ and $B = [\begin{array}{ccc} \frac{1}{3} & \frac{1}{6} & - \frac{2}{3} \\ \frac{1}{2} & \frac{5}{6} & 0 \end{array}]$ , compute $4 A - 6 B$ .
Show solution
Answer $[\begin{array}{ccc} 0 & 2 & 0 \\ 2 & - 5 & 6 \end{array}]$
$\begin{array}{c} 4 A = [\begin{array}{ccc} 2 & 3 & - 4 \\ 5 & 0 & 6 \end{array}] \\ 6 B = [\begin{array}{ccc} 2 & 1 & - 4 \\ 3 & 5 & 0 \end{array}] \end{array}$
Subtracting, $4 A - 6 B = [\begin{array}{ccc} 0 & 2 & 0 \\ 2 & - 5 & 6 \end{array}]$ .
Short answerMediumNCERT Exercise 3.2, Q6
Simplify $\sec θ [\begin{array}{cc} \sec θ & \tan θ \\ \tan θ & \sec θ \end{array}] - \tan θ [\begin{array}{cc} \tan θ & \sec θ \\ \sec θ & \tan θ \end{array}]$ .
Show solution
Answer $I_{2} = [\begin{array}{cc} 1 & 0 \\ 0 & 1 \end{array}]$
$\begin{array}{c} [\begin{array}{cc} \sec^{2} θ - \tan^{2} θ & \sec θ \tan θ - \tan θ \sec θ \\ \sec θ \tan θ - \tan θ \sec θ & \sec^{2} θ - \tan^{2} θ \end{array}] \\ = [\begin{array}{cc} 1 & 0 \\ 0 & 1 \end{array}] \end{array}$
7
Short answerMediumNCERT Exercise 3.2, Q7
Find $X$ and $Y$ .
1. $X + Y = [\begin{array}{cc} 6 & - 2 \\ 4 & 8 \end{array}]$ and $X - Y = [\begin{array}{cc} 2 & 4 \\ - 2 & 0 \end{array}]$
  Show solution
  Answer $X = [\begin{array}{cc} 4 & 1 \\ 1 & 4 \end{array}]$ , $Y = [\begin{array}{cc} 2 & - 3 \\ 3 & 4 \end{array}]$
  Adding, $2 X = [\begin{array}{cc} 8 & 2 \\ 2 & 8 \end{array}]$ , so $X = [\begin{array}{cc} 4 & 1 \\ 1 & 4 \end{array}]$ . Subtracting, $2 Y = [\begin{array}{cc} 4 & - 6 \\ 6 & 8 \end{array}]$ , so $Y = [\begin{array}{cc} 2 & - 3 \\ 3 & 4 \end{array}]$ .
2. $3 X + 2 Y = [\begin{array}{cc} 7 & 0 \\ 4 & 11 \end{array}]$ and $2 X + Y = [\begin{array}{cc} 4 & 1 \\ 3 & 6 \end{array}]$
  Show solution
  Answer $X = [\begin{array}{cc} 1 & 2 \\ 2 & 1 \end{array}]$ , $Y = [\begin{array}{cc} 2 & - 3 \\ - 1 & 4 \end{array}]$
  Double the second equation: $4 X + 2 Y = [\begin{array}{cc} 8 & 2 \\ 6 & 12 \end{array}]$ . Subtract the first: $X = [\begin{array}{cc} 8 - 7 & 2 - 0 \\ 6 - 4 & 12 - 11 \end{array}] = [\begin{array}{cc} 1 & 2 \\ 2 & 1 \end{array}]$ .
  Then $Y = [\begin{array}{cc} 4 & 1 \\ 3 & 6 \end{array}] - 2 X = [\begin{array}{cc} 2 & - 3 \\ - 1 & 4 \end{array}]$ .
Short answerEasyNCERT Exercise 3.2, Q8
Find $X$ if $Y = [\begin{array}{cc} 1 & - 2 \\ 2 & 4 \end{array}]$ and $3 X - Y = [\begin{array}{cc} 2 & 1 \\ - 4 & 5 \end{array}]$ .
Show solution
Answer $[\begin{array}{cc} 1 & - \frac{1}{3} \\ - \frac{2}{3} & 3 \end{array}]$
$3 X = [\begin{array}{cc} 2 & 1 \\ - 4 & 5 \end{array}] + [\begin{array}{cc} 1 & - 2 \\ 2 & 4 \end{array}] = [\begin{array}{cc} 3 & - 1 \\ - 2 & 9 \end{array}]$ , so $X = [\begin{array}{cc} 1 & - \frac{1}{3} \\ - \frac{2}{3} & 3 \end{array}]$ .
Short answerEasyNCERT Exercise 3.2, Q9
Find $x$ and $y$ if $3 [\begin{array}{cc} 1 & x \\ y & 0 \end{array}] + [\begin{array}{cc} 2 & 1 \\ - 1 & 4 \end{array}] = [\begin{array}{cc} 5 & 7 \\ 8 & 4 \end{array}]$ .
Show solution
Answer $x = 2$ , $y = 3$ .
Position (1, 2): $3 x + 1 = 7$ , so $x = 2$ . Position (2, 1): $3 y - 1 = 8$ , so $y = 3$ . The other positions agree: $3 + 2 = 5$ , $0 + 4 = 4$ .
Short answerEasyNCERT Exercise 3.2, Q10
Solve for $x$ , $y$ , $z$ , $t$ : $4 [\begin{array}{cc} x & y \\ z & t \end{array}] - 2 [\begin{array}{cc} 1 & 3 \\ - 2 & 1 \end{array}] = 2 [\begin{array}{cc} 3 & 3 \\ 4 & 5 \end{array}]$ .
Show solution
Answer $x = 2$ , $y = 3$ , $z = 1$ , $t = 3$ .
$4 x - 2 = 6 \Rightarrow x = 2$ ; $4 y - 6 = 6 \Rightarrow y = 3$ ; $4 z + 4 = 8 \Rightarrow z = 1$ ; $4 t - 2 = 10 \Rightarrow t = 3$ .
Short answerEasyNCERT Exercise 3.2, Q11
If $x [\begin{array}{c} 3 \\ 1 \end{array}] + y [\begin{array}{c} - 1 \\ 2 \end{array}] = [\begin{array}{c} 7 \\ 7 \end{array}]$ , find $x$ and $y$ .
Show solution
Answer $x = 3$ , $y = 2$ .
$3 x - y = 7$ and $x + 2 y = 7$ . Doubling the first and adding: $7 x = 21$ , so $x = 3$ and $y = 2$ .
Short answerMediumNCERT Exercise 3.2, Q12
Given $2 [\begin{array}{cc} x & y \\ z & w \end{array}] = [\begin{array}{cc} x & 5 \\ 3 & w \end{array}] + [\begin{array}{cc} 3 & y - x \\ w + z & - 4 \end{array}]$ , find $x$ , $y$ , $z$ and $w$ .
Show solution
Answer $x = 3$ , $y = 2$ , $z = - 1$ , $w = - 4$ .
(1, 1): $2 x = x + 3$ , so $x = 3$ . (1, 2): $2 y = 5 + y - x = y + 2$ , so $y = 2$ . (2, 2): $2 w = w - 4$ , so $w = - 4$ . (2, 1): $2 z = 3 + w + z$ , so $z = 3 + w = - 1$ .
ProofMediumNCERT Exercise 3.2, Q13
For a real number $t$ let $G (t) = [\begin{array}{ccc} 1 & t & \frac{1}{2} t^{2} \\ 0 & 1 & t \\ 0 & 0 & 1 \end{array}]$ . Show that $G (s) G (t) = G (s + t)$ .
Show solution
$G (s) G (t) = [\begin{array}{ccc} 1 & t + s & \frac{1}{2} t^{2} + s t + \frac{1}{2} s^{2} \\ 0 & 1 & t + s \\ 0 & 0 & 1 \end{array}]$
Since $\frac{1}{2} t^{2} + s t + \frac{1}{2} s^{2} = \frac{1}{2} (s + t)^{2}$ , the product is exactly $G (s + t)$ . (The other entries are 0 or 1 because both factors are upper triangular with 1s on the diagonal.)
14
Short answerMediumNCERT Exercise 3.2, Q14
Show that $A B \neq B A$ in each case.
1. $A = [\begin{array}{cc} 2 & 1 \\ - 3 & 4 \end{array}]$ , $B = [\begin{array}{cc} 0 & 1 \\ 1 & 2 \end{array}]$
  Show solution
  Answer $A B = [\begin{array}{cc} 1 & 4 \\ 4 & 5 \end{array}]$ , $B A = [\begin{array}{cc} - 3 & 4 \\ - 4 & 9 \end{array}]$
  $A B = [\begin{array}{cc} 1 & 4 \\ 4 & 5 \end{array}]$ while $B A = [\begin{array}{cc} - 3 & 4 \\ - 4 & 9 \end{array}]$ ; already the (1, 1) entries differ (1 and $- 3$ ).
2. $A = [\begin{array}{ccc} 1 & 0 & 2 \\ 0 & 1 & - 1 \\ 3 & 1 & 0 \end{array}]$ , $B = [\begin{array}{ccc} 0 & 1 & 1 \\ 2 & 0 & 1 \\ 1 & - 1 & 0 \end{array}]$
  Show solution
  Answer $A B = [\begin{array}{ccc} 2 & - 1 & 1 \\ 1 & 1 & 1 \\ 2 & 3 & 4 \end{array}]$ , $B A = [\begin{array}{ccc} 3 & 2 & - 1 \\ 5 & 1 & 4 \\ 1 & - 1 & 3 \end{array}]$
  $\begin{array}{c} A B = [\begin{array}{ccc} 2 & - 1 & 1 \\ 1 & 1 & 1 \\ 2 & 3 & 4 \end{array}] \\ B A = [\begin{array}{ccc} 3 & 2 & - 1 \\ 5 & 1 & 4 \\ 1 & - 1 & 3 \end{array}] \end{array}$
  The (1, 1) entries are 2 and 3, so $A B \neq B A$ .
Short answerMediumNCERT Exercise 3.2, Q15
Find $A^{2} - 4 A + 3 I$ for $A = [\begin{array}{ccc} 1 & 2 & 0 \\ 0 & 1 & 3 \\ 2 & 0 & 1 \end{array}]$ .
Show solution
Answer $[\begin{array}{ccc} 0 & - 4 & 6 \\ 6 & 0 & - 6 \\ - 4 & 4 & 0 \end{array}]$
$A^{2} = [\begin{array}{ccc} 1 & 4 & 6 \\ 6 & 1 & 6 \\ 4 & 4 & 1 \end{array}]$
$A^{2} - 4 A + 3 I = [\begin{array}{ccc} 1 & 4 & 6 \\ 6 & 1 & 6 \\ 4 & 4 & 1 \end{array}] - [\begin{array}{ccc} 4 & 8 & 0 \\ 0 & 4 & 12 \\ 8 & 0 & 4 \end{array}] + [\begin{array}{ccc} 3 & 0 & 0 \\ 0 & 3 & 0 \\ 0 & 0 & 3 \end{array}] = [\begin{array}{ccc} 0 & - 4 & 6 \\ 6 & 0 & - 6 \\ - 4 & 4 & 0 \end{array}]$ .
ProofHardNCERT Exercise 3.2, Q16
If $A = [\begin{array}{ccc} 1 & 0 & 1 \\ 2 & 1 & 0 \\ 0 & 1 & 2 \end{array}]$ , prove that $A^{3} - 4 A^{2} + 5 A - 4 I = O$ .
Show solution
$\begin{array}{c} A^{2} = [\begin{array}{ccc} 1 & 1 & 3 \\ 4 & 1 & 2 \\ 2 & 3 & 4 \end{array}] \\ \begin{aligned} A^{3} & = A \cdot A^{2} \\ = [\begin{array}{ccc} 3 & 4 & 7 \\ 6 & 3 & 8 \\ 8 & 7 & 10 \end{array}] \end{aligned} \end{array}$
$\begin{aligned} A^{3} - 4 A^{2} + 5 A - 4 I & = [\begin{array}{ccc} 3 & 4 & 7 \\ 6 & 3 & 8 \\ 8 & 7 & 10 \end{array}] - [\begin{array}{ccc} 4 & 4 & 12 \\ 16 & 4 & 8 \\ 8 & 12 & 16 \end{array}] + [\begin{array}{ccc} 5 & 0 & 5 \\ 10 & 5 & 0 \\ 0 & 5 & 10 \end{array}] - [\begin{array}{ccc} 4 & 0 & 0 \\ 0 & 4 & 0 \\ 0 & 0 & 4 \end{array}] \\ = O \end{aligned}$
Every position cancels. In position (1, 1), for example, $3 - 4 (1) + 5 (1) - 4 = 0$ ; in position (2, 3), $8 - 4 (2) + 5 (0) - 0 = 0$ .
Short answerMediumNCERT Exercise 3.2, Q17
For $A = [\begin{array}{cc} 4 & 2 \\ - 1 & 1 \end{array}]$ , find numbers $p$ and $q$ such that $A^{2} = p A + q I$ .
Show solution
Answer $p = 5$ , $q = - 6$ .
$A^{2} = [\begin{array}{cc} 14 & 10 \\ - 5 & - 1 \end{array}]$ and $p A + q I = [\begin{array}{cc} 4 p + q & 2 p \\ - p & p + q \end{array}]$ .
Position (1, 2): $2 p = 10$ , so $p = 5$ . Position (1, 1): $20 + q = 14$ , so $q = - 6$ . Check (2, 2): $p + q = - 1$ , which matches.
18
ProofHardNCERT Exercise 3.2, Q18
Let $t = \tan β$ (with $\cos β \neq 0$ ) and $A = [\begin{array}{cc} 0 & - t \\ t & 0 \end{array}]$ .
1. Show that $(I + A) (I - A) = (I - A) (I + A) = \sec^{2} β I$ .
  Show solution
  $A^{2} = [\begin{array}{cc} - t^{2} & 0 \\ 0 & - t^{2} \end{array}] = - t^{2} I$ . Since $A$ commutes with $I$ , both products equal $I - A^{2} = (1 + t^{2}) I = \sec^{2} β I$ .
2. Show that $\cos^{2} β (I + A)^{2} = [\begin{array}{cc} \cos 2 β & - \sin 2 β \\ \sin 2 β & \cos 2 β \end{array}]$ .
  Show solution
  $(I + A)^{2} = I + 2 A + A^{2} = (1 - t^{2}) I + 2 A = [\begin{array}{cc} 1 - t^{2} & - 2 t \\ 2 t & 1 - t^{2} \end{array}]$ .
  Multiply by $\cos^{2} β$ : $\cos^{2} β (1 - t^{2}) = \cos^{2} β - \sin^{2} β = \cos 2 β$ and $2 t \cos^{2} β = 2 \sin β \cos β = \sin 2 β$ , which gives the rotation matrix through $2 β$ .
3. Check part (b) numerically for $β = \frac{π}{4}$ .
  Show solution
  AnswerBoth sides equal $[\begin{array}{cc} 0 & - 1 \\ 1 & 0 \end{array}]$ .
  Here $t = 1$ and $\cos^{2} β = \frac{1}{2}$ , so $(I + A)^{2} = {[\begin{array}{cc} 1 & - 1 \\ 1 & 1 \end{array}]}^{2} = [\begin{array}{cc} 0 & - 2 \\ 2 & 0 \end{array}]$ and half of it is $[\begin{array}{cc} 0 & - 1 \\ 1 & 0 \end{array}]$ , the rotation through $\frac{π}{2}$ .
19
NumericalMediumNCERT Exercise 3.2, Q19
A cooperative society has ₹50,000 to deposit in two schemes paying 6 % and 8 % simple interest a year. Writing the amounts as the row $[x y]$ and the rates as the column $[\begin{array}{c} 0.06 \\ 0.08 \end{array}]$ , find how to split the money to earn a yearly interest of (a) ₹3400, (b) ₹3700.
1. Interest ₹3400: amount at 8 % (in ₹)?
  ₹
  Show answer
  Answer 20000 ₹
  $0.06 x + 0.08 y = 3400$ with $x = 50000 - y$ gives $3000 + 0.02 y = 3400$ , so $y = 20000$ at 8 % and $x = 30000$ at 6 %.
2. Interest ₹3700: amount at 8 % (in ₹)?
  ₹
  Show answer
  Answer 35000 ₹
  $3000 + 0.02 y = 3700$ gives $y = 35000$ at 8 % and $x = 15000$ at 6 %.
NumericalEasyNCERT Exercise 3.2, Q20
Before a new term, a stationery shop sells its whole stock: 7 packs of 10 geometry boxes at ₹120 per box, 12 packs of 10 notebooks at ₹45 per notebook and 5 packs of 10 water bottles at ₹150 per bottle. Use a row matrix of quantities and a column matrix of prices to find its total takings (in ₹).
₹
Show answer
Answer 21300 ₹
$[70 120 50] [\begin{array}{c} 120 \\ 45 \\ 150 \end{array}] = [8400 + 5400 + 7500] = [21300]$ , so ₹21,300.
Quick answerMediumNCERT Exercise 3.2, Q21
$A$ is $2 \times 3$ , $B$ is $3 \times n$ and $C$ is $m \times 2$ . The product $C A B$ is defined and square exactly when
$m = 2$ and $n = 3$ $m = n$ (any value) $m = 3$ and $n = 2$ never
Show answer
Answer (B) $m = n$ (any value)
$C A$ is $(m \times 2) (2 \times 3) = m \times 3$ , and $(C A) B$ is $(m \times 3) (3 \times n) = m \times n$ . Both products exist for every $m$ and $n$ ; the result is square when $m = n$ .
Quick answerMediumNCERT Exercise 3.2, Q22
With $A$ of order $2 \times 3$ and $B$ of order $3 \times 2$ , which expression is defined?
$A B + I_{2}$ $A + B$ $B A + I_{2}$ $A B + B A$
Show answer
Answer (A) $A B + I_{2}$
$A B$ is $2 \times 2$ , so $A B + I_{2}$ is defined. $B A$ is $3 \times 3$ , so it cannot be added to $I_{2}$ or to $A B$ , and $A + B$ adds a $2 \times 3$ to a $3 \times 2$ matrix.

3.5 Transpose of a Matrix

In Section 3.2 the canteen record could be written with the days as rows or with the days as columns. Switching from one arrangement to the other is an operation in its own right, and it leads to two special kinds of square matrix.

Definition

Transpose

The transpose of an $m \times n$ matrix $A = [a_{i j}]$ is the $n \times m$ matrix $A^{'} = [a_{j i}]$ , also written $A^{T}$ . Its rows are the columns of $A$ , taken in the same order.

Example ${[\begin{array}{ccc} 4 & - 1 & 7 \\ 0 & 5 & 2 \end{array}]}^{'} = [\begin{array}{cc} 4 & 0 \\ - 1 & 5 \\ 7 & 2 \end{array}]$

3/3Row 2 becomes column 2: $A^{'}$ is $3 \times 2$ .

Fig. 3.10Transposing turns row

i

into column

i

: the

(i, j)

entry moves to position

(j, i)

.NCERT Definition 3 · p. 61

On the left is the matrix A with row 1 reading 4, minus 1, 7 and row 2 reading 0, 5, 2, the rows in two different colours. On the right is an empty bracket for A prime. Step 1 shows only A. In step 2 the three entries of row 1 slide across and stand as the first column of A prime, top to bottom. In step 3 the entries of row 2 slide across to form the second column, completing the 3 by 2 transpose with rows 4, 0; minus 1, 5; 7, 2. Faint copies stay behind in A.

In symbols, the $(i, j)$ entry of $A^{'}$ is the $(j, i)$ entry of $A$ . A row matrix transposes into a column matrix and the other way round, which is why a point can be written either way. For a square matrix the transpose has the same order, and its diagonal stays exactly where it was.

3.5.1 Properties of transpose of the matrices

PropertyRules for transposes

Let $A$ and $B$ have orders for which each expression below makes sense, and let $k$ be a scalar. Transposing passes through sums, $(A + B)^{'} = A^{'} + B^{'}$ , and through scalar multiples, $(k A)^{'} = k A^{'}$ . Done twice it changes nothing, $(A^{'})^{'} = A$ . On a product it reverses the order: $(A B)^{'} = B^{'} A^{'}$ .

For the first three rules compare entries: adding or scaling entries does not depend on where they sit, and transposing twice returns every entry to its own place.

For the product rule let $A$ be $m \times n$ and $B$ be $n \times p$ , so that $(A B)^{'}$ and $B^{'} A^{'}$ are both $p \times m$ . The $(k, i)$ entry of $(A B)^{'}$ is the $(i, k)$ entry of $A B$ , which is $\sum_{j} a_{i j} b_{j k}$ . The $(k, i)$ entry of $B^{'} A^{'}$ is $\sum_{j} (B^{'})_{k j} (A^{'})_{j i} = \sum_{j} b_{j k} a_{i j}$ , the same sum. Hence $(A B)^{'} = B^{'} A^{'}$ .

The product rule reverses the order, and it has to: the orders in $A^{'} B^{'}$ are $(n \times m) (p \times n)$ , which usually do not even fit together. The rules extend to longer products, for example $(A B C)^{'} = C^{'} B^{'} A^{'}$ .

Example 20NCERT Example 20 · p. 61

Checking three transpose rules

For $A = [\begin{array}{ccc} 1 & - 2 & 0 \\ 3 & 5 & - 4 \end{array}]$ and $B = [\begin{array}{ccc} 2 & 1 & 6 \\ - 1 & 0 & 3 \end{array}]$ , check three rules: the transpose of $A + B$ is $A^{'} + B^{'}$ ; the transpose of $- 3 A$ is $- 3 A^{'}$ ; and transposing $A$ twice gives back $A$ .

Try it yourself first. Then reveal the solution one step at a time.

Example 21NCERT Example 21 · p. 63

The reversal rule in action

For $A = [\begin{array}{cc} 2 & - 1 \\ 0 & 3 \end{array}]$ and $B = [\begin{array}{ccc} 1 & 4 & 0 \\ - 2 & 1 & 5 \end{array}]$ , verify that $(A B)^{'} = B^{'} A^{'}$ . Is $A^{'} B^{'}$ defined?

Try it yourself first. Then reveal the solution one step at a time.

3.6 Symmetric and Skew Symmetric Matrices

Picture a mirror laid along the main diagonal of a square matrix. The transpose is exactly the mirror image: $a_{i j}$ swaps places with $a_{j i}$ , and the diagonal stays put. Two kinds of square matrix respond to this mirror in a special way (Fig. 3.11).

Definition

Symmetric matrix

A square matrix $A$ with $A^{'} = A$ , that is, $a_{j i} = a_{i j}$ for all $i$ and $j$ .

Example $[\begin{array}{ccc} 2 & 5 & - 1 \\ 5 & 0 & 3 \\ - 1 & 3 & 4 \end{array}]$

Definition

Skew symmetric matrix

A square matrix $A$ with $A^{'} = - A$ , that is, $a_{j i} = - a_{i j}$ for all $i$ and $j$ .

Example $[\begin{array}{ccc} 0 & 5 & - 1 \\ - 5 & 0 & 3 \\ 1 & - 3 & 0 \end{array}]$

3/3Partners opposite: $a_{j i} = - a_{i j}$ , and every diagonal entry is 0.

Fig. 3.11Symmetric: mirror partners are equal. Skew symmetric: partners are negatives, so the diagonal must be zero.NCERT Definitions 4 and 5 · p. 63

A 3 by 3 matrix with a dashed mirror line from the top-left to the bottom-right corner. Step 1 shows a general matrix 3, 8, minus 2; 1, 6, 5; 4, 0, 7 with curved arrows joining mirror partners such as 8 and 1. Step 2 shows a symmetric matrix 2, 5, minus 1; 5, 0, 3; minus 1, 3, 4 where each pair of partners is equal and shaded alike. Step 3 shows a skew symmetric matrix 0, 5, minus 1; minus 5, 0, 3; 1, minus 3, 0 where partners are negatives of each other and the diagonal is all zeros.

Look at a diagonal position, where the row and column numbers agree. There the skew condition demands $a_{i i} = - a_{i i}$ , and the only number equal to its own negative is 0, so a skew symmetric matrix always has a zero diagonal. Off the diagonal, mirror partners are negatives of each other. A matrix that is both symmetric and skew symmetric has $A = A^{'} = - A$ , so $2 A = O$ and $A$ must be the zero matrix.

Symmetric matrices appear wherever a relationship works the same both ways. A table of road distances between cities is symmetric, because the distance from Pune to Nashik equals the distance from Nashik to Pune; so is a table recording which pairs of students share a class.

Theorem 1

For every square matrix $A$ with real entries, $A + A^{'}$ is symmetric and $A - A^{'}$ is skew symmetric.

Theorem 2

Every square matrix can be written as the sum of a symmetric matrix and a skew symmetric matrix, and in only one way.

Splitting a square matrix

A = \underset{symmetric}{\underset{⏟}{\frac{1}{2} (A + A^{'})}} + \underset{skew symmetric}{\underset{⏟}{\frac{1}{2} (A - A^{'})}}

Example 22NCERT Example 22 · p. 65

Splitting a matrix into two halves

Split $A = [\begin{array}{ccc} 3 & 5 & - 1 \\ 1 & - 2 & 4 \\ 7 & 0 & 6 \end{array}]$ into a symmetric part $P$ and a skew symmetric part $Q$ with $A = P + Q$ .

Try it yourself first. Then reveal the solution one step at a time.

Practice · 12 questions

Practice 3.3: questions with answers

Practice for Sections 3.5 and 3.6 (matches NCERT Exercise 3.3).

1
Short answerEasyNCERT Exercise 3.3, Q1
Write the transpose of each matrix.
1. $[\begin{array}{cccc} - 2 & 4 & \frac{1}{3} & 7 \end{array}]$
  Show solution
  Answer $[\begin{array}{c} - 2 \\ 4 \\ \frac{1}{3} \\ 7 \end{array}]$
  A $1 \times 4$ row becomes the $4 \times 1$ column $[\begin{array}{c} - 2 \\ 4 \\ \frac{1}{3} \\ 7 \end{array}]$ .
2. $[\begin{array}{cc} 4 & - 2 \\ 0 & 7 \end{array}]$
  Show solution
  Answer $[\begin{array}{cc} 4 & 0 \\ - 2 & 7 \end{array}]$
  The diagonal 4, 7 stays; the off-diagonal entries swap: $[\begin{array}{cc} 4 & 0 \\ - 2 & 7 \end{array}]$ .
3. $[\begin{array}{cc} 1 & - 4 \\ 0 & 6 \\ - 5 & 2 \end{array}]$
  Show solution
  Answer $[\begin{array}{ccc} 1 & 0 & - 5 \\ - 4 & 6 & 2 \end{array}]$
  A $3 \times 2$ matrix gives the $2 \times 3$ matrix $[\begin{array}{ccc} 1 & 0 & - 5 \\ - 4 & 6 & 2 \end{array}]$ .
Short answerMediumNCERT Exercise 3.3, Q2
For $A = [\begin{array}{ccc} 1 & 0 & - 3 \\ 2 & 4 & 1 \\ - 1 & 5 & 2 \end{array}]$ and $B = [\begin{array}{ccc} 3 & - 2 & 0 \\ 1 & 1 & 4 \\ 0 & - 3 & 5 \end{array}]$ , verify that (a) $(A + B)^{'} = A^{'} + B^{'}$ and (b) $(A - B)^{'} = A^{'} - B^{'}$ .
Show solution
$\begin{array}{c} A + B = [\begin{array}{ccc} 4 & - 2 & - 3 \\ 3 & 5 & 5 \\ - 1 & 2 & 7 \end{array}] \\ \begin{aligned} (A + B)^{'} & = [\begin{array}{ccc} 4 & 3 & - 1 \\ - 2 & 5 & 2 \\ - 3 & 5 & 7 \end{array}] \\ = A^{'} + B^{'} \end{aligned} \end{array}$
$\begin{array}{c} A - B = [\begin{array}{ccc} - 2 & 2 & - 3 \\ 1 & 3 & - 3 \\ - 1 & 8 & - 3 \end{array}] \\ \begin{aligned} (A - B)^{'} & = [\begin{array}{ccc} - 2 & 1 & - 1 \\ 2 & 3 & 8 \\ - 3 & - 3 & - 3 \end{array}] \\ = A^{'} - B^{'} \end{aligned} \end{array}$
Here $A^{'} = [\begin{array}{ccc} 1 & 2 & - 1 \\ 0 & 4 & 5 \\ - 3 & 1 & 2 \end{array}]$ and $B^{'} = [\begin{array}{ccc} 3 & 1 & 0 \\ - 2 & 1 & - 3 \\ 0 & 4 & 5 \end{array}]$ ; adding and subtracting them gives the same two matrices.
Short answerMediumNCERT Exercise 3.3, Q3
If $A^{'} = [\begin{array}{cc} 2 & 0 \\ - 1 & 3 \\ 4 & 1 \end{array}]$ and $B = [\begin{array}{ccc} 1 & 5 & - 2 \\ 0 & - 3 & 6 \end{array}]$ , verify that (a) $(A + B)^{'} = A^{'} + B^{'}$ and (b) $(A - B)^{'} = A^{'} - B^{'}$ .
Show solution
$A = (A^{'})^{'} = [\begin{array}{ccc} 2 & - 1 & 4 \\ 0 & 3 & 1 \end{array}]$ and $B^{'} = [\begin{array}{cc} 1 & 0 \\ 5 & - 3 \\ - 2 & 6 \end{array}]$ .
$\begin{array}{c} \begin{aligned} (A + B)^{'} & = {[\begin{array}{ccc} 3 & 4 & 2 \\ 0 & 0 & 7 \end{array}]}^{'} \\ = [\begin{array}{cc} 3 & 0 \\ 4 & 0 \\ 2 & 7 \end{array}] \\ = A^{'} + B^{'} \end{aligned} \\ \begin{aligned} (A - B)^{'} & = [\begin{array}{cc} 1 & 0 \\ - 6 & 6 \\ 6 & - 5 \end{array}] \\ = A^{'} - B^{'} \end{aligned} \end{array}$
Short answerEasyNCERT Exercise 3.3, Q4
If $A^{'} = [\begin{array}{cc} 1 & - 2 \\ 3 & 0 \end{array}]$ and $B = [\begin{array}{cc} 2 & 1 \\ - 1 & 4 \end{array}]$ , find $(2 A - B)^{'}$ .
Show solution
Answer $[\begin{array}{cc} 0 & - 3 \\ 5 & - 4 \end{array}]$
$(2 A - B)^{'} = 2 A^{'} - B^{'} = [\begin{array}{cc} 2 & - 4 \\ 6 & 0 \end{array}] - [\begin{array}{cc} 2 & - 1 \\ 1 & 4 \end{array}] = [\begin{array}{cc} 0 & - 3 \\ 5 & - 4 \end{array}]$ .
5
Short answerMediumNCERT Exercise 3.3, Q5
Verify that $(A B)^{'} = B^{'} A^{'}$ in each case.
1. $A = [\begin{array}{c} 3 \\ 0 \\ - 2 \end{array}]$ , $B = [\begin{array}{ccc} 4 & - 1 & 5 \end{array}]$
  Show solution
  $\begin{array}{c} A B = [\begin{array}{ccc} 12 & - 3 & 15 \\ 0 & 0 & 0 \\ - 8 & 2 & - 10 \end{array}] \\ \begin{aligned} (A B)^{'} & = [\begin{array}{ccc} 12 & 0 & - 8 \\ - 3 & 0 & 2 \\ 15 & 0 & - 10 \end{array}] \\ = [\begin{array}{c} 4 \\ - 1 \\ 5 \end{array}] [\begin{array}{ccc} 3 & 0 & - 2 \end{array}] \\ = B^{'} A^{'} \end{aligned} \end{array}$
2. $A = [\begin{array}{cc} 1 & 2 \\ 0 & - 1 \end{array}]$ , $B = [\begin{array}{ccc} 3 & 1 & 0 \\ 2 & - 2 & 4 \end{array}]$
  Show solution
  $\begin{array}{c} A B = [\begin{array}{ccc} 7 & - 3 & 8 \\ - 2 & 2 & - 4 \end{array}] \\ (A B)^{'} = [\begin{array}{cc} 7 & - 2 \\ - 3 & 2 \\ 8 & - 4 \end{array}] \end{array}$
  $\begin{aligned} B^{'} A^{'} & = [\begin{array}{cc} 3 & 2 \\ 1 & - 2 \\ 0 & 4 \end{array}] [\begin{array}{cc} 1 & 0 \\ 2 & - 1 \end{array}] \\ = [\begin{array}{cc} 7 & - 2 \\ - 3 & 2 \\ 8 & - 4 \end{array}] \end{aligned}$
6
ProofMediumNCERT Exercise 3.3, Q6
Verify that $A^{'} A = I$ for each matrix.
1. $A = [\begin{array}{cc} \cos φ & \sin φ \\ \sin φ & - \cos φ \end{array}]$ (a reflection)
  Show solution
  $A$ is symmetric, so $A^{'} = A$ and $A^{'} A = A^{2}$ :
  $\begin{aligned} A^{2} & = [\begin{array}{cc} \cos^{2} φ + \sin^{2} φ & \cos φ \sin φ - \sin φ \cos φ \\ \sin φ \cos φ - \cos φ \sin φ & \sin^{2} φ + \cos^{2} φ \end{array}] \\ = I \end{aligned}$
2. $A = [\begin{array}{cc} \frac{3}{5} & - \frac{4}{5} \\ \frac{4}{5} & \frac{3}{5} \end{array}]$
  Show solution
  $A^{'} A = [\begin{array}{cc} \frac{3}{5} & \frac{4}{5} \\ - \frac{4}{5} & \frac{3}{5} \end{array}] [\begin{array}{cc} \frac{3}{5} & - \frac{4}{5} \\ \frac{4}{5} & \frac{3}{5} \end{array}] = [\begin{array}{cc} \frac{9 + 16}{25} & \frac{- 12 + 12}{25} \\ \frac{- 12 + 12}{25} & \frac{16 + 9}{25} \end{array}] = I$ .
7
Short answerEasyNCERT Exercise 3.3, Q7
Classify each matrix.
1. Show that $[\begin{array}{ccc} 2 & - 4 & 7 \\ - 4 & 0 & 1 \\ 7 & 1 & - 3 \end{array}]$ is symmetric.
  Show solution
  Partners agree: $a_{12} = a_{21} = - 4$ , $a_{13} = a_{31} = 7$ , $a_{23} = a_{32} = 1$ . So $A^{'} = A$ .
2. Show that $[\begin{array}{ccc} 0 & 2 & - 5 \\ - 2 & 0 & 3 \\ 5 & - 3 & 0 \end{array}]$ is skew symmetric.
  Show solution
  The diagonal is all 0, and partners are opposite: $2$ and $- 2$ , $- 5$ and $5$ , $3$ and $- 3$ . So $A^{'} = - A$ .
Short answerEasyNCERT Exercise 3.3, Q8
For $A = [\begin{array}{cc} 2 & - 3 \\ 4 & 1 \end{array}]$ , verify that (a) $A + A^{'}$ is symmetric and (b) $A - A^{'}$ is skew symmetric.
Show solution
$A^{'} = [\begin{array}{cc} 2 & 4 \\ - 3 & 1 \end{array}]$ , so $A + A^{'} = [\begin{array}{cc} 4 & 1 \\ 1 & 2 \end{array}]$ , which equals its transpose (partners 1 and 1).
$A - A^{'} = [\begin{array}{cc} 0 & - 7 \\ 7 & 0 \end{array}]$ : zero diagonal and partners $- 7$ and $7$ , so it is skew symmetric.
Short answerMediumNCERT Exercise 3.3, Q9
Find $\frac{1}{2} (A + A^{'})$ and $\frac{1}{2} (A - A^{'})$ for the general $2 \times 2$ matrix $A = [\begin{array}{cc} a & b \\ c & d \end{array}]$ . When is the second one $O$ ?
Show solution
Answer $[\begin{array}{cc} a & \frac{b + c}{2} \\ \frac{b + c}{2} & d \end{array}]$ and $[\begin{array}{cc} 0 & \frac{b - c}{2} \\ \frac{c - b}{2} & 0 \end{array}]$ ; the second is O exactly when $b = c$ .
Averaging $A$ and $A^{'}$ keeps the diagonal and replaces each off-diagonal pair by its mean $\frac{b + c}{2}$ . Half the difference has zero diagonal and entries $\pm \frac{b - c}{2}$ .
The skew part vanishes exactly when $b = c$ , that is, when $A$ is already symmetric.
10
Short answerMediumNCERT Exercise 3.3, Q10
Write each matrix as $P + Q$ , where $P$ is symmetric and $Q$ is skew symmetric.
1. $[\begin{array}{cc} 4 & 1 \\ - 3 & 2 \end{array}]$
  Show solution
  Answer $[\begin{array}{cc} 4 & 1 \\ - 3 & 2 \end{array}] = [\begin{array}{cc} 4 & - 1 \\ - 1 & 2 \end{array}] + [\begin{array}{cc} 0 & 2 \\ - 2 & 0 \end{array}]$
  $\begin{array}{c} \begin{aligned} \frac{1}{2} (A + A^{'}) & = \frac{1}{2} [\begin{array}{cc} 8 & - 2 \\ - 2 & 4 \end{array}] \\ = [\begin{array}{cc} 4 & - 1 \\ - 1 & 2 \end{array}] \end{aligned} \\ \begin{aligned} \frac{1}{2} (A - A^{'}) & = \frac{1}{2} [\begin{array}{cc} 0 & 4 \\ - 4 & 0 \end{array}] \\ = [\begin{array}{cc} 0 & 2 \\ - 2 & 0 \end{array}] \end{aligned} \end{array}$
  The first part equals its transpose, the second is the negative of its transpose, and they add to $A$ .
2. $[\begin{array}{ccc} 1 & 4 & - 2 \\ 0 & 3 & 5 \\ 6 & - 1 & 2 \end{array}]$
  Show solution
  Answer $[\begin{array}{ccc} 1 & 4 & - 2 \\ 0 & 3 & 5 \\ 6 & - 1 & 2 \end{array}] = [\begin{array}{ccc} 1 & 2 & 2 \\ 2 & 3 & 2 \\ 2 & 2 & 2 \end{array}] + [\begin{array}{ccc} 0 & 2 & - 4 \\ - 2 & 0 & 3 \\ 4 & - 3 & 0 \end{array}]$
  $\begin{array}{c} \begin{aligned} \frac{1}{2} (A + A^{'}) & = \frac{1}{2} [\begin{array}{ccc} 2 & 4 & 4 \\ 4 & 6 & 4 \\ 4 & 4 & 4 \end{array}] \\ = [\begin{array}{ccc} 1 & 2 & 2 \\ 2 & 3 & 2 \\ 2 & 2 & 2 \end{array}] \end{aligned} \\ \begin{aligned} \frac{1}{2} (A - A^{'}) & = \frac{1}{2} [\begin{array}{ccc} 0 & 4 & - 8 \\ - 4 & 0 & 6 \\ 8 & - 6 & 0 \end{array}] \\ = [\begin{array}{ccc} 0 & 2 & - 4 \\ - 2 & 0 & 3 \\ 4 & - 3 & 0 \end{array}] \end{aligned} \end{array}$
  The first part equals its transpose, the second is the negative of its transpose, and they add to $A$ .
3. $[\begin{array}{ccc} 2 & - 1 & 3 \\ 5 & 0 & - 4 \\ 1 & 2 & - 3 \end{array}]$
  Show solution
  Answer $[\begin{array}{ccc} 2 & - 1 & 3 \\ 5 & 0 & - 4 \\ 1 & 2 & - 3 \end{array}] = [\begin{array}{ccc} 2 & 2 & 2 \\ 2 & 0 & - 1 \\ 2 & - 1 & - 3 \end{array}] + [\begin{array}{ccc} 0 & - 3 & 1 \\ 3 & 0 & - 3 \\ - 1 & 3 & 0 \end{array}]$
  $\begin{array}{c} \begin{aligned} \frac{1}{2} (A + A^{'}) & = \frac{1}{2} [\begin{array}{ccc} 4 & 4 & 4 \\ 4 & 0 & - 2 \\ 4 & - 2 & - 6 \end{array}] \\ = [\begin{array}{ccc} 2 & 2 & 2 \\ 2 & 0 & - 1 \\ 2 & - 1 & - 3 \end{array}] \end{aligned} \\ \begin{aligned} \frac{1}{2} (A - A^{'}) & = \frac{1}{2} [\begin{array}{ccc} 0 & - 6 & 2 \\ 6 & 0 & - 6 \\ - 2 & 6 & 0 \end{array}] \\ = [\begin{array}{ccc} 0 & - 3 & 1 \\ 3 & 0 & - 3 \\ - 1 & 3 & 0 \end{array}] \end{aligned} \end{array}$
  The first part equals its transpose, the second is the negative of its transpose, and they add to $A$ .
4. $[\begin{array}{cc} 3 & - 2 \\ 6 & 1 \end{array}]$
  Show solution
  Answer $[\begin{array}{cc} 3 & - 2 \\ 6 & 1 \end{array}] = [\begin{array}{cc} 3 & 2 \\ 2 & 1 \end{array}] + [\begin{array}{cc} 0 & - 4 \\ 4 & 0 \end{array}]$
  $\begin{array}{c} \begin{aligned} \frac{1}{2} (A + A^{'}) & = \frac{1}{2} [\begin{array}{cc} 6 & 4 \\ 4 & 2 \end{array}] \\ = [\begin{array}{cc} 3 & 2 \\ 2 & 1 \end{array}] \end{aligned} \\ \begin{aligned} \frac{1}{2} (A - A^{'}) & = \frac{1}{2} [\begin{array}{cc} 0 & - 8 \\ 8 & 0 \end{array}] \\ = [\begin{array}{cc} 0 & - 4 \\ 4 & 0 \end{array}] \end{aligned} \end{array}$
  The first part equals its transpose, the second is the negative of its transpose, and they add to $A$ .
Quick answerMediumNCERT Exercise 3.3, Q11
If $A$ is symmetric and $B$ is skew symmetric, of the same order, then $A B - B A$ is
symmetricskew symmetricthe zero matrixthe identity matrix
Show answer
Answer (A) symmetric
$(A B - B A)^{'} = B^{'} A^{'} - A^{'} B^{'} = (- B) A - A (- B) = A B - B A$ , so it equals its own transpose: symmetric.
Quick answerMediumNCERT Exercise 3.3, Q12
If $A = [\begin{array}{cc} \cos α & - \sin α \\ \sin α & \cos α \end{array}]$ and $A + A^{'} = \sqrt{2} I$ , then a value of $α$ is
$\frac{π}{6}$ $\frac{π}{3}$ $\frac{π}{2}$ $\frac{π}{4}$
Show answer
Answer (D) $\frac{π}{4}$
$A + A^{'} = [\begin{array}{cc} 2 \cos α & 0 \\ 0 & 2 \cos α \end{array}] = \sqrt{2} I$ gives $\cos α = \frac{\sqrt{2}}{2}$ , so $α = \frac{π}{4}$ .

3.7 Invertible Matrices

Every non-zero number $a$ has a reciprocal $a^{- 1}$ with $a \cdot a^{- 1} = a^{- 1} \cdot a = 1$ . The matrix version replaces 1 by the identity $I$ and asks for a matrix that undoes $A$ .

Definition

Invertible matrix

A square matrix $A$ of order $m$ is invertible when some square matrix $B$ of the same order gives $A B = B A = I$ . Such a $B$ is the inverse of $A$ , written $A^{- 1}$ .

Example ${[\begin{array}{cc} 3 & 5 \\ 1 & 2 \end{array}]}^{- 1} = [\begin{array}{cc} 2 & - 5 \\ - 1 & 3 \end{array}]$

ExampleNCERT example after Definition 6 · p. 68

Checking a pair of inverses

Show that $A = [\begin{array}{cc} 3 & 5 \\ 1 & 2 \end{array}]$ and $B = [\begin{array}{cc} 2 & - 5 \\ - 1 & 3 \end{array}]$ are inverses of each other.

Try it yourself first. Then reveal the solution one step at a time.

Fig. 3.12 shows the idea geometrically: one matrix shears a square into a parallelogram, and its inverse shears it straight back.

2/3 $A$ slides each point right by its height: $P \to (3, 2)$ .

Fig. 3.12

A^{- 1}

undoes

A

: applying

A

and then

A^{- 1}

leaves every point where it started.NCERT idea of Definition 6 · p. 68

A grid with a 2 by 2 square drawn from the origin, with a small coloured corner to show orientation and a marked point P at (1, 2) on its top edge. Step 1 shows the square and P. Step 2 applies A, the shear with rows 1, 1 and 0, 1: each point moves right by its height, so the square becomes a parallelogram leaning right and P moves to (3, 2). Step 3 applies A inverse, with rows 1, minus 1 and 0, 1, which moves each point back left by its height, restoring the square and P at (1, 2).

Two remarks follow directly from the definition. First, only square matrices can have inverses: if $A$ is $m \times n$ with $m \neq n$ , then $A B$ and $B A$ have different orders, so they cannot both equal one identity matrix. Second, the relationship is mutual: the equations $A B = B A = I$ say equally that $A$ undoes $B$ , so $B = A^{- 1}$ goes with $A = B^{- 1}$ .

Being square and non-zero is not enough. The matrix $A = [\begin{array}{cc} 1 & 2 \\ 2 & 4 \end{array}]$ of Example 15 has no inverse: there $A B = O$ with $B \neq O$ , and an inverse $C$ would give $B = I B = (C A) B = C (A B) = C O = O$ , which is false. Chapter 4 gives a test for invertibility (a non-zero determinant) and a formula for the inverse.

Theorem 3Uniqueness of the inverse

A square matrix has at most one inverse.

Theorem 4Inverse of a product

The product of two invertible square matrices of equal order is invertible, and its inverse is the product of the inverses in reverse order: $(A B)^{- 1} = B^{- 1} A^{- 1}$ .

Practice · 3 questions

Practice 3.4: questions with answers

Practice for Section 3.7 (Q1 matches NCERT Exercise 3.4; Q2 and Q3 are extra).

Quick answerEasyNCERT Exercise 3.4, Q1
For square matrices $A$ and $B$ of order 2, which condition guarantees that $B = A^{- 1}$ ?
$A B = O$ $A B = B A$ $A + B = I$ $A B = B A = I$
Show answer
Answer (D) $A B = B A = I$
The definition needs both products to equal $I$ . $A B = B A$ alone holds, for example, when $B = A$ .
Short answerEasy
Check whether $[\begin{array}{cc} 2 & 1 \\ 5 & 3 \end{array}]$ and $[\begin{array}{cc} 3 & - 1 \\ - 5 & 2 \end{array}]$ are inverses of each other.
Show solution
AnswerYes: both products equal $I$ .
$[\begin{array}{cc} 2 & 1 \\ 5 & 3 \end{array}] [\begin{array}{cc} 3 & - 1 \\ - 5 & 2 \end{array}] = [\begin{array}{cc} 6 - 5 & - 2 + 2 \\ 15 - 15 & - 5 + 6 \end{array}] = I$ , and the product in the other order is also $I$ .
Short answerMedium
Let $A = [\begin{array}{cc} 1 & 2 \\ 0 & 1 \end{array}]$ and $B = [\begin{array}{cc} 1 & 0 \\ 3 & 1 \end{array}]$ , with $A^{- 1} = [\begin{array}{cc} 1 & - 2 \\ 0 & 1 \end{array}]$ and $B^{- 1} = [\begin{array}{cc} 1 & 0 \\ - 3 & 1 \end{array}]$ . Find $B^{- 1} A^{- 1}$ and check that it is the inverse of $A B$ . Is $A^{- 1} B^{- 1}$ also the inverse of $A B$ ?
Show solution
Answer $B^{- 1} A^{- 1} = [\begin{array}{cc} 1 & - 2 \\ - 3 & 7 \end{array}]$ is the inverse of $A B = [\begin{array}{cc} 7 & 2 \\ 3 & 1 \end{array}]$ ; $A^{- 1} B^{- 1} = [\begin{array}{cc} 7 & - 2 \\ - 3 & 1 \end{array}]$ is not.
$\begin{array}{c} A B = [\begin{array}{cc} 7 & 2 \\ 3 & 1 \end{array}] \\ B^{- 1} A^{- 1} = [\begin{array}{cc} 1 & - 2 \\ - 3 & 7 \end{array}] \end{array}$
$(A B) (B^{- 1} A^{- 1}) = [\begin{array}{cc} 1 & 0 \\ 0 & 1 \end{array}]$ and $(B^{- 1} A^{- 1}) (A B) = [\begin{array}{cc} 1 & 0 \\ 0 & 1 \end{array}]$ , as Theorem 4 predicts.
But $A^{- 1} B^{- 1} = [\begin{array}{cc} 7 & - 2 \\ - 3 & 1 \end{array}]$ , which differs from $B^{- 1} A^{- 1}$ , so by uniqueness (Theorem 3) it is not the inverse of $A B$ .

Miscellaneous examples

The last three examples combine several ideas from the chapter: powers of a matrix with mathematical induction, transposes with products, and matrix equations with an unknown matrix.

Example 23NCERT Example 23 · p. 70

Powers of a rotation

For a real number $θ$ let $R (θ) = [\begin{array}{cc} \cos θ & - \sin θ \\ \sin θ & \cos θ \end{array}]$ . Prove that $R (θ)^{n} = R (n θ)$ for every natural number $n$ .

Try it yourself first. Then reveal the solution one step at a time.

Geometrically, $R (θ)$ turns every point of the plane through the angle $θ$ anticlockwise about the origin, so applying it $n$ times turns through $n θ$ (Fig. 3.13). For $θ = 30 °$ , twelve turns make a full circle: $R (30 °)^{12} = R (360 °) = I$ .

4/4 $R (θ)^{3} = R (3 θ)$ : a turn of $90 °$ .

Fig. 3.13Each multiplication by

R (θ)

adds

θ

to the angle, so

R (θ)^{n} = R (n θ)

.NCERT idea of Example 23 · p. 70

A circle centred at the origin with axes. Step 1 shows an arrow along the positive x-axis. Step 2 shows it turned anticlockwise through 30 degrees by R of theta, with a faint copy of the start. Step 3 shows it turned through 60 degrees, labelled R of theta squared equals R of 2 theta. Step 4 shows it at 90 degrees, pointing straight up, labelled R of theta cubed equals R of 3 theta. Small arcs mark each 30 degree turn.

Example 24NCERT Example 24 · p. 70

When is a product of symmetric matrices symmetric?

Let $A$ and $B$ be symmetric matrices of the same order. Prove that $A B$ is symmetric if and only if $A B = B A$ . Illustrate with $A = [\begin{array}{cc} 1 & 2 \\ 2 & 0 \end{array}]$ , first with $B = [\begin{array}{cc} 0 & 1 \\ 1 & 3 \end{array}]$ and then with $C = A + I$ .

Try it yourself first. Then reveal the solution one step at a time.

Example 25NCERT Example 25 · p. 71

Finding an unknown matrix

Find the matrix $X$ such that $[\begin{array}{cc} 2 & 1 \\ 5 & 3 \end{array}] X - [\begin{array}{cc} 1 & 0 \\ 2 & 1 \end{array}] [\begin{array}{cc} 3 & 1 \\ 1 & 2 \end{array}] = O$ .

Try it yourself first. Then reveal the solution one step at a time.

Quick check

1 of 2

Using

R (θ)^{n} = R (n θ)

, what is the

(1, 1)

entry of

R (30 °)^{6}

Show answer

Answer -1

R (30 °)^{6} = R (180 °)

, whose

(1, 1)

entry is

\cos 180 ° = - 1

. (Six turns of

30 °

make half a turn.)

Recap

Matrices: key points and summary

A matrix sets out numbers or functions in a rectangle; with $m$ rows and $n$ columns its order is $m \times n$ and it holds $m n$ entries $a_{i j}$ (Section 3.2.1).
Row, column, square, diagonal, scalar, identity and zero matrices are named by their shape or pattern; identity ⊂ scalar ⊂ diagonal ⊂ square (Section 3.3).
Two matrices are equal when they have the same order and equal corresponding entries (Section 3.3.1).
Sums and differences exist only for matrices of the same order and work entry by entry; $k A$ multiplies every entry by $k$ (Section 3.4.1, Section 3.4.2).
Addition is commutative and associative, with identity O and inverse $- A$ ; scalars distribute over sums (Section 3.4.3, Section 3.4.4).
$A B$ exists when the columns of $A$ equal the rows of $B$ : $(m \times n) (n \times p)$ gives $m \times p$ , with $c_{i k} = \sum_{j} a_{i j} b_{j k}$ (Section 3.4.5).
In general $A B \neq B A$ , and $A B = O$ is possible with $A \neq O$ , $B \neq O$ (Order matters: AB and BA, A zero product from non-zero matrices).
Multiplication is associative and distributive over addition, and $I A = A I = A$ (Section 3.4.6).
The transpose $A^{'} = [a_{j i}]$ satisfies $(A^{'})^{'} = A$ , $(k A)^{'} = k A^{'}$ , $(A + B)^{'} = A^{'} + B^{'}$ and $(A B)^{'} = B^{'} A^{'}$ (Section 3.5.1).
$A$ is symmetric if $A^{'} = A$ and skew symmetric if $A^{'} = - A$ , in which case its diagonal is zero (Section 3.6).
Every square matrix splits uniquely as $\frac{1}{2} (A + A^{'}) + \frac{1}{2} (A - A^{'})$ , symmetric plus skew symmetric (Theorem 2).
$B = A^{- 1}$ when $A B = B A = I$ ; only square matrices can have inverses, an inverse is unique, and $(A B)^{- 1} = B^{- 1} A^{- 1}$ (Section 3.7).
Powers of a square matrix can be found by induction, as in $R (θ)^{n} = R (n θ)$ (Example 23).

Flash cards · 10

Rules to remember

1 / 10

Miscellaneous exercises

Find your NCERT exercise question

Our practice questions are written for ncert.ai and follow the NCERT exercises. Find the one that matches the question in your textbook; its answer is with it.

In your NCERT textbook	Practice question here
NCERT Exercise 3.1, Q1	Practice 3.1 · Q1
NCERT Exercise 3.1, Q2	Practice 3.1 · Q2
NCERT Exercise 3.1, Q3	Practice 3.1 · Q3
NCERT Exercise 3.1, Q4	Practice 3.1 · Q4
NCERT Exercise 3.1, Q5	Practice 3.1 · Q5
NCERT Exercise 3.1, Q6	Practice 3.1 · Q6
NCERT Exercise 3.1, Q7	Practice 3.1 · Q7
NCERT Exercise 3.1, Q8	Practice 3.1 · Q8
NCERT Exercise 3.1, Q9	Practice 3.1 · Q9
NCERT Exercise 3.1, Q10	Practice 3.1 · Q10
NCERT Exercise 3.2, Q1	Practice 3.2 · Q1
NCERT Exercise 3.2, Q2	Practice 3.2 · Q2
NCERT Exercise 3.2, Q3	Practice 3.2 · Q3
NCERT Exercise 3.2, Q4	Practice 3.2 · Q4
NCERT Exercise 3.2, Q5	Practice 3.2 · Q5
NCERT Exercise 3.2, Q6	Practice 3.2 · Q6
NCERT Exercise 3.2, Q7	Practice 3.2 · Q7
NCERT Exercise 3.2, Q8	Practice 3.2 · Q8
NCERT Exercise 3.2, Q9	Practice 3.2 · Q9
NCERT Exercise 3.2, Q10	Practice 3.2 · Q10
NCERT Exercise 3.2, Q11	Practice 3.2 · Q11
NCERT Exercise 3.2, Q12	Practice 3.2 · Q12
NCERT Exercise 3.2, Q13	Practice 3.2 · Q13
NCERT Exercise 3.2, Q14	Practice 3.2 · Q14
NCERT Exercise 3.2, Q15	Practice 3.2 · Q15
NCERT Exercise 3.2, Q16	Practice 3.2 · Q16
NCERT Exercise 3.2, Q17	Practice 3.2 · Q17
NCERT Exercise 3.2, Q18	Practice 3.2 · Q18
NCERT Exercise 3.2, Q19	Practice 3.2 · Q19
NCERT Exercise 3.2, Q20	Practice 3.2 · Q20
NCERT Exercise 3.2, Q21	Practice 3.2 · Q21
NCERT Exercise 3.2, Q22	Practice 3.2 · Q22
NCERT Exercise 3.3, Q1	Practice 3.3 · Q1
NCERT Exercise 3.3, Q2	Practice 3.3 · Q2
NCERT Exercise 3.3, Q3	Practice 3.3 · Q3
NCERT Exercise 3.3, Q4	Practice 3.3 · Q4
NCERT Exercise 3.3, Q5	Practice 3.3 · Q5
NCERT Exercise 3.3, Q6	Practice 3.3 · Q6
NCERT Exercise 3.3, Q7	Practice 3.3 · Q7
NCERT Exercise 3.3, Q8	Practice 3.3 · Q8
NCERT Exercise 3.3, Q9	Practice 3.3 · Q9
NCERT Exercise 3.3, Q10	Practice 3.3 · Q10
NCERT Exercise 3.3, Q11	Practice 3.3 · Q11
NCERT Exercise 3.3, Q12	Practice 3.3 · Q12
NCERT Exercise 3.4, Q1	Practice 3.4 · Q1
NCERT Misc. Exercise 3, Q1	Miscellaneous exercises · Q1
NCERT Misc. Exercise 3, Q2	Miscellaneous exercises · Q2
NCERT Misc. Exercise 3, Q3	Miscellaneous exercises · Q3
NCERT Misc. Exercise 3, Q4	Miscellaneous exercises · Q4
NCERT Misc. Exercise 3, Q5	Miscellaneous exercises · Q5
NCERT Misc. Exercise 3, Q6	Miscellaneous exercises · Q6
NCERT Misc. Exercise 3, Q7	Miscellaneous exercises · Q7
NCERT Misc. Exercise 3, Q8	Miscellaneous exercises · Q8
NCERT Misc. Exercise 3, Q9	Miscellaneous exercises · Q9
NCERT Misc. Exercise 3, Q10	Miscellaneous exercises · Q10
NCERT Misc. Exercise 3, Q11	Miscellaneous exercises · Q11

Quick check: MCQs on Matrices

3.1 Introduction2 questions
3.2.1 Order of a matrix4 questions
3.3.1 Equality of matrices4 questions
3.4.4 Properties of scalar multiplication of a matrix4 questions
3.4.6 Properties of multiplication of matrices4 questions
3.5.1 Properties of transpose of the matrices4 questions
3.6 Symmetric and Skew Symmetric Matrices4 questions
3.7 Invertible Matrices3 questions
Miscellaneous examples2 questions

Glossary · 14

Key terms (English–Hindi)

diagonal matrixविकर्ण आव्यूहvikarn aavyooh: A square matrix whose entries off the main diagonal are all 0.First seen in the chapter →

elementअवयवavayav: One number or function in a matrix; $a_{i j}$ sits in row $i$ and column $j$ .First seen in the chapter →
equal matricesसमान आव्यूहsamaan aavyooh: Matrices of the same order whose corresponding entries are all equal.

identity matrixतत्समक आव्यूहtatsamak aavyooh: The square matrix $I$ with 1 on the diagonal and 0 elsewhere; $A I = I A = A$ .First seen in the chapter →
invertible matrixव्युत्क्रमणीय आव्यूहvyutkramaneey aavyooh: A square matrix $A$ with some $B$ such that $A B = B A = I$ ; then $B = A^{- 1}$ .

matrixआव्यूहaavyooh: A rectangle of numbers or functions, set out in rows and columns and handled as one object.First seen in the chapter →
matrix productआव्यूहों का गुणनफलaavyoohon ka gunanphal: For $A$ of order $m \times n$ and $B$ of order $n \times p$ , the $m \times p$ matrix whose $(i, k)$ entry is row $i$ of $A$ times column $k$ of $B$ .

orderकोटिkoti: The size $m \times n$ of a matrix: $m$ rows and $n$ columns.First seen in the chapter →

scalar matrixअदिश आव्यूहadish aavyooh: A diagonal matrix whose diagonal entries all equal one number $k$ .First seen in the chapter →
skew symmetric matrixविषम सममित आव्यूहvisham sammit aavyooh: A square matrix with $A^{'} = - A$ ; its diagonal entries are 0.
square matrixवर्ग आव्यूहvarg aavyooh: A matrix with as many rows as columns.First seen in the chapter →
symmetric matrixसममित आव्यूहsammit aavyooh: A square matrix equal to its transpose: $A^{'} = A$ .

transposeपरिवर्तparivart: The matrix $A^{'}$ (or $A^{T}$ ) whose rows are the columns of $A$ .

zero matrixशून्य आव्यूहshoonya aavyooh: A matrix, written O, with every entry 0.First seen in the chapter →

Matrices: Class 12 Maths · Chapter 3

What you will learn in Matrices

3.1 Introduction

3.2 Matrix

3.2.1 Order of a matrix

3.3 Types of Matrices

3.3.1 Equality of matrices

Practice 3.1: questions with answers

3.4 Operations on Matrices

3.4.1 Addition of matrices

3.4.2 Multiplication of a matrix by a scalar

3.4.3 Properties of matrix addition

3.4.4 Properties of scalar multiplication of a matrix

3.4.5 Multiplication of matrices

Order matters: AB and BA

A zero product from non-zero matrices

3.4.6 Properties of multiplication of matrices

Practice 3.2: questions with answers

3.5 Transpose of a Matrix

3.5.1 Properties of transpose of the matrices

3.6 Symmetric and Skew Symmetric Matrices

Practice 3.3: questions with answers

3.7 Invertible Matrices

Practice 3.4: questions with answers

Miscellaneous examples

Matrices: key points and summary

Miscellaneous exercises

Practice questions with answers (NCERT “Miscellaneous exercises”)

Find your NCERT exercise question

Quick check: MCQs on Matrices

Key terms (English–Hindi)

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