Chapter 3 · Matrices · Go for it

Matrices, the quick way

A matrix is a table of numbers. Learn nine moves, and one rule for multiplying, and you can answer most questions in this chapter.

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The chapter in one minute

What a matrix is, and what the exam asks.

A matrix is a table of numbers in rows and columns, written inside square brackets.

In this chapter you add matrices, multiply them by a number, multiply two of them together, and flip them over (the transpose).

The exam asks short calculations: find a product, find the unknowns, show that $A^2 - 5A + 7I = O$ ⁠, or split a matrix into a symmetric and a skew symmetric part.

Most questions on this chapter are one of nine kinds, and each kind has a fixed recipe. You will learn all nine below.

9question types

14formula cards

1rule for every product: row times column

Orderrows by columns: check it first, every time

Formulas

The formulas you must know

Learn these cards. Almost everything in the chapter is built from them.

Order and entries

A = [a_{ij}]_{m\times n}

$m$ rows, $n$ columns, so $mn$ entries. The entry $a_{ij}$ sits in row $i$ ⁠, column $j$ ⁠.

Equal matrices

a_{ij} = b_{ij} \text{ for all } i, j

Same order, and every pair of entries in the same place is equal.

Sum and difference

A \pm B = [a_{ij} \pm b_{ij}]

Only for matrices of the same order. Work place by place.

A number times a matrix

kA = [k\,a_{ij}]

Multiply every entry by $k$ ⁠.

Product, one entry

c_{ik} = \sum_{j=1}^{n} a_{ij}\,b_{jk}

Row $i$ of $A$ times column $k$ of $B$ ⁠: pair up, multiply, add.

Order of a product

(m\times n)(n\times p) \to m\times p

The inside numbers must match. The outside numbers give the answer's order.

Product laws

\begin{aligned} (AB)C &= A(BC) \\ A(B + C) &= AB + AC \end{aligned}

But the order matters: $AB$ and $BA$ are usually different.

Identity

AI = IA = A

For a square $A$ ⁠, with $I$ of the same order. $I$ has 1 on the diagonal and 0 elsewhere. Also $A^2 = AA$ ⁠.

Transpose rules

\begin{aligned} (A')' &= A \\ (A + B)' &= A' + B' \\ (kA)' &= kA' \end{aligned}

$A'$ is $A$ with its rows written as columns.

Transpose of a product

(AB)' = B'A'

The order reverses.

Symmetric and skew

\begin{aligned} &\text{Symmetric: } A' = A \\ &\text{Skew: } A' = -A \end{aligned}

A skew symmetric matrix has 0 everywhere on its diagonal.

Split any square matrix

\begin{aligned} A &= \tfrac{1}{2}(A + A') \\ &\quad + \tfrac{1}{2}(A - A') \end{aligned}

The first part is symmetric, the second is skew symmetric.

Inverse

AB = BA = I

Then $B$ is the inverse of $A$ ⁠, written $A^{-1}$ ⁠. Both are square, of the same order.

Only one inverse

\begin{aligned} B &= BI = B(AC) \\ &= (BA)C = IC = C \end{aligned}

Say $B$ and $C$ are both inverses of $A$ ⁠. Use $AC = I$ ⁠, then $BA = I$ ⁠. So a matrix has only one inverse. The syllabus asks for this short proof.

Question types

The question types and how to crack each one

For each type: how to spot it, what to do, and one real NCERT question worked the exam way.

1. Build a matrix from a rule for $a_{ij}$

Spot it: "Construct a $2\times2$ matrix whose elements are given by" a rule for $a_{ij}$ ⁠.

List the places. A $2\times2$ matrix has $a_{11}$ ⁠, $a_{12}$ ⁠, $a_{21}$ and $a_{22}$ ⁠.
In the rule, put $i$ as the row number and $j$ as the column number.
Work out each entry and write it in its place.

Exercise 3.1 Q4 (i). Construct a $2\times2$ matrix $A = [a_{ij}]$ ⁠, where $a_{ij} = \frac{(i + j)^2}{2}$ ⁠.

\begin{aligned} a_{11} &= \frac{(1 + 1)^2}{2} = 2 \\ a_{12} &= \frac{(1 + 2)^2}{2} = \frac{9}{2} \\ a_{21} &= \frac{(2 + 1)^2}{2} = \frac{9}{2} \\ a_{22} &= \frac{(2 + 2)^2}{2} = 8 \end{aligned}

∴ $A = \begin{bmatrix} 2 & \frac{9}{2} \\ \frac{9}{2} & 8 \end{bmatrix}$

See a full exam answer · Go deeper: see it visually · Basics: the address $a_{ij}$

In past papers (2): click to see

Sample 2023-24 Q1 · Sample 2020-21 Q5(a)

2. Equal matrices: match entry with entry

Spot it: two matrices joined by an equals sign, with unknowns such as $x$ or $a - b$ inside.

Check that both sides have the same order.
Write "Equating corresponding entries" and one equation for each place.
Solve one unknown at a time. If you meet a quadratic, find both roots and check each one in all the equations.

Exercise 3.1 Q7. Find $a$ ⁠, $b$ ⁠, $c$ and $d$ from

\begin{aligned} &\begin{bmatrix} a - b & 2a + c \\ 2a - b & 3c + d \end{bmatrix} \\ &= \begin{bmatrix} -1 & 5 \\ 0 & 13 \end{bmatrix} \end{aligned}

Equating corresponding entries,

\begin{aligned} a - b &= -1 \quad\ldots(1) \\ 2a + c &= 5 \quad\ldots(2) \\ 2a - b &= 0 \quad\ldots(3) \\ 3c + d &= 13 \quad\ldots(4) \end{aligned}

(3) − (1) gives $a = 1$ ⁠. Then from (3), $b = 2a = 2$ ⁠.

From (2), $c = 5 - 2a = 3$ ⁠. From (4), $d = 13 - 3c = 4$ ⁠.

∴ $a = 1$ ⁠, $b = 2$ ⁠, $c = 3$ ⁠, $d = 4$

See a full exam answer · Go deeper: see it visually · Basics: solving equations

In past papers (2): click to see

Board 2024 Q5 · Board 2024 Q32(b)

3. Add, subtract, multiply by a number

Spot it: $A + B$ ⁠, $A - B$ or $3A - C$ ⁠, with the matrices given.

Check the orders match. Only matrices of the same order can be added or subtracted.
For a number times a matrix, multiply every entry by the number.
Add or subtract the entries in the same place.

Exercise 3.2 Q1 (iii). Find $3A - C$ ⁠, where

A = \begin{bmatrix} 2 & 4 \\ 3 & 2 \end{bmatrix},\quad C = \begin{bmatrix} -2 & 5 \\ 3 & 4 \end{bmatrix}

\begin{aligned} 3A &= \begin{bmatrix} 6 & 12 \\ 9 & 6 \end{bmatrix} \\ 3A - C &= \begin{bmatrix} 6 + 2 & 12 - 5 \\ 9 - 3 & 6 - 4 \end{bmatrix} \end{aligned}

∴ $3A - C = \begin{bmatrix} 8 & 7 \\ 6 & 2 \end{bmatrix}$

See a full exam answer · Go deeper: see it visually · Basics: adding only like with like

4. Find the unknown matrix $X$

Spot it: a matrix equation with an unknown matrix $X$ in it, such as $2X + Y$ equal to a given matrix.

Get $X$ on its own, as in ordinary algebra: $2X = (2X + Y) - Y$ ⁠.
Subtract the matrices place by place.
Divide every entry by the number in front of $X$ ⁠.

Exercise 3.2 Q8. Find $X$ ⁠, if

Y = \begin{bmatrix} 3 & 2 \\ 1 & 4 \end{bmatrix}

2X + Y = \begin{bmatrix} 1 & 0 \\ -3 & 2 \end{bmatrix}

\begin{aligned} 2X &= \begin{bmatrix} 1 & 0 \\ -3 & 2 \end{bmatrix} - \begin{bmatrix} 3 & 2 \\ 1 & 4 \end{bmatrix} \\ &= \begin{bmatrix} 1 - 3 & 0 - 2 \\ -3 - 1 & 2 - 4 \end{bmatrix} \\ &= \begin{bmatrix} -2 & -2 \\ -4 & -2 \end{bmatrix} \end{aligned}

∴ $X = \begin{bmatrix} -1 & -1 \\ -2 & -1 \end{bmatrix}$

See a full exam answer · Go deeper: see it visually

In past papers (1): click to see

Board 2023 Q4

5. Multiply two matrices: row times column

Spot it: two matrices written side by side, or $AB$ ⁠, $BA$ or $A^2$ ⁠.

Write the orders. In $(m\times n)(n\times p)$ the inside numbers must match, and the answer is $m\times p$ ⁠.
For the entry in row $i$ ⁠, column $k$ ⁠: go along row $i$ of the first and down column $k$ of the second. Multiply the pairs and add.
Fill in the answer one row at a time.

Exercise 3.2 Q3 (iii). Compute the product

P = \begin{bmatrix} 1 & -2 \\ 2 & 3 \end{bmatrix}\begin{bmatrix} 1 & 2 & 3 \\ 2 & 3 & 1 \end{bmatrix}

The orders are $(2\times2)(2\times3)$ ⁠, so $P$ is $2\times3$ ⁠.

P = \begin{bmatrix} 1 - 4 & 2 - 6 & 3 - 2 \\ 2 + 6 & 4 + 9 & 6 + 3 \end{bmatrix}

∴ $P = \begin{bmatrix} -3 & -4 & 1 \\ 8 & 13 & 9 \end{bmatrix}$

See a full exam answer · Go deeper: see it visually · Basics: pair up, multiply, add

In past papers (7): click to see

Board 2024 Q32(b) · Sample 2026-27 Q6 · Sample 2026-27 Q26 · Sample 2025-26 Q32 · Sample 2023-24 Q1 · Sample 2020-21 Q5(a) · Sample 2020-21 Q36(b)

6. One-mark MCQs: orders and definitions

Spot it: four options, asking when a sum or product is defined, what order it has, or what a word such as symmetric or inverse means.

Write the order under every matrix.
A product needs the inside numbers to match. A sum needs the same order.
Pick the option that fits, and write its letter with its value.

Exercise 3.2 Q21 (MCQ). $P$ ⁠, $Y$ and $W$ have orders $p\times k$ ⁠, $3\times k$ and $n\times3$ ⁠. $PY + WY$ is defined when: (A) $k = 3$ ⁠, $p = n$ (B) $k$ is arbitrary, $p = 2$ (C) $p$ is arbitrary, $k = 3$ (D) $k = 2$ ⁠, $p = 3$

$PY$ is $(p\times k)(3\times k)$ ⁠. The inside numbers must match, so $k = 3$ ⁠, and then $PY$ is $p\times3$ ⁠.

$WY$ is $(n\times3)(3\times k)$ ⁠, so it is $n\times k$ ⁠, that is $n\times3$ ⁠.

A sum needs the same order, so $p = n$ ⁠.

∴ Answer: (A) $k = 3$ ⁠, $p = n$

See a full exam answer · Go deeper: see it visually

In past papers (8): click to see

Board 2026 Q2 · Board 2025 Q9 · Board 2025 Q13 · Board 2024 Q2 · Sample 2025-26 Q2 · Sample 2024-25 Q2 · Sample 2022-23 Q1 · Sample 2020-21 Q4

7. Polynomials in $A$ ⁠: show that it gives $O$

Spot it: $A^2$ or $A^3$ mixed with numbers and $I$ ⁠, as in "show that $A^2 - 5A + 7I = O$ " or "find $k$ ".

Find $A^2 = AA$ by row times column. Never square the entries.
Multiply $A$ by its number. A plain number $c$ becomes $cI$ ⁠: $c$ on the diagonal, 0 elsewhere.
Combine the entries place by place, and end with "Hence proved".

Miscellaneous Q5. If $A = \begin{bmatrix} 3 & 1 \\ -1 & 2 \end{bmatrix}$ ⁠, show that $A^2 - 5A + 7I = O$ ⁠.

\begin{aligned} A^2 &= \begin{bmatrix} 3 & 1 \\ -1 & 2 \end{bmatrix}\begin{bmatrix} 3 & 1 \\ -1 & 2 \end{bmatrix} \\ &= \begin{bmatrix} 9 - 1 & 3 + 2 \\ -3 - 2 & -1 + 4 \end{bmatrix} \\ &= \begin{bmatrix} 8 & 5 \\ -5 & 3 \end{bmatrix} \end{aligned}

Also $5A = \begin{bmatrix} 15 & 5 \\ -5 & 10 \end{bmatrix}$ and $7I = \begin{bmatrix} 7 & 0 \\ 0 & 7 \end{bmatrix}$ ⁠.

\begin{aligned} &A^2 - 5A + 7I \\ &= \begin{bmatrix} 8 - 15 + 7 & 5 - 5 + 0 \\ -5 + 5 + 0 & 3 - 10 + 7 \end{bmatrix} \\ &= \begin{bmatrix} 0 & 0 \\ 0 & 0 \end{bmatrix} = O \end{aligned}

∴ $A^2 - 5A + 7I = O$ ⁠. Hence proved.

See a full exam answer · Go deeper: see it visually · Basics: the identity matrix

In past papers (1): click to see

Board 2023 Q14

8. Transpose: symmetric or skew symmetric?

Spot it: "show that $A$ is symmetric", "show that $A$ is skew symmetric", or "find $A'$ ".

Write $A'$ ⁠: the rows of $A$ become its columns.
Compare. If $A' = A$ ⁠, it is symmetric. If $A' = -A$ ⁠, it is skew symmetric.
End with the conclusion in words.

Exercise 3.3 Q7 (ii). Show that $A$ is a skew symmetric matrix, where

A = \begin{bmatrix} 0 & 1 & -1 \\ -1 & 0 & 1 \\ 1 & -1 & 0 \end{bmatrix}

\begin{aligned} A' &= \begin{bmatrix} 0 & -1 & 1 \\ 1 & 0 & -1 \\ -1 & 1 & 0 \end{bmatrix} \\ &= -\begin{bmatrix} 0 & 1 & -1 \\ -1 & 0 & 1 \\ 1 & -1 & 0 \end{bmatrix} \\ &= -A \end{aligned}

∴ $A' = -A$ ⁠, so $A$ is skew symmetric.

See a full exam answer · Go deeper: see it visually · Basics: reading a table sideways

In past papers (5): click to see

Board 2026 Q6 · Sample 2026-27 Q4 · Sample 2025-26 Q3 · Sample 2024-25 Q7 · Sample 2023-24 Q10

9. Split a matrix: symmetric plus skew symmetric

Spot it: "Express the matrix as the sum of a symmetric and a skew symmetric matrix."

Write $A'$ ⁠.
Work out $P = \frac{1}{2}(A + A')$ ⁠, the symmetric part, and $Q = \frac{1}{2}(A - A')$ ⁠, the skew symmetric part.
Check that $P' = P$ and $Q' = -Q$ ⁠, then write $A = P + Q$ ⁠.

Exercise 3.3 Q10 (i). Express $A = \begin{bmatrix} 3 & 5 \\ 1 & -1 \end{bmatrix}$ as the sum of a symmetric and a skew symmetric matrix.

A' = \begin{bmatrix} 3 & 1 \\ 5 & -1 \end{bmatrix}

\begin{aligned} P &= \frac{1}{2}(A + A') \\ &= \frac{1}{2}\begin{bmatrix} 6 & 6 \\ 6 & -2 \end{bmatrix} = \begin{bmatrix} 3 & 3 \\ 3 & -1 \end{bmatrix} \end{aligned}

\begin{aligned} Q &= \frac{1}{2}(A - A') \\ &= \frac{1}{2}\begin{bmatrix} 0 & 4 \\ -4 & 0 \end{bmatrix} = \begin{bmatrix} 0 & 2 \\ -2 & 0 \end{bmatrix} \end{aligned}

$P' = P$ ⁠, so $P$ is symmetric. $Q' = -Q$ ⁠, so $Q$ is skew symmetric.

∴ $A = \begin{bmatrix} 3 & 3 \\ 3 & -1 \end{bmatrix} + \begin{bmatrix} 0 & 2 \\ -2 & 0 \end{bmatrix}$

See a full exam answer · Go deeper: see it visually · Basics: mirror patterns

Shortcuts

Shortcuts and quick tips

Small habits that save time and marks.

Write the order under each matrix before you start. It tells you whether a sum or product exists, and the size of the answer.
For each entry of a product, run one finger along the row and another down the column.
Check one entry of every product a second time. Sign slips are easy to make, so watch lines like $-1 - (-5) = 4$ ⁠.
$6I$ has 6 on the diagonal and 0 everywhere else. It is how a plain number joins a matrix sum.
Only a square matrix has a square: a $2\times3$ matrix has no $A^2$ ⁠.
A skew symmetric matrix always has 0s on its diagonal. One nonzero diagonal entry, and it is not skew.
Two equations in $X$ and $Y$ ⁠? Add or subtract them, just as in simultaneous equations.
After splitting a matrix, add the two parts back: you must get $A$ ⁠. Also check $P' = P$ and $Q' = -Q$ ⁠.
$AB = O$ does not mean $A = O$ or $B = O$ ⁠. Example 15 multiplies two nonzero matrices and gets $O$ ⁠.
Know the names for one-mark questions: square ( $m = n$ ⁠), row, column, diagonal (square, with 0 off the diagonal), scalar (a diagonal matrix with equal diagonal entries), identity $I$ and zero $O$ ⁠.

Watch out

Mistakes that cost marks

Each one is common, and each has an easy fix.

Squaring each entry to get $A^2$ → $A^2 = AA$ ⁠: multiply row by column.
Multiplying the entries in the same place → a product is always row times column.
Writing $AB = BA$ → keep the order; the two products are usually different.
Writing $(AB)' = A'B'$ → the order reverses: $(AB)' = B'A'$ ⁠.
Adding a plain number, as in $A + 7$ → write $7I$ instead; it changes only the diagonal.
Adding matrices of different orders → check the orders first; such a sum is not defined.
Forgetting the $\frac{1}{2}$ in $P = \frac{1}{2}(A + A')$ → check that $P + Q = A$ at the end.
Stopping at one answer when the equations lead to a quadratic → solve it fully; there may be two answers.

Marks

How the examiner gives marks

From CBSE's own marking schemes. Write these lines and the marks follow.

Show the working line of each product, such as $2 - 8$ and $6 + 20$ ⁠. Examiners mark method, and a slip is penalised only once, so keep going.
Write conclusions in words, such as " $A' = -A$ ⁠, so $A$ is skew symmetric". Lines like this carry marks.
In a "show that" question, end with the target written out and "Hence proved". That last line carries half a mark.
In a CBSE sample paper (2025-26), showing $AB = I$ by multiplying out earned 1 mark, and "hence $A^{-1} = B$ " earned another half.
A 1-mark MCQ earns its mark for the right option only. Write the letter and the value, such as "(B) $2\times n$ ".

Practice

Practise these first

Easiest first. Together they cover every type above.

Exercise 3.1 Q8 (MCQ) when a matrix is square
Exercise 3.4 Q1 (MCQ) what makes two matrices inverses
Example 3 a $3\times2$ matrix from a rule for $a_{ij}$
Exercise 3.3 Q1 transposes: rows become columns
Exercise 3.2 Q22 (MCQ) the order of a difference
Exercise 3.1 Q6 equal matrices, with one part that has two answers
Exercise 3.2 Q9 a number times a matrix, then equal matrices
Exercise 3.2 Q7 find $X$ and $Y$ by adding and subtracting
Exercise 3.2 Q1 sums and products, and $AB$ is not $BA$
Exercise 3.2 Q17 find $k$ in an equation with $A^2$ and $I$
Example 22 a $3\times3$ matrix split into symmetric plus skew
Exercise 3.2 Q15 a polynomial in a $3\times3$ matrix

You can do this

Nine moves, and one rule behind most of them: row times column. Work through the practice list, check each product twice, and you are ready for the exam.

Ready to go deeper? Open the interactive lesson

Past papers

24 CBSE board and sample-paper questions on Matrices, each with a full-marks exam answer. Practise them →