Matrices become reliable when order and position are checked before any operation. I will help you separate element-wise rules from multiplication rules, verify dimensions at each step and connect inverse methods cleanly with determinants.
Take a breath. Matrices is one of those chapters that looks frightening for about twenty minutes and then becomes one of the friendliest things in your whole Class 12 Maths course. There is no limit to evaluate, no tricky substitution, no long integration. A matrix is just numbers arranged neatly in rows and columns — like a class timetable, or the marks sheet your school prints, or the price list stuck on the wall of the canteen. Everything you will learn here is about handling those tidy arrangements: adding them, scaling them, flipping them, and multiplying them in one slightly unusual way that we will build up very slowly together.
I want to say something honestly at the start. Most students who struggle with Matrices do not struggle with the ideas. They struggle because they rushed the very first definitions — order, notation, and what aij means — and then everything after that felt slippery. So we are going to go slowly at the beginning and then speed up. By the end of this page you will be able to look at any board question on Matrices and know exactly which of about six moves it is asking for.
This chapter sits in Unit II: Algebra of the CBSE 2026-27 Class 12 Maths syllabus. Unit II is worth 10 marks in the board paper, and Matrices shares those marks with Determinants. That is a comfortable, reliable 10 marks — the kind you should be walking into the exam hall planning to collect in full.
What You’ll Learn
Here is everything on this page. Tap any line to jump straight to it, and come back here whenever you want to move around.
Chapters go badly when students open the book at a random page. Work through this chapter in the order below and it will feel like walking down a staircase instead of climbing a wall.
Nail the vocabulary first. Order, element, aij, row, column. Spend fifteen honest minutes here. Everything later depends on it.
Learn the types by picture, not by definition. Use the family map below — square, diagonal, scalar, identity, zero, row, column. Say each one out loud.
Do addition and scalar multiplication next. They behave exactly like ordinary algebra, so they cost you almost nothing and they build confidence.
Then transpose, then symmetric and skew symmetric. Transpose is one physical action — flipping rows into columns. Symmetric and skew symmetric are just questions you ask after you flip.
Give matrix multiplication a full sitting of its own. This is the one genuinely new skill in the chapter. Do not squeeze it into the last ten minutes of a study session.
Finish with invertible matrices and the uniqueness proof. It is a short, elegant proof and it is very much askable in the board exam.
Then the worksheet at the bottom. Write your answers on paper before you open the reveal. Reading a solution is not the same as producing one.
Study Notes
What A Matrix Really Is
Imagine your school prints a small table of marks for three students across two subjects. You would draw three rows, one per student, and two columns, one per subject, and drop the marks into the boxes. Strip away the student names and the subject headings and keep only the numbers inside a big pair of brackets — that is a matrix.
Key Rule — What a matrix is
A matrix is a rectangular arrangement of numbers in rows and columns, written inside square brackets. Each number inside is called an element (or entry). In this whole chapter, every element is a real number.
A =
62
71
48
55
80
66
three rows, two columns
Notice that a matrix by itself is not a number and it does not have a value. You cannot “evaluate” a matrix the way you evaluate 3 + 4. A matrix is an object — a container that holds numbers in a fixed arrangement. (Later, in the Determinants chapter, you will meet a different bracket shape that does boil down to a single number. Keep the two apart in your head right from today: square brackets mean matrix, vertical bars mean determinant.)
Order of a matrix. The order tells you the shape. If a matrix has m rows and n columns, we say it is a matrix of order m × n, read aloud as “m by n”. The matrix above is of order 3 × 2, because it has 3 rows and 2 columns.
Exam Tip — Rows first, always
Order is rows × columns, in that order, every single time. A useful memory hook: R comes before C in the alphabet, and in “Rows × Columns”. Students who mix this up lose marks not because they cannot do matrices but because they described the shape backwards.
Notation. We write a general matrix as A = [aij], where aij is the element sitting in the i-th row and the j-th column. So a23 is the element in row 2, column 3. Again: the first subscript is the row, the second is the column. Same rule as the order, so you only have one thing to remember.
A =
a11
a12
a13
a21
a22
a23
a general 2 × 3 matrix
How many elements does an m × n matrix hold? Exactly mn of them — every row contributes n boxes and there are m rows. That small fact turns into an exam question surprisingly often, so let us make it a rule.
Key Rule — Counting elements
A matrix of order m × n has exactly mn elements. Turn it around: if you are told a matrix has 12 elements, then m × n = 12, so the possible orders are all the factor pairs of 12 — 1×12, 2×6, 3×4, 4×3, 6×2 and 12×1. Six possible orders, one for each divisor of 12.
Example 1 — Building a matrix from a rule
Question. Construct the 2 × 3 matrix A = [aij] whose elements are given by aij = 2i + 3j.
Working. The order is 2 × 3, so i runs 1 to 2 and j runs 1 to 3. Just walk the boxes one at a time and substitute:
Why it works. A rule like aij = 2i + 3j is a machine: you feed it a position and it hands you back the number that belongs there. The only discipline required is going in a fixed order — finish row 1 completely, then row 2 — so that you never drop a box or fill one twice.
Example 2 — Reading order and elements off a table
Question. A stationery shop records how many notebooks, pens and erasers it sold on Monday, Tuesday and Wednesday:
S =
14
22
9
11
18
15
20
25
7
State the order of S, and write down s23 and s31.
Working. There are 3 rows (the three days) and 3 columns (the three items), so S is of order 3 × 3. Now s23 means row 2, column 3 — go down to the second row, across to the third entry: s23 = 15. And s31 means row 3, column 1: s31 = 20.
Why it works. Reading a matrix is a two-step physical action: down for the first subscript, then across for the second. Say it under your breath the first ten times — “down two, across three” — and it becomes automatic. In real life those numbers mean something (15 erasers on Tuesday), and that is exactly why matrices are used everywhere from timetables to computer graphics.
Common Mistake — Reading aij backwards
In a hurry, students read a23 as “column 2, row 3”. In a square matrix this gives you a wrong-but-plausible number, so you never notice. Fix it now with one sentence you repeat every time: row first, column second. It is the same order as the order of the matrix, and the same order as reading English — you pick a line, then move along it.
When are two matrices the same? Your instinct might be “when they contain the same numbers”, but that is not quite strict enough. Two matrices are equal only when they are the same shape and every single corresponding box carries the same number. Think of two identical seating plans for the same classroom: same number of rows, same number of desks per row, and the same student in every seat.
Key Rule — Equality of matrices
Two matrices A = [aij] and B = [bij] are equal, written A = B, if and only if: (i) they have the same order, and (ii) aij = bij for every i and j. Both conditions must hold. Same numbers in a different shape is not equality.
3
5
1
4
≠
3
5
1
4
(same numbers, different order — not equal)
This definition is far more useful than it first looks, because it turns one matrix equation into a whole set of ordinary equations. If someone hands you an equality between two 2 × 2 matrices, they have quietly handed you four simultaneous equations. That is the entire trick behind every “find x and y” question in this chapter.
Example 3 — Finding unknowns by comparing entries
Question. Find x, y, z and w if
x + 3
2y
z − 1
5w
=
0
6
3
20
Working. Both matrices are 2 × 2, so the orders already match. Now compare box by box:
x + 3 = 0 → x = −3 2y = 6 → y = 3 z − 1 = 3 → z = 4 5w = 20 → w = 4
Why it works. Equality of matrices is defined entry by entry, so a single matrix statement splits into four completely independent little equations. Each one has one unknown, so each one falls in a single line. Do not over-think these — they are gifts.
Example 4 — When the entries are tangled together
Question. Find x and y if
2x + y
5
3
x − y
=
7
5
3
2
Working. Comparing entries gives 2x + y = 7 and x − y = 2 (the other two boxes just confirm 5 = 5 and 3 = 3, which is a useful sign you have set it up right). Add the two equations: (2x + y) + (x − y) = 7 + 2, so 3x = 9 and x = 3. Put that back into x − y = 2: 3 − y = 2, so y = 1.
Check. 2(3) + 1 = 7 ✓ and 3 − 1 = 2 ✓.
Why it works. Sometimes the corresponding entries give you equations that share unknowns, and then you are simply back in the simultaneous-equations world you have known since Class 9. Nothing about matrices makes that harder. Always take ten seconds to substitute your answers back into the original matrix — it is free marks insurance.
Common Mistake — Forgetting to check the order first
Students dive straight into comparing entries without noticing that one matrix is 2 × 3 and the other is 3 × 2. If the orders differ, the matrices are simply not equal and no values of x and y can fix that — the correct answer is “no solution, the orders do not match”. Glance at the shapes before you write a single equation.
Read the map downwards as “is a special kind of”: every identity matrix is a scalar matrix, every scalar matrix is diagonal, and every diagonal matrix is square.
Mathematicians give special names to matrices with special shapes, in the same way we name special triangles. You do not need to memorise these as dry definitions — look at the map above and you will see they mostly answer one question: what is the relationship between the number of rows and the number of columns, and where are the zeros?
Name
Condition
A quick example
Row matrix
Exactly one row, so order 1 × n
4
−1
7
Column matrix
Exactly one column, so order m × 1
2
5
−3
Rectangular matrix
Number of rows ≠ number of columns
1
0
6
3
2
−4
Square matrix
Number of rows = number of columns, order n × n
1
2
5
0
Diagonal matrix
Square, and every off-diagonal element is 0
7
0
0
0
−2
0
0
0
5
Scalar matrix
Diagonal, and all the diagonal entries are equal
3
0
0
0
3
0
0
0
3
Identity matrix I
Diagonal, and every diagonal entry is 1
1
0
0
0
1
0
0
0
1
Zero (null) matrix O
Every single element is 0 — any order allowed
0
0
0
0
0
0
The phrase principal diagonal (or main diagonal) means the entries a11, a22, a33, and so on — the ones running from the top-left corner down towards the bottom-right corner, where the row number equals the column number. Only a square matrix has a proper principal diagonal, which is why diagonal, scalar and identity matrices are all square by definition.
Key Rule — The nesting of the square family
Every identity matrix is a scalar matrix (all its diagonal entries are equal — they all happen to be 1). Every scalar matrix is a diagonal matrix. Every diagonal matrix is a square matrix. The arrows only point one way: a diagonal matrix such as diag(7, −2, 5) is certainly not scalar, and a scalar matrix such as diag(3, 3, 3) is certainly not the identity.
Example 5 — Naming matrices correctly
Question. Give the order of each matrix and name it as precisely as you can.
P =
0
0
0
Q =
6
0
0
6
R =
2
0
0
0
0
0
0
0
9
Working. P is 1 × 3. It has one row, so it is a row matrix; but every entry is 0, so the most precise name is the zero matrix of order 1 × 3. Both descriptions are true, and in an exam you should say the sharper one. Q is 2 × 2, square, all off-diagonal entries are 0, and both diagonal entries equal 6 — so it is a scalar matrix. It is not the identity, because the diagonal entries are 6, not 1. R is 3 × 3, square, and every off-diagonal entry is 0, so it is a diagonal matrix. The fact that one diagonal entry happens to be 0 does not disqualify it — the definition says nothing about the diagonal entries, only about the off-diagonal ones.
Why it works. Notice how R catches people out. The rule for a diagonal matrix is a rule about the off-diagonal boxes only. Diagonal entries are allowed to be anything at all, zero included.
Example 6 — Constructing a diagonal matrix from a rule
Question. Write the 3 × 3 matrix D = [dij] where dij = i2 when i = j, and dij = 0 when i ≠ j.
Working. The condition i ≠ j fills every off-diagonal box with 0 straight away. The three diagonal boxes get 12 = 1, 22 = 4 and 32 = 9.
D =
1
0
0
0
4
0
0
0
9
Why it works. A two-part rule like this is really just the definition of a diagonal matrix wearing a disguise. Whenever you see “= something when i = j, = 0 when i ≠ j”, you can write down all the zeros immediately and then fill in only three boxes. It should take you under a minute.
Common Mistake — Calling every square matrix with zeros “diagonal”
A matrix such as
1
0
4
2
has zeros in it and is square, but it is not diagonal — the entry 4 sits off the diagonal and is not zero. The test is total: every off-diagonal entry must be zero, without exception.
Two matrices deserve their own section because they play the same role for matrices that 0 and 1 play for ordinary numbers. Once you see that parallel, half the properties in this chapter stop needing to be memorised.
Key Rule — The zero matrix O
The zero matrix (or null matrix), written O, is a matrix in which every element is 0. There is one for each order — a 2 × 3 zero matrix, a 3 × 3 zero matrix, and so on. It behaves like the number 0: A + O = O + A = A for any matrix A of the same order, and A + (−A) = O.
Key Rule — The identity matrix I
The identity matrix of order n, written I or In, is the square matrix with 1 in every position on the principal diagonal and 0 everywhere else. It behaves like the number 1 under multiplication: AI = IA = A whenever the multiplication makes sense.
I2 =
1
0
0
1
I3 =
1
0
0
0
1
0
0
0
1
Example 7 — The identity really does leave a matrix alone
Question. With A =
7
−3
2
9
, verify that AI = IA = A, and that A + O = A.
Working. (You will learn the mechanics of multiplication properly a little later on this page; for now just trust the arithmetic and enjoy the result.)
AI =
7
−3
2
9
1
0
0
1
=
7
−3
2
9
= A
IA =
1
0
0
1
7
−3
2
9
=
7
−3
2
9
= A
A + O =
7
−3
2
9
+
0
0
0
0
=
7
−3
2
9
= A
Why it works. This is worth pausing on, because I is the one matrix that commutes with everything. Almost nothing else in this chapter satisfies XY = YX, so the identity is genuinely special. It is also the matrix that the whole idea of an inverse is built on — A−1 is defined as the matrix that turns A back into I.
Exam Tip — O and I are not interchangeable
A quick sanity phrase for the exam hall: O is the additive identity, I is the multiplicative identity. Adding O changes nothing; multiplying by I changes nothing. Multiplying by O, on the other hand, destroys everything — AO = OA = O. Keep the two letters and the two operations paired correctly and you will not slip.
Here is a completely physical idea. Take your matrix, and tip it over about its main diagonal so that the first row stands up and becomes the first column, the second row becomes the second column, and so on. That new matrix is called the transpose. If you have ever turned a portrait photo sideways into landscape, you already have the picture in your head.
Key Rule — Transpose
If A = [aij] is a matrix of order m × n, then its transpose, written A′ or AT, is the matrix of order n × m obtained by interchanging the rows and columns of A. In symbols, the (i, j) entry of A′ is aji.
Notice what happens to the order: a 3 × 2 matrix transposes into a 2 × 3 matrix. The numbers swap places. Only a square matrix keeps the same order after transposing, and that single observation is the seed of the whole next section.
Example 8 — Transposing, and watching the order flip
Question. Write down A′ for A =
1
2
3
4
5
6
and state both orders.
Working. A has 3 rows and 2 columns, so A is of order 3 × 2. Row 1 of A is (1, 2), and it becomes column 1 of A′. Row 2 of A is (3, 4), and it becomes column 2 of A′. Row 3 of A is (5, 6), and it becomes column 3 of A′.
A′ =
1
3
5
2
4
6
of order 2 × 3
Why it works. Do it one row at a time and say what you are doing out loud: “row one goes down as column one”. Students who try to transpose a matrix by staring at it and hoping usually misplace one entry. Students who move one row at a time never do.
Four properties of the transpose come up again and again, and three of them are exactly what you would guess. The fourth one has a twist that examiners love.
Property
In symbols
In words
Transposing twice does nothing
(A′)′ = A
Flip it, flip it back — you are home again.
Transpose of a sum
(A + B)′ = A′ + B′
Add first or flip first; the answer is the same.
Transpose of a scalar multiple
(kA)′ = k A′
A constant can pass straight through a transpose.
Transpose of a product
(AB)′ = B′A′
The order reverses. This is the one to remember.
Example 9 — Verifying the reversal law (AB)′ = B′A′
Question. With A =
1
2
3
4
5
6
and B =
1
0
2
1
0
3
, verify that (AB)′ = B′A′.
Working. First the product. A is 2 × 3 and B is 3 × 2, so AB is 2 × 2.
Why it works. Look at the orders and the reversal becomes obvious rather than magical. AB is 2 × 2, so (AB)′ is 2 × 2. If you tried A′B′ instead, you would be multiplying a 3 × 2 by a 3 × 2, and the inner numbers (2 and 3) do not match — the product would not even exist. The reversal is not an arbitrary rule; it is the only order that fits.
Common Mistake — Writing (AB)′ = A′B′
This is probably the single most common transpose error in board papers. The correct law reverses the order: (AB)′ = B′A′. A phrase that sticks: “the last shall be first”. The same reversal appears again with inverses later on, so learning it properly now pays you twice.
Now that you can transpose a matrix, you can ask a natural question: what happens if a matrix is unchanged by transposing? Or what if transposing simply flips every sign? Those two questions give us the two most important special square matrices in the syllabus.
Key Rule — Symmetric and skew symmetric
A square matrix A is symmetric if A′ = A, that is, aij = aji for all i and j. A square matrix A is skew symmetric if A′ = −A, that is, aij = −aji for all i and j. Both ideas only make sense for square matrices, because otherwise A′ would not even have the same order as A.
A symmetric matrix is a mirror: whatever sits above the diagonal is copied exactly below it. A skew symmetric matrix is a mirror with the sign flipped: whatever sits above the diagonal appears below it with the opposite sign.
Symmetric:
2
−1
3
−1
0
5
3
5
−4
Skew symmetric:
0
4
−2
−4
0
7
2
−7
0
Key Rule — Every skew symmetric matrix has zeros on its diagonal
Put i = j in the definition aij = −aji. You get aii = −aii, so 2aii = 0, so aii = 0. Every diagonal entry of a skew symmetric matrix is forced to be zero. This is a two-line proof that boards genuinely ask for, and it is also the fastest way to spot that a matrix is not skew symmetric — one non-zero number on the diagonal and it is disqualified instantly.
Example 10 — Testing a matrix for symmetry
Question. Decide whether each matrix is symmetric, skew symmetric, or neither.
P =
2
−1
3
−1
0
5
3
5
−4
Q =
0
4
−2
−4
0
7
2
−7
0
R =
1
2
3
1
Working. P: transpose it and you get back exactly P (check the pairs: p12 = −1 = p21, p13 = 3 = p31, p23 = 5 = p32). So P′ = P and P is symmetric. Q: the diagonal is all zeros, which is promising. Check the pairs: q12 = 4 and q21 = −4, q13 = −2 and q31 = 2, q23 = 7 and q32 = −7. Every pair is equal and opposite, so Q′ = −Q and Q is skew symmetric. R: r12 = 2 but r21 = 3. They are neither equal nor opposite, so R is neither.
Why it works. You do not have to write out the full transpose to test this. Just check the mirror pairs across the diagonal: (1,2) against (2,1), (1,3) against (3,1), (2,3) against (3,2). Three quick checks for a 3 × 3 matrix and you are done.
Now the big result of this section, and one of the most reliably examined ideas in the whole chapter. Any square matrix at all — however messy — can be split into a symmetric part plus a skew symmetric part, and there is only one way to do it.
Key Rule — The symmetric plus skew symmetric decomposition
For any square matrix A, write P = ½(A + A′) and Q = ½(A − A′). Then P is always symmetric, Q is always skew symmetric, and A = P + Q. Moreover this splitting is unique.
Why P is symmetric: P′ = ½(A + A′)′ = ½(A′ + (A′)′) = ½(A′ + A) = P. Why Q is skew symmetric: Q′ = ½(A − A′)′ = ½(A′ − A) = −½(A − A′) = −Q. And P + Q = ½(A + A′) + ½(A − A′) = ½(2A) = A.
Example 11 — Splitting a matrix into symmetric and skew symmetric parts
Question. Express A =
2
5
3
1
4
7
−1
3
6
as the sum of a symmetric and a skew symmetric matrix.
Step 1 — write A′.
A′ =
2
1
−1
5
4
3
3
7
6
Step 2 — add and subtract.
A + A′ =
4
6
2
6
8
10
2
10
12
A − A′ =
0
4
4
−4
0
4
−4
−4
0
Step 3 — halve each one.
P = ½(A + A′) =
2
3
1
3
4
5
1
5
6
Q = ½(A − A′) =
0
2
2
−2
0
2
−2
−2
0
Step 4 — check. P is symmetric (mirror pairs 3 and 3, 1 and 1, 5 and 5 ✓). Q is skew symmetric (zero diagonal, and pairs 2 and −2 ✓). Adding them entry by entry: 2 + 0 = 2, 3 + 2 = 5, 1 + 2 = 3 in the first row, and so on all the way through — you recover A exactly.
Why it works. The trick is a pure piece of algebra you have used before without noticing: any object x can be written as ½(x + y) + ½(x − y) for any y. Choosing y = A′ is what makes the two halves land in the symmetric and skew symmetric camps. Follow the four steps in the same order every time and this question is four guaranteed marks.
Exam Tip — Halves are normal here, so do not panic
In this example the halves cancelled neatly and every entry came out a whole number, but that is luck, not law. If A =
1
5
2
3
then A + A′ =
2
7
7
6
and P =
1
7/2
7/2
3
, with Q =
0
3/2
−3/2
0
. Fractions are perfectly correct answers. Leave them as fractions, never as rounded decimals, and always finish by checking that P + Q gives back A.
Common Mistake — Forgetting the ½ or checking only one part
Two errors cost marks here. First, writing P = A + A′ without the ½ — the sum would then come to 2A, not A. Second, verifying that P is symmetric but never checking Q. Examiners award marks for the verification lines, so write them: state P′ = P, state Q′ = −Q, and state P + Q = A.
Exam Tip — A bonus fact worth knowing
For any matrix A (square or not), the products A′A and AA′ are always square and always symmetric. Quick proof for the first: (A′A)′ = A′(A′)′ = A′A, using the reversal law. That is a favourite one-mark or two-mark question.
Good news: this is the easiest operation in the chapter and it works exactly the way your instinct says it should. To add two matrices, add the numbers that are sitting in the same position. Think of two shopkeepers each keeping a sales table with the same rows and the same columns, and you want the combined totals — you simply add the matching boxes.
Key Rule — Addition of matrices
If A = [aij] and B = [bij] are two matrices of the same order m × n, then A + B is the m × n matrix whose (i, j) entry is aij + bij. If the orders are different, A + B is not defined — there is no answer to give.
Example 12 — Adding and subtracting matrices
Question. With A =
3
−1
2
0
4
5
and B =
−2
6
1
7
−3
0
, find A + B and A − B.
Working. Both are 2 × 3, so both operations are defined and both answers will be 2 × 3. Add (or subtract) box by box:
A + B =
3 + (−2)
−1 + 6
2 + 1
0 + 7
4 + (−3)
5 + 0
=
1
5
3
7
1
5
A − B =
3 − (−2)
−1 − 6
2 − 1
0 − 7
4 − (−3)
5 − 0
=
5
−7
1
−7
7
5
Why it works. Subtraction is not really a new operation: A − B just means A + (−1)B. The only thing that trips people is signs, especially when both numbers are negative. Write the intermediate step with the brackets in, exactly as shown above, at least until the sign handling feels automatic.
Example 13 — Solving a matrix equation with addition
Question. Find the matrix X such that A + X = B, where A and B are the matrices from the previous example.
Working. Treat it exactly like the number equation 3 + x = 7. Subtract A from both sides: X = B − A.
X =
−2
6
1
7
−3
0
−
3
−1
2
0
4
5
=
−5
7
−1
7
−7
−5
Check. A + X =
3
−1
2
0
4
5
+
−5
7
−1
7
−7
−5
=
−2
6
1
7
−3
0
= B ✓
Why it works. Matrix addition obeys all the same rearrangement rules as ordinary addition, so you can move terms across the equals sign in the usual way. Notice that X automatically has the same order as A and B — it could not be anything else.
Because addition is done entry by entry, and ordinary numbers already obey the commutative and associative laws, all the familiar properties come along for free:
Property
Statement
Condition
Commutative
A + B = B + A
A and B of the same order
Associative
(A + B) + C = A + (B + C)
All three of the same order
Additive identity
A + O = O + A = A
O is the zero matrix of the same order
Additive inverse
A + (−A) = (−A) + A = O
−A is A with every sign changed
Cancellation
If A + B = A + C then B = C
All of the same order
Common Mistake — Trying to add matrices of different orders
You cannot add a 2 × 3 matrix to a 3 × 2 matrix, even though both contain six numbers. There is no rule that pairs up the boxes, so the sum simply does not exist. In an exam the full-marks answer to such a question is a clear sentence: “A + B is not defined because A is of order 2 × 3 and B is of order 3 × 2.” Do not invent an answer.
A scalar is just an ordinary number, as opposed to a matrix. Multiplying a matrix by a scalar means scaling every single entry by that number — like doubling every quantity in a recipe, or raising every price in a menu by the same factor.
Key Rule — Scalar multiplication
If A = [aij] is an m × n matrix and k is any real number, then kA = [k aij]. The order does not change: kA is still m × n. Notice there is no compatibility condition at all — you can multiply any matrix by any scalar.
3
2
−1
0
4
=
3(2)
3(−1)
3(0)
3(4)
=
6
−3
0
12
Example 14 — Combining scalar multiples
Question. If A =
2
−3
4
1
and B =
1
5
−2
0
, find 2A − 3B.
Working. Scale each matrix first, then subtract.
2A =
4
−6
8
2
3B =
3
15
−6
0
2A − 3B =
4 − 3
−6 − 15
8 − (−6)
2 − 0
=
1
−21
14
2
Why it works. Because scalar multiplication distributes over the entries, you can safely scale first and combine afterwards. The riskiest box is the one where you subtract a negative, here 8 − (−6) = 14. Write the bracket. Every year, students lose marks for turning that into 2.
Example 15 — Solving for an unknown matrix
Question. Find X if 2X + 3A = B, where A =
1
2
3
−1
and B =
7
8
11
5
.
Working. Rearrange exactly as you would with numbers: 2X = B − 3A, so X = ½(B − 3A).
3A =
3
6
9
−3
B − 3A =
4
2
2
8
X = ½
4
2
2
8
=
2
1
1
4
Check. 2X =
4
2
2
8
, and 2X + 3A =
4
2
2
8
+
3
6
9
−3
=
7
8
11
5
= B ✓
Why it works. Dividing a matrix by 2 is the same as multiplying it by the scalar ½, which is allowed for any matrix. Note that we never “divide by a matrix” — that operation does not exist. We only ever multiply by scalars, or later, by inverses.
Property
Statement
Distributive over matrix addition
k(A + B) = kA + kB
Distributive over scalar addition
(k + l)A = kA + lA
Associative for scalars
k(lA) = (kl)A
Multiplying by 1
1A = A
Multiplying by 0
0A = O, the zero matrix
Negative of a matrix
(−1)A = −A
Exam Tip — Scalars pass through everything
A scalar can be pulled out of a sum, a transpose or a product freely: (kA)′ = kA′, and k(AB) = (kA)B = A(kB). This is a genuinely useful simplification — if a question gives you 2A and 5B, take the 10 outside straight away and multiply the small numbers instead of the big ones.
Before you multiply, write both orders side by side. The two inside numbers must be equal, and the two outside numbers tell you the size of the answer.
Before you learn how to multiply, you must learn whether you are allowed to. Unlike addition, where the two matrices must be exactly the same shape, multiplication has a lopsided condition that catches everyone the first time. Look at the picture above and burn it into your memory, because it answers about a third of all the multiplication questions you will ever be asked.
Key Rule — The compatibility condition
The product AB is defined only when the number of columns of A equals the number of rows of B. If A is of order m × n and B is of order n × p, then AB exists and is of order m × p. Write the two orders side by side: the inner numbers must match, and the outer numbers give the order of the answer.
Say it as a sentence and it is much easier to hold on to: columns of the first must equal rows of the second. When that holds, we say A and B are conformable for the product AB. And note carefully that AB being defined tells you nothing about whether BA is defined — those are two separate questions.
Example 16 — Deciding which products exist
Question. A is of order 3 × 4, B is of order 4 × 2, C is of order 2 × 3 and D is of order 3 × 4. State which of AB, BA, BC, CA, AD and (AB)C are defined, and give the order of each one that is.
Working. Write the two orders down and look at the inner pair each time.
AB: (3 × 4)(4 × 2) — inner 4 and 4 match → defined, order 3 × 2 BA: (4 × 2)(3 × 4) — inner 2 and 3 do not match → not defined BC: (4 × 2)(2 × 3) — match → defined, order 4 × 3 CA: (2 × 3)(3 × 4) — match → defined, order 2 × 4 AD: (3 × 4)(3 × 4) — 4 and 3 do not match → not defined (AB)C: AB is 3 × 2, so (3 × 2)(2 × 3) — match → defined, order 3 × 3
Why it works. Look at AB and BA together. AB exists, BA does not. That is your first real warning that matrix multiplication is not going to behave like ordinary multiplication, and we will meet that theme properly two sections from now. Also notice AD: two matrices of exactly the same order cannot always be multiplied. Same shape is the rule for addition, not for multiplication.
Exam Tip — Do the order check before any arithmetic
In the exam, the very first line of any multiplication answer should be the order check. It takes five seconds, it earns you method marks, and it stops you from spending six minutes producing an answer to a product that does not exist. If it is not defined, say so in a full sentence and move on with a clear conscience.
Every entry of AB is one row of A walked along one column of B: multiply the pairs, then add. Row number picks the row of the answer, column number picks the column.
Here is the one genuinely new skill in this chapter, and it is worth taking slowly. To find a single entry of the product, you take one row from the first matrix and one column from the second matrix, multiply them together term by term, and add up the results. That single number then goes into the answer at the position given by which row and which column you used.
Key Rule — How to build the product AB
If A is m × n and B is n × p, then the entry in row i, column j of AB is
ai1b1j + ai2b2j + ai3b3j + … + ainbnj
In words: row i of A, dotted with column j of B. Walk along the row and down the column at the same pace, multiply each pair you meet, then add everything up.
A picture that helps a lot of students: imagine your left index finger crawling along a row of A while your right index finger crawls down a column of B. Every time both fingers land, multiply that pair and note it down. When your fingers run out, add all the notes together. That total is one entry of the answer. Then move a finger and do it again.
Example 17 — A first full multiplication
Question. Find AB where A =
1
−2
3
4
0
5
and B =
2
1
−1
3
0
4
.
Step 1 — the order check. A is 2 × 3 and B is 3 × 2. The inner numbers are 3 and 3, so AB exists and has order 2 × 2. You are looking for four numbers.
Why it works. Watch how the labels do the filing for you. “Row 1 with column 2” produces the number that belongs in row 1, column 2 — you never have to wonder where a number goes. Keep the four calculations in a neat list like this and your working stays checkable, both by you and by the examiner.
Example 18 — The same two matrices, multiplied the other way round
Question. For the same A and B, find BA.
Working. Now B comes first: B is 3 × 2 and A is 2 × 3, inner numbers 2 and 2 match, so BA exists and is of order 3 × 3. Nine entries this time.
Row 1 of B is (2, 1): (2)(1)+(1)(4) = 6, (2)(−2)+(1)(0) = −4, (2)(3)+(1)(5) = 11 Row 2 of B is (−1, 3): (−1)(1)+(3)(4) = 11, (−1)(−2)+(3)(0) = 2, (−1)(3)+(3)(5) = 12 Row 3 of B is (0, 4): (0)(1)+(4)(4) = 16, (0)(−2)+(4)(0) = 0, (0)(3)+(4)(5) = 20
BA =
6
−4
11
11
2
12
16
0
20
Why it works. Compare this with the previous example. AB was a 2 × 2 matrix and BA is a 3 × 3 matrix. They are not merely different numbers — they are different sizes. This is the clearest possible proof that AB and BA are two entirely separate objects, and it is why the order of the letters in a matrix product is never decoration.
Example 19 — A 3 × 3 product
Question. Find AB where A =
1
2
0
0
1
3
2
0
1
and B =
1
0
2
3
1
0
0
2
1
.
Working. Both are 3 × 3, so AB is 3 × 3 and there are nine entries to find. Take the rows of A one at a time and run each against all three columns of B.
Row 1 of A is (1, 2, 0): (1)(1)+(2)(3)+(0)(0) = 7, (1)(0)+(2)(1)+(0)(2) = 2, (1)(2)+(2)(0)+(0)(1) = 2 Row 2 of A is (0, 1, 3): (0)(1)+(1)(3)+(3)(0) = 3, (0)(0)+(1)(1)+(3)(2) = 7, (0)(2)+(1)(0)+(3)(1) = 3 Row 3 of A is (2, 0, 1): (2)(1)+(0)(3)+(1)(0) = 2, (2)(0)+(0)(1)+(1)(2) = 2, (2)(2)+(0)(0)+(1)(1) = 5
AB =
7
2
2
3
7
3
2
2
5
Why it works. Nothing new has happened — it is the same row-meets-column move done nine times instead of four. The discipline that keeps you accurate is to finish an entire row of the answer before you start the next one. Jumping around the grid is how entries end up in the wrong boxes.
Example 20 — Matrix multiplication doing a real job
Question. Two school shops sell notebooks, pens and geometry boxes. The numbers sold last week were:
N =
12
8
5
10
15
4
where row 1 is Shop A, row 2 is Shop B, and the columns are notebooks, pens and geometry boxes in that order. The prices in rupees are 40, 15 and 120 respectively. Use matrix multiplication to find each shop’s total revenue.
Working. Put the prices in a column matrix so that the orders fit: N is 2 × 3 and the price matrix P is 3 × 1, so NP is 2 × 1 — one total per shop, which is exactly what we want.
NP =
12
8
5
10
15
4
40
15
120
=
12(40) + 8(15) + 5(120)
10(40) + 15(15) + 4(120)
=
1200
1105
So Shop A took in ₹1200 and Shop B took in ₹1105.
Why it works. This is the reason matrix multiplication is defined in that odd row-times-column way in the first place. Multiplying quantities by prices and adding them up is exactly what a bill is, and the row-meets-column rule does every bill in the table at once. If you ever wonder why anyone invented this operation, remember this example.
Common Mistake — Multiplying entry by entry
The tempting mistake is to do what you did for addition — multiply the matching boxes. That is not matrix multiplication and it will earn zero marks. Matrix multiplication always involves a row of the first matrix meeting a whole column of the second, with a sum at the end. If your working has no plus signs in it, something has gone wrong.
Exam Tip — Check the order of your answer before you start
Write the expected order of the product at the top of your working, for instance “AB will be 2 × 4”. Then count the entries you actually produced. If you expected eight numbers and wrote six, you know immediately that you dropped something, and you know exactly where to look.
Matrix multiplication keeps most of the properties you are used to. It is a short list, and the honest way to learn it is to notice which familiar property is missing — that missing one gets its own section next.
Property
Statement
Condition on orders
Associative
(AB)C = A(BC)
All the products must be defined
Distributive from the left
A(B + C) = AB + AC
B and C of the same order
Distributive from the right
(A + B)C = AC + BC
A and B of the same order
Multiplicative identity
AI = IA = A
I of the matching order
Multiplying by the zero matrix
AO = OA = O
O of the matching order
Scalars slide through
k(AB) = (kA)B = A(kB)
k any real number
Transpose of a product
(AB)′ = B′A′
Order reverses
Example 21 — Verifying associativity and distributivity
Question. With A =
1
2
3
1
, B =
0
1
2
−1
and C =
1
−1
1
2
, verify that (AB)C = A(BC) and that A(B + C) = AB + AC.
Part 1 — associativity.
AB =
4
−1
2
2
so (AB)C =
4
−1
2
2
1
−1
1
2
=
3
−6
4
2
BC =
1
2
1
−4
so A(BC) =
1
2
3
1
1
2
1
−4
=
3
−6
4
2
The two agree, so (AB)C = A(BC) ✓.
Part 2 — distributivity. First B + C =
1
0
3
1
, so
A(B + C) =
1
2
3
1
1
0
3
1
=
7
2
6
1
And separately AB =
4
−1
2
2
and AC =
3
3
4
−1
, so AB + AC =
7
2
6
1
. The two agree ✓.
Why it works. Associativity is genuinely useful, not just decorative. It means that in a product of three matrices you may bracket wherever it is easiest — and very often one pairing gives you much smaller numbers than the other. What you may never do is change the left-to-right order of the letters. Associativity moves brackets; it does not move matrices.
Common Mistake — Confusing associativity with commutativity
(AB)C = A(BC) says you can re-bracket. It absolutely does not say ABC = BAC or CBA. In matrix algebra the sequence of the letters is fixed for life; only the brackets are free to move. Keep that distinction sharp and you will avoid the most expensive error in this chapter.
This is the section where matrices stop pretending to be numbers. Two of the deepest habits you have from ordinary arithmetic simply stop being true, and the CBSE syllabus names both of them explicitly. Read this section slowly; it produces short, high-value board questions almost every year.
Key Rule — Matrix multiplication is not commutative
For ordinary numbers, 3 × 5 = 5 × 3 always. For matrices, AB and BA are generally different. Sometimes only one of them is even defined; sometimes both are defined but have different orders; and sometimes both are defined, both are the same order, and they are still different matrices. So AB ≠ BA in general.
Example 22 — Two matrices that refuse to commute
Question. Show that AB ≠ BA for A =
1
2
3
4
and B =
0
1
1
0
.
Working. Both are 2 × 2, so both products exist and both are 2 × 2.
AB =
1
2
3
4
0
1
1
0
=
2
1
4
3
BA =
0
1
1
0
1
2
3
4
=
3
4
1
2
Since
2
1
4
3
≠
3
4
1
2
, we conclude AB ≠ BA.
Why it works. This B is the matrix that swaps things around. Multiplying on the right swaps the columns of A; multiplying on the left swaps the rows of A. Swapping the columns of a matrix and swapping its rows are obviously two different actions, so of course the two products differ. Whenever a question asks you to “show that AB ≠ BA”, you only need one example worked out in full — compute both, display both, and state clearly that they are unequal.
Key Rule — Non-zero matrices can multiply to give the zero matrix
For ordinary numbers, if xy = 0 then x = 0 or y = 0. For matrices this fails completely: there exist non-zero matrices A and B with AB = O. The syllabus asks you to know this for square matrices of order 2, so a 2 × 2 example is all you ever need to produce.
Example 23 — A product of two non-zero matrices that vanishes
Question. Show that there exist non-zero 2 × 2 matrices A and B with AB = O.
Working. Take A =
1
−1
−1
1
and B =
1
1
1
1
. Neither is the zero matrix — every entry of both is −1 or 1.
Here BA also happens to be O, but that is a coincidence of this particular pair, not a rule.
Why it works. Every entry of AB is a row of A dotted with a column of B, and each of those dot products came out as 1 − 1 = 0. The two matrices are not zero, but they are arranged so that every row of the first cancels itself against every column of the second. A second pair worth remembering: A =
2
4
1
2
and B =
2
−4
−1
2
also give AB = O.
Example 24 — A matrix whose square is zero
Question. Let N =
0
1
0
0
. Show that N ≠ O but N² = O.
Working. N is clearly not the zero matrix, because n12 = 1. Now
N² =
0
1
0
0
0
1
0
0
=
(0)(0)+(1)(0)
(0)(1)+(1)(0)
(0)(0)+(0)(0)
(0)(1)+(0)(0)
=
0
0
0
0
= O
Why it works. For an ordinary number, x² = 0 forces x = 0. For matrices it does not. This is the sharpest possible illustration that you cannot “take square roots” of a matrix equation or cancel matrices the way you cancel numbers. Keep this tiny matrix N in your memory — it is the standard counterexample and it fits on one line.
Common Mistake — Cancelling a matrix from both sides
Because AB = O does not force A = O or B = O, you also cannot cancel: AB = AC does not imply B = C. A concrete counterexample: with A =
2
4
1
2
, B =
1
1
1
1
and C =
3
3
0
0
, both AB and AC come out as
6
6
3
3
, yet B ≠ C. Cancellation only becomes legal when A is invertible, which is exactly what the next section is about.
With ordinary numbers, the reciprocal of 5 is the number that multiplies with 5 to give 1. The inverse of a matrix is exactly the same idea, with the identity matrix I playing the role of 1. And just as 0 has no reciprocal, some matrices have no inverse at all — which turns out to be the interesting part of the story.
Key Rule — Invertible matrix and inverse
A square matrix A of order n is called invertible if there exists a square matrix B of the same order n such that
AB = BA = In
In that case B is called the inverse of A and we write B = A−1. Both products must equal I — you cannot check only one of them and declare victory.
Two things follow immediately from that definition, and both are worth saying out loud. First, only square matrices can be invertible, because AB and BA both have to make sense and both have to come out as the identity. Second, being square is not enough — plenty of square matrices have no inverse. A square matrix with no inverse is called singular; one that does have an inverse is called non-singular.
Exam Tip — Where the actual computing of an inverse lives
Read this carefully, because it saves a lot of confusion. In the Matrices chapter your job is the definition of invertibility, the verification that a given B really is A−1, and the proof that the inverse is unique. Actually finding A−1 from scratch belongs to the Determinants chapter, where you will learn the adjoint formula A−1 = (1/|A|) adj A, and the test that A is invertible exactly when |A| ≠ 0. So if a question here asks you to “show that B is the inverse of A”, it wants two multiplications, not a formula.
Example 25 — Verifying that one matrix is the inverse of another
Question. Show that B =
3
−5
−1
2
is the inverse of A =
2
5
1
3
.
Working. Both are 2 × 2, so both products exist and both should come out as I2.
Why it works. This is a two-multiplication question and nothing more. Write both products, show both equal I, and finish with the conclusion sentence. Students lose a mark surprisingly often by computing only AB — the definition demands both, so give both.
Now the theorem the syllabus asks for by name. It is short, it is elegant, and once you have seen the trick you will never forget it.
Key Rule — Uniqueness of the inverse
If a matrix has an inverse, that inverse is unique.
Proof. Let A be an invertible matrix of order n, and suppose both B and C are inverses of A. By definition that means
AB = BA = I and AC = CA = I
Now watch what happens to B:
B = BI = B(AC) = (BA)C = IC = C
Reading the chain left to right: I is the multiplicative identity so BI = B; we replaced I by AC because C is an inverse of A; we re-bracketed using associativity; we replaced BA by I because B is an inverse of A; and finally IC = C. Therefore B = C, so the two supposed inverses were the same matrix all along. Hence the inverse, when it exists, is unique. □
Exam Tip — How to earn every mark on the uniqueness proof
Examiners look for four things: (1) the assumption “let B and C both be inverses of A”, (2) writing down AB = BA = I and AC = CA = I, (3) the chain B = BI = B(AC) = (BA)C = IC = C, and (4) the conclusion. Notice that the proof leans entirely on two properties you already know — the identity property and associativity. It never uses commutativity, which is just as well, since matrices do not have it.
Example 26 — A square matrix with no inverse at all
Question. Show that A =
2
4
1
2
is not invertible.
Working. Take the non-zero matrix C =
2
−4
−1
2
and multiply:
AC =
2
4
1
2
2
−4
−1
2
=
(2)(2)+(4)(−1)
(2)(−4)+(4)(2)
(1)(2)+(2)(−1)
(1)(−4)+(2)(2)
=
0
0
0
0
= O
Now suppose, for contradiction, that A had an inverse A−1. Multiply AC = O on the left by A−1:
A−1(AC) = A−1O → (A−1A)C = O → IC = O → C = O
But C is plainly not the zero matrix. That is a contradiction, so no such A−1 can exist and A is singular.
Why it works. This little argument is worth understanding rather than memorising, because it links the two big surprises of this chapter. A matrix that can multiply something non-zero down to O can never be undone — the information is destroyed and no matrix can bring it back. (In the Determinants chapter you will meet the quick test: here |A| = (2)(2) − (4)(1) = 0, and zero determinant means singular.)
Example 27 — Why cancellation becomes legal once A is invertible
Question. Earlier we saw that AB = AC does not force B = C. Show that it does force B = C when A is invertible.
Working. Suppose A is invertible and AB = AC. Multiply both sides on the left by A−1 — and be careful to multiply on the same side of both, since matrices do not commute:
A−1(AB) = A−1(AC) → (A−1A)B = (A−1A)C → IB = IC → B = C
A concrete check. Take the invertible A =
2
5
1
3
with A−1 =
3
−5
−1
2
, and B =
1
0
2
3
. Then AB =
12
15
7
9
, and multiplying that on the left by A−1 returns
1
0
2
3
= B, exactly as the argument promises.
Why it works. Compare this with the counterexample in the previous section, where A =
2
4
1
2
was singular and cancellation failed. Invertibility is precisely the property that makes matrix algebra behave like ordinary algebra again. That is the whole reason inverses matter.
Exam Tip — A bonus result: the inverse of a product also reverses
If A and B are both invertible matrices of the same order, then AB is invertible and (AB)−1 = B−1A−1. The check is one line: (AB)(B−1A−1) = A(BB−1)A−1 = AIA−1 = AA−1 = I, and similarly the other way round. Notice the reversal again, exactly as with the transpose — the everyday version is that to undo “socks then shoes” you must take off the shoes first.
Common Mistake — Saying a non-square matrix has an inverse
Only square matrices can be invertible. If a question hands you a 2 × 3 matrix and asks for its inverse, the correct answer is that the inverse does not exist because the matrix is not square. Also resist writing A/B or B/A — there is no division of matrices anywhere in mathematics. Write AB−1 or B−1A instead, and be clear which side you meant.
If you are working from an older book, a senior’s notes, or a video made a few years ago, you will run into a topic that is no longer part of your course. I want to name it clearly so that you neither waste time on it nor panic when you see it.
Common Mistake — Removed from the CBSE 2026-27 Class 12 Maths syllabus
Elementary row and column operations, and finding the inverse of a matrix by elementary operations, are not in the current syllabus. You will still find them in older editions, in second-hand notes and in plenty of videos online. They are not examinable for your board paper. If you see a question asking you to “find A−1 using elementary transformations”, it is from an out-of-date paper.
So how are you expected to find an inverse? Through the Determinants chapter, using the adjoint: A−1 = (1/|A|) adj A, valid whenever |A| ≠ 0. That is the method your board paper will use. Within the Matrices chapter itself, inverses appear only as a definition, a verification exercise, and the uniqueness proof you met above.
Key Rule — Exactly what Matrices covers in 2026-27
Concept, notation, order, equality and types of matrices; zero and identity matrices; transpose; symmetric and skew symmetric matrices; addition, scalar multiplication and matrix multiplication with their simple properties; non-commutativity of multiplication and the existence of non-zero matrices whose product is the zero matrix (restricted to square matrices of order 2); and invertible matrices with the proof of uniqueness of the inverse. All matrices have real entries. This chapter sits in Unit II (Algebra), which carries 10 marks shared with Determinants.
Use this the night before a test. If any line here makes you pause, scroll back up to that section rather than trying to re-learn it from the summary — summaries are for reminding, not for teaching.
Idea
What you must be able to say or do
Order
Rows × columns, in that order. An m × n matrix has mn elements.
aij
Row i, column j. Row first, always.
Equality
Same order and every corresponding entry equal.
Square family
Identity ⊂ scalar ⊂ diagonal ⊂ square. Diagonal means every off-diagonal entry is 0.
Zero matrix O
All entries 0, any order. A + O = A and AO = OA = O.
Identity I
1s on the principal diagonal, 0s elsewhere. AI = IA = A.
Transpose
Rows become columns. (A′)′ = A, (A + B)′ = A′ + B′, (kA)′ = kA′, and (AB)′ = B′A′.
Symmetric
A′ = A, so aij = aji. Square only.
Skew symmetric
A′ = −A, so aij = −aji and every diagonal entry is 0.
Decomposition
A = ½(A + A′) + ½(A − A′); the first part is symmetric, the second skew symmetric, and the split is unique.
Addition
Same order required. Entry by entry. Commutative and associative.
Scalar multiple
kA scales every entry. Order unchanged. No condition needed.
Product defined?
Columns of the first = rows of the second. (m × n)(n × p) gives m × p.
Product entry
Row i of A dotted with column j of B, then summed.
Multiplication properties
Associative and distributive; AI = IA = A; AO = OA = O. Not commutative.
Two big surprises
AB ≠ BA in general, and AB = O is possible with A ≠ O and B ≠ O.
Invertible
A is invertible if there is B with AB = BA = I. Only square matrices qualify; the inverse is unique.
Uniqueness proof
B = BI = B(AC) = (BA)C = IC = C.
Not in the syllabus
Elementary row and column operations, and inverse by elementary operations.
Ten questions, arranged roughly in the order you met the ideas above. Work each one on paper first — properly, with the working written out — and only then open the answer. If you read the solution before attempting the question, you will feel like you understand it and then freeze in the exam. That gap is real, and writing first is the only thing that closes it.
Q1.
Construct the 3 × 2 matrix A = [aij] whose elements are given by aij = (i + 2j)2.
Show Answer
The order is 3 × 2, so i runs 1 to 3 and j runs 1 to 2. Substituting position by position:
(a) A matrix has 24 elements. What are all its possible orders? (b) What if it has 13 elements instead?
Show Answer
(a) The order m × n must satisfy mn = 24, so every factor pair of 24 gives a valid order. The divisors of 24 are 1, 2, 3, 4, 6, 8, 12 and 24, so there are 8 possible orders: 1 × 24, 2 × 12, 3 × 8, 4 × 6, 6 × 4, 8 × 3, 12 × 2 and 24 × 1.
(b) 13 is prime, so its only divisors are 1 and 13. There are 2 possible orders: 1 × 13 and 13 × 1. In general the number of possible orders equals the number of divisors of the number of elements.
Q3.
Find x and y if
3x − y
8
2
x + 3y
=
5
8
2
15
.
Show Answer
Both matrices are 2 × 2, so compare corresponding entries. The entries 8 and 2 match automatically, and the other two give
3x − y = 5 …(i) x + 3y = 15 …(ii)
From (i), y = 3x − 5. Substituting into (ii): x + 3(3x − 5) = 15, so 10x − 15 = 15, giving 10x = 30 and x = 3. Then y = 3(3) − 5 = 4.
Check: 3(3) − 4 = 5 ✓ and 3 + 3(4) = 15 ✓.
Q4.
If A =
1
−2
4
3
0
−1
and B =
2
1
−3
0
5
2
, find 3A − 2B.
Show Answer
Scale each matrix first:
3A =
3
−6
12
9
0
−3
2B =
4
2
−6
0
10
4
Now subtract entry by entry, taking care with the double negatives (12 − (−6) = 18 and −3 − 4 = −7):
3A − 2B =
−1
−8
18
9
−10
−7
Q5.
For A =
2
1
3
−1
and B =
1
4
2
0
, find AB and BA. What do your answers show?
Show Answer
Both are 2 × 2, so both products exist and both are 2 × 2.
, which is not (AB)′. The reversal in the law is essential, not cosmetic.
Q7.
(a) Express A =
4
3
−1
1
5
7
3
1
2
as the sum of a symmetric and a skew symmetric matrix. (b) Prove that for any square matrix A, the matrix A + A′ is symmetric.
Show Answer
(a) First write the transpose:
A′ =
4
1
3
3
5
1
−1
7
2
A + A′ =
8
4
2
4
10
8
2
8
4
A − A′ =
0
2
−4
−2
0
6
4
−6
0
P = ½(A + A′) =
4
2
1
2
5
4
1
4
2
Q = ½(A − A′) =
0
1
−2
−1
0
3
2
−3
0
P is symmetric (its mirror pairs match), Q is skew symmetric (zero diagonal and opposite pairs), and adding them entry by entry returns A exactly. So A = P + Q.
(b) Let S = A + A′. Then
S′ = (A + A′)′ = A′ + (A′)′ = A′ + A = A + A′ = S
using the transpose-of-a-sum law, the fact that transposing twice returns the original, and the commutativity of matrix addition. Since S′ = S, the matrix A + A′ is symmetric. □
Q8.
(a) Find a non-zero 2 × 2 matrix B such that AB = O, where A =
1
2
2
4
. (b) Show that N =
0
2
0
0
satisfies N ≠ O but N² = O, and explain why N cannot be invertible.
(Any B whose two columns are multiples of the column
2
−1
will work; for instance
2
2
−1
−1
does too.)
(b) N ≠ O because n12 = 2. And
N² =
0
2
0
0
0
2
0
0
=
0
0
0
0
= O
If N were invertible, then multiplying N² = O on the left by N−1 would give (N−1N)N = N−1O, that is IN = O, that is N = O — contradicting N ≠ O. So N is not invertible.
Q9.
Show that B =
2
−1
−5
3
is the inverse of A =
3
1
5
2
, and state what the uniqueness theorem then tells you.
Show Answer
Compute both products, as the definition requires.
So A is invertible and A−1 = B. By the uniqueness theorem, this is the only matrix with that property — no other matrix can satisfy AC = CA = I. (The proof: if C were another inverse, C = CI = C(AB) = (CA)B = IB = B.)
Q10.
Three canteens sell samosas, sandwiches and juice boxes. Last Friday they sold:
Q =
25
10
6
18
14
9
30
8
4
where the rows are Canteen 1, 2 and 3 and the columns are samosas, sandwiches and juice boxes. The prices are ₹30, ₹45 and ₹110 respectively. Use matrix multiplication to find each canteen’s takings.
Show Answer
Write the prices as a 3 × 1 column matrix P so that the orders are conformable: Q is 3 × 3 and P is 3 × 1, so QP is 3 × 1 — one total per canteen.
QP =
25
10
6
18
14
9
30
8
4
30
45
110
=
25(30) + 10(45) + 6(110)
18(30) + 14(45) + 9(110)
30(30) + 8(45) + 4(110)
=
1860
2160
1700
So Canteen 1 took ₹1860, Canteen 2 took ₹2160 and Canteen 3 took ₹1700. Notice that putting the prices in a column is what makes the multiplication legal; a row matrix of prices would not be conformable with Q on the right.
That is the whole chapter. If you got seven or more of those right on the first attempt, you are in genuinely good shape for the board paper. If you got three, that is fine too — go back to the two sections that tripped you up, re-read the worked examples with a pen in your hand, and try again tomorrow.
And here is the only study rule I really believe in. Do not aim to “finish matrices” tonight. Aim for one more correct question than you managed yesterday. One extra, every day. It sounds far too small to matter, and that is exactly why it works — it is small enough that you will actually do it, and by the time the exam arrives you will have quietly become the student who finds this chapter easy. Be patient with yourself, and keep going.
Kaizen Close
Redo one matrix product and write the dimensions beside every factor. One dimension check today will prevent most avoidable multiplication errors.
Continue Learning
More chapters from this subject — keep the momentum going.
