Matrices ICSE Class 10: Order, Operations and Matrix Multiplication

What Matrices is about

Weightage: CISCE publishes no chapter or unit weightage for ICSE Mathematics. Matrices is in the 2. Algebra (v) Matrices part of the 2027 syllabus and can be asked in both Section A and Section B of the 80-mark paper.Source: CISCE ICSE Mathematics syllabus, examination year 2027

Key concepts

Order; row, column and square matrices

The order of a matrix is (number of rows) × (number of columns). A matrix with one row is a row matrix (order $1\times n$); one with one column is a column matrix ($m\times 1$); one with equal numbers of rows and columns is a square matrix.

The entry in row $i$ and column $j$ is written $a_{ij}$. Two matrices are equal only if they have the same order and every pair of corresponding entries is equal; this gives one equation per entry.

Worked example

State the order of $\left[\begin{array}{c}3\\ -2\end{array}\right]$ and $\left[\begin{array}{ccc}1& 0& 5\\ 2& 7& -3\end{array}\right]$. Find $x$ and $y$ if $\left[\begin{array}{cc}x+2& 3\\ 4& y-1\end{array}\right]=\left[\begin{array}{cc}5& 3\\ 4& 6\end{array}\right]$.

Try it first, then show the solutionSolution

1. The first has 2 rows and 1 column: order $2\times 1$ (a column matrix).
2. The second has 2 rows and 3 columns: order $2\times 3$.
3. Equal matrices: $x+2=5$ and $y-1=6$, so $x=3$, $y=7$.

Answer: 2 × 1 and 2 × 3; x = 3, y = 7

Null and identity matrices

A null (zero) matrix $O$ has every entry 0. Adding it changes nothing: $A+O=A$.

The identity matrix of order 2 is $I=\left[\begin{array}{cc}1& 0\\ 0& 1\end{array}\right]$: 1s on the leading diagonal, 0s elsewhere. Multiplying by it changes nothing: $AI=IA=A$. So $I$ behaves like the number 1 and $O$ like the number 0.

Careful: unlike numbers, $A^2=O$ does not force $A=O$; for example ${\left[\begin{array}{cc}0& 1\\ 0& 0\end{array}\right]}^2=O$.

Worked example

Find $X$ if $2X+\left[\begin{array}{cc}1& 2\\ 3& 4\end{array}\right]=\left[\begin{array}{cc}5& 0\\ -1& 8\end{array}\right]$.

Try it first, then show the solutionSolution

1. $2X=\left[\begin{array}{cc}5& 0\\ -1& 8\end{array}\right]-\left[\begin{array}{cc}1& 2\\ 3& 4\end{array}\right]=\left[\begin{array}{cc}4& -2\\ -4& 4\end{array}\right]$.
2. $X=\frac{1}{2}\left[\begin{array}{cc}4& -2\\ -4& 4\end{array}\right]=\left[\begin{array}{cc}2& -1\\ -2& 2\end{array}\right]$.

Answer: $X=\left[\begin{array}{cc}2& -1\\ -2& 2\end{array}\right]$

Addition and subtraction

Two matrices can be added or subtracted only if they have the same order (they are then compatible for addition). Add or subtract the corresponding entries.

Addition is commutative, $A+B=B+A$, and associative. Subtraction is not commutative: $A-B=-(B-A)$.

Worked example

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

Try it first, then show the solutionSolution

1. $A+B=\left[\begin{array}{cc}2+1& -1+5\\ 3-2& 4+0\end{array}\right]=\left[\begin{array}{cc}3& 4\\ 1& 4\end{array}\right]$.
2. $A-B=\left[\begin{array}{cc}2-1& -1-5\\ 3+2& 4-0\end{array}\right]=\left[\begin{array}{cc}1& -6\\ 5& 4\end{array}\right]$.

Answer: $A+B=\left[\begin{array}{cc}3& 4\\ 1& 4\end{array}\right],A-B=\left[\begin{array}{cc}1& -6\\ 5& 4\end{array}\right]$

Multiplying by a number (scalar multiplication)

To multiply a matrix by a number $k$, multiply every entry by $k$. This is how expressions like $3A-2B$ are built: find $3A$ and $2B$ first, then subtract.

The examiners noted in 2025 that some candidates multiplied only some entries of $B$ by 5. Write the full matrix of the multiple before combining.

Worked example

For the same $A$ and $B$, find $3A-2B$.

Try it first, then show the solutionSolution

1. $3A=\left[\begin{array}{cc}6& -3\\ 9& 12\end{array}\right]$, $2B=\left[\begin{array}{cc}2& 10\\ -4& 0\end{array}\right]$.
2. $3A-2B=\left[\begin{array}{cc}4& -13\\ 13& 12\end{array}\right]$.

Answer: $\left[\begin{array}{cc}4& -13\\ 13& 12\end{array}\right]$

Compatibility for multiplication and the order of a product

The product $AB$ is defined only if the **number of columns of $A$ equals the number of rows of $B$**. If $A$ is $m\times n$ and $B$ is $n\times p$, then $AB$ is $m\times p$ (the outer numbers).

So a $2\times 2$ matrix can multiply a $2\times 1$ column on its right, giving a $2\times 1$ matrix; a $1\times 2$ row can multiply a $2\times 2$ matrix on its right, giving $1\times 2$. But a $2\times 1$ matrix times a $2\times 2$ matrix is not defined, because 1 ≠ 2.

Worked example

A is 2 × 3 and B is 3 × 1. Is AB defined? Is BA defined? Give the order where defined.

Try it first, then show the solutionSolution

1. AB: columns of A (3) = rows of B (3), so AB is defined, of order 2 × 1.
2. BA: columns of B (1) ≠ rows of A (2), so BA is not defined.

Answer: AB is 2 × 1; BA is not defined.

Multiplying 2 × 2 matrices: row by column

For $A=\left[\begin{array}{cc}a& b\\ c& d\end{array}\right]$ and $B=\left[\begin{array}{cc}p& q\\ r& s\end{array}\right]$,
$AB=\left[\begin{array}{cc}ap+br& aq+bs\\ cp+dr& cq+ds\end{array}\right].$
Entry (row $i$, column $j$) of $AB$ = row $i$ of $A$ times column $j$ of $B$: multiply in pairs and add.

In general $AB\ne BA$, so the order of multiplication matters. $A^2$ means $A\times A$, found by the same rule, never by squaring the entries.

Worked example

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

Try it first, then show the solutionSolution

1. $AB=\left[\begin{array}{cc}2(1)+(-1)(-2)& 2(5)+(-1)(0)\\ 3(1)+4(-2)& 3(5)+4(0)\end{array}\right]=\left[\begin{array}{cc}4& 10\\ -5& 15\end{array}\right]$.
2. $BA=\left[\begin{array}{cc}1(2)+5(3)& 1(-1)+5(4)\\ -2(2)+0(3)& -2(-1)+0(4)\end{array}\right]=\left[\begin{array}{cc}17& 19\\ -4& 2\end{array}\right]$.
3. So $AB\ne BA$.

Answer: $AB=\left[\begin{array}{cc}4& 10\\ -5& 15\end{array}\right],BA=\left[\begin{array}{cc}17& 19\\ -4& 2\end{array}\right]$

Matrix equations and expressions like A² − kA

Many questions ask you to show that a matrix satisfies an equation such as $A^2-3A-4I=O$, or to find a number $k$ that makes one true. Work out $A^2$ by row-by-column multiplication, then the scalar multiples, then combine entry by entry. Remember that a number on its own must be written as a multiple of $I$ (e.g. $4I$), because you cannot add a number to a matrix.

To find unknown entries from a product, multiply out, then equate corresponding entries.

Worked example

If $A=\left[\begin{array}{cc}1& 2\\ 3& 2\end{array}\right]$, show that $A^2-3A=4I$.

Try it first, then show the solutionSolution

1. $A^2=\left[\begin{array}{cc}1+6& 2+4\\ 3+6& 6+4\end{array}\right]=\left[\begin{array}{cc}7& 6\\ 9& 10\end{array}\right]$.
2. $3A=\left[\begin{array}{cc}3& 6\\ 9& 6\end{array}\right]$.
3. $A^2-3A=\left[\begin{array}{cc}4& 0\\ 0& 4\end{array}\right]=4I$.

Answer: Shown.

Maths in real life: Matrices

Each concept, met in an everyday situation. Read the situation, try the problem, then check the solution. The one-line hook is what to remember.

Topics in this chapter

• Multiplication of Matrices
• Matrix Equations and Equality of Matrices
• Addition, Subtraction and Scalar Multiplication of Matrices

Matrices formula sheet

Order

$m\times n=\text{rows}\times \text{columns}$

Row matrix 1 × n; column matrix m × 1.

Identity and null matrices (order 2)

$I=\left[\begin{array}{cc}1& 0\\ 0& 1\end{array}\right];O=\left[\begin{array}{cc}0& 0\\ 0& 0\end{array}\right]$

AI = IA = A; A + O = A.

Addition

$\left[\begin{array}{cc}a& b\\ c& d\end{array}\right]+\left[\begin{array}{cc}p& q\\ r& s\end{array}\right]=\left[\begin{array}{cc}a+p& b+q\\ c+r& d+s\end{array}\right]$

Same order only.

Scalar multiplication

$k\left[\begin{array}{cc}a& b\\ c& d\end{array}\right]=\left[\begin{array}{cc}ka& kb\\ kc& kd\end{array}\right]$

Every entry is multiplied.

Product of 2 × 2 matrices

$\left[\begin{array}{cc}a& b\\ c& d\end{array}\right]\left[\begin{array}{cc}p& q\\ r& s\end{array}\right]=\left[\begin{array}{cc}ap+br& aq+bs\\ cp+dr& cq+ds\end{array}\right]$

Row by column.

2 × 2 times 2 × 1

$\left[\begin{array}{cc}a& b\\ c& d\end{array}\right]\left[\begin{array}{c}x\\ y\end{array}\right]=\left[\begin{array}{c}ax+by\\ cx+dy\end{array}\right]$

Result is 2 × 1.

Compatibility for AB

$(m\times n)(n\times p)=m\times p$

Inner numbers must match.

How the board asks Matrices

The question patterns to expect, and the method that works for each.

Tricks to remember Matrices

Short memory aids. Say each one aloud once, then use it on a question the same day.

• Run along the row, dive down the column

  For each entry of AB, your left finger runs along a row of A while your right finger dives down a column of B.

  Use it when: Every multiplication.

• Inner numbers meet, outer numbers stay

  (2 × 3)(3 × 1): the inner 3s meet, so it works; the outer 2 and 1 give the order 2 × 1.

  Use it when: Compatibility and order questions.

• A squared is A times A

  Squaring a matrix is multiplying it by itself, not squaring its numbers.

  Use it when: A² MCQs (the most common lost mark).

• Numbers become nI

  In a matrix equation, a lone number like 7 must be written 7I.

  Use it when: A² − 5A + 7I = O type questions.

Matrices: board-level solved questions

Ordered from board level to HOTS. Cover the solution, write your own, then compare step by step.

Case-study practice (Section B style)

Read the situation, then answer every part. In the board paper this is a 10-mark question split into 3 + 3 + 4 marks.

Where the marks are

How a typical answer is split into step marks. Show every step: a right final answer alone rarely earns full marks.

Where students lose marks in Matrices

The slips examiners see most often, and what to write instead.

• ✗Mistake: Squaring each entry instead of multiplying the matrix by itself

  CISCE examiners’ comments, 2024 and 2025: for $A^2$, candidates squared each element, choosing the wrong option in both years.

  Wrong working

  ${\left[\begin{array}{cc}0& 1\\ 1& 0\end{array}\right]}^2=\left[\begin{array}{cc}{0}^2& 1^2\\ 1^2& 0^2\end{array}\right]=\left[\begin{array}{cc}0& 1\\ 1& 0\end{array}\right]$

  Right working

  $\left[\begin{array}{cc}0& 1\\ 1& 0\end{array}\right]\left[\begin{array}{cc}0& 1\\ 1& 0\end{array}\right]=\left[\begin{array}{cc}0+1& 0+0\\ 0+0& 1+0\end{array}\right]=\left[\begin{array}{cc}1& 0\\ 0& 1\end{array}\right]$

  ✓Do this: Write $A^2=A\times A$ and multiply row by column.

  Usually costs: 1 mark in Section A.

• ✗Mistake: Errors in a product used to find unknowns

  CISCE examiners’ comments, 2024: in $AB=C$, candidates made errors in the product (for example getting $4y=1$, so $y=\frac{1}{4}$).

  ✓Do this: Write each entry of the product as a sum in brackets before simplifying, then equate entries one by one.

  Usually costs: Most of a 4-mark question.

• ✗Mistake: Wrong matrices used, or wrong order of operations

  CISCE examiners’ comments, 2025: candidates found $A+B$ instead of $A+C$, and made errors in $B(A+C)$; they also did not multiply all elements of $B$ by 5.

  ✓Do this: Copy the expression, underline each matrix name, and do one operation per line, writing every matrix in full.

  Usually costs: 1 mark per wrong part.

• ✗Mistake: Assuming AB = BA

  Matrix multiplication is not commutative, but students reverse the order in expressions like B(A + C).

  ✓Do this: Keep the left matrix on the left; B(A + C) means B times (A + C).

  Usually costs: 1 to 2 marks.

• ✗Mistake: Adding a number to a matrix

  In $A^2-5A+7=O$ students add 7 to every entry.

  ✓Do this: Write the number as $7I$, which adds 7 only to the diagonal entries.

  Usually costs: 1 mark.

Practice questions on Matrices

Write each answer on paper first. Open the hint only if you are stuck, and the answer only when you have finished.

1. Q1 · 1 mark · MCQBoard level

   A matrix with 3 rows and 2 columns has order:
   (a) 2 × 3
   (b) 3 × 2
   (c) 6
   (d) 3 × 3

   Hint

   Rows first.

   Answer

   (b)

2. Q2 · 1 mark · MCQBoard level

   The matrix $\left[\begin{array}{cc}1& 0\\ 0& 1\end{array}\right]$ is called:
   (a) a null matrix
   (b) a row matrix
   (c) the identity matrix
   (d) a column matrix

   Hint

   1s on the diagonal.

   Answer

   (c)

3. Q3 · 1 mark · MCQHarder board level

   If $A=\left[\begin{array}{cc}1& 1\\ 0& 1\end{array}\right]$, then $A^2$ is:
   (a) $\left[\begin{array}{cc}1& 1\\ 0& 1\end{array}\right]$
   (b) $\left[\begin{array}{cc}1& 2\\ 0& 1\end{array}\right]$
   (c) $\left[\begin{array}{cc}2& 2\\ 0& 2\end{array}\right]$
   (d) $\left[\begin{array}{cc}1& 0\\ 0& 1\end{array}\right]$

   Hint

   Row by column.

   Answer

   (b) $\left[\begin{array}{cc}1& 2\\ 0& 1\end{array}\right]$

4. Q4 · 1 mark · Assertion-reasonBoard level

   Assertion (A): If $A$ is $2\times 2$ and $B$ is $2\times 1$, then $AB$ is defined.
   Reason (R): $AB$ is defined when the number of columns of $A$ equals the number of rows of $B$.
   (a) (A) is true and (R) is false.
   (b) (A) is false and (R) is true.
   (c) Both (A) and (R) are true, and (R) is the correct explanation for (A).
   (d) Both (A) and (R) are true, but (R) is not the correct explanation for (A).

   Hint

   Compare 2 and 2.

   Answer

   (c)

5. Q5 · 4 marks · Short answerHarder board level

   Find $X$ if $3X-\left[\begin{array}{cc}1& 2\\ 0& -3\end{array}\right]=\left[\begin{array}{cc}5& -8\\ 6& 9\end{array}\right]$.

   Hint

   Add, then divide every entry by 3.

   Answer

   $3X=\left[\begin{array}{cc}6& -6\\ 6& 6\end{array}\right]$, so $X=\left[\begin{array}{cc}2& -2\\ 2& 2\end{array}\right]$.

6. Q6 · 4 marks · Short answerHarder board level

   Find $x$ and $y$ if $\left[\begin{array}{cc}x& 2\\ 3& y\end{array}\right]\left[\begin{array}{c}1\\ -1\end{array}\right]=\left[\begin{array}{c}4\\ -2\end{array}\right]$.

   Hint

   The product is 2 × 1.

   Answer

   $x-2=4\Rightarrow x=6$; $3-y=-2\Rightarrow y=5$.

7. Q7 · 3 marks · Long answerBoard level

   If $A=\left[\begin{array}{cc}2& 1\\ 0& 3\end{array}\right]$ and $B=\left[\begin{array}{cc}1& 0\\ 2& -1\end{array}\right]$, find $AB$ and $BA$, and state whether $AB=BA$.

   Hint

   Row by column, in the given order.

   Answer

   $AB=\left[\begin{array}{cc}4& -1\\ 6& -3\end{array}\right]$, $BA=\left[\begin{array}{cc}2& 1\\ 4& -1\end{array}\right]$; $AB\ne BA$.

8. Q8 · 4 marks · Long answerHarder board level

   If $A=\left[\begin{array}{cc}3& -2\\ 4& -2\end{array}\right]$, find the value of $k$ for which $A^2=kA-2I$.

   Hint

   Compute A² and compare one convenient entry.

   Answer

   $A^2=\left[\begin{array}{cc}1& -2\\ 4& -4\end{array}\right]$; $kA-2I=\left[\begin{array}{cc}3k-2& -2k\\ 4k& -2k-2\end{array}\right]$. From $-2k=-2$, $k=1$; check: $3-2=1$, $4=4$, $-4=-4$. So $k=1$.

9. Q9 · 4 marks · Long answerHOTS / competency

   A $2\times 2$ matrix $M$ satisfies $M\left[\begin{array}{c}1\\ 1\end{array}\right]=\left[\begin{array}{c}3\\ 7\end{array}\right]$ and $M\left[\begin{array}{c}1\\ -1\end{array}\right]=\left[\begin{array}{c}-1\\ 1\end{array}\right]$. Find $M$.

   Hint

   Let $M=\left[\begin{array}{cc}a& b\\ c& d\end{array}\right]$ and form four equations.

   Answer

   $a+b=3$, $a-b=-1$: $a=1$, $b=2$. $c+d=7$, $c-d=1$: $c=4$, $d=3$. $M=\left[\begin{array}{cc}1& 2\\ 4& 3\end{array}\right]$.

10. Q10 · 4 marks · Long answerHarder board level

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

    Hint

    One matrix per line.

    Answer

    $A^2=\left[\begin{array}{cc}1& -3\\ 9& -2\end{array}\right]$, $AB=\left[\begin{array}{cc}2& 10\\ 3& 10\end{array}\right]$, $2B=\left[\begin{array}{cc}2& 8\\ 0& -4\end{array}\right]$. Result $=\left[\begin{array}{cc}1& -5\\ 6& -16\end{array}\right]$.

11. Q11 · 1 mark · MCQHOTS / competency

    For $A=\left[\begin{array}{cc}0& 1\\ 0& 0\end{array}\right]$, which statement is true?
    (a) $A^2=A$
    (b) $A^2=O$
    (c) $A^2=I$
    (d) $A^2=2A$

    Hint

    Multiply it out; does A² = O force A = O?

    Answer

    (b): $A^2$ is the null matrix even though $A$ is not, which never happens with ordinary numbers.

12. Q12 · 3 marks · Short answerHOTS / competency

    Find $x$ if $\left[\begin{array}{cc}x& 1\end{array}\right]\left[\begin{array}{cc}1& 0\\ -2& 0\end{array}\right]\left[\begin{array}{c}x\\ 3\end{array}\right]=O$. First state the order of the final product.

    Hint

    (1 × 2)(2 × 2)(2 × 1) is 1 × 1.

    Answer

    Order $1\times 1$. $\left[\begin{array}{cc}x& 1\end{array}\right]\left[\begin{array}{cc}1& 0\\ -2& 0\end{array}\right]=\left[\begin{array}{cc}x-2& 0\end{array}\right]$; times $\left[\begin{array}{c}x\\ 3\end{array}\right]$ gives $x(x-2)$. $x(x-2)=0\Rightarrow x=0$ or $2$.

Test your recall

Answer in your head or on paper first, then open the card. Pulling an answer out of memory makes it stick far better than re-reading.

1. 1.What is the order of a matrix with 3 rows and 2 columns?Show answer

   3 × 2

2. 2.When can two matrices be added?Show answer

   When they have the same order.

3. 3.When is the product AB defined?Show answer

   When the number of columns of A equals the number of rows of B.

4. 4.Write the 2 × 2 identity matrix.Show answer

   $\left[\begin{array}{cc}1& 0\\ 0& 1\end{array}\right]$

5. 5.Is AB always equal to BA?Show answer

   No.

6. 6.Order of a (2 × 2)(2 × 1) product?Show answer

   2 × 1

7. 7.How do you find kA?Show answer

   Multiply every entry of A by k.

8. 8.What is AI equal to?Show answer

   A

Quick revision checklist

The night before the exam: can you do each of these without looking?

• ☐Can you state the order of any matrix?
• ☐Can you decide whether a sum or product is defined, and the order of the product?
• ☐Can you add, subtract and multiply by a number without missing an entry?
• ☐Can you multiply two 2 × 2 matrices row by column, and a 2 × 2 by a 2 × 1?
• ☐Do you find A² by A × A, never by squaring entries?
• ☐Can you find unknowns from AB = C and check with the remaining entry?
• ☐Can you show identities like A² − 5A + 7I = O?

Matrices: frequently asked questions