Matrices ISC Class 12: Notes, Properties and Solved Questions

What Matrices is about

Weightage: Matrices is part of the unit Algebra, which carries 10 of the 80 marks in the ISC Mathematics, shared with 1 other chapter. Questions on it can appear in any section of the paper.Source: CISCE ISC Mathematics syllabus, examination year 2027, distribution of marks

Key concepts

Order, types and equality of matrices

$A=[a_{ij}{]}_{m\times n}$ has $a_{ij}$ in row i and column j. Types: row (1 × n), column (m × 1), square (m = n), zero (all entries 0), diagonal (square, non-diagonal entries 0), scalar (diagonal with all diagonal entries equal), identity I (scalar with diagonal entries 1), upper/lower triangular (zeros below/above the diagonal). Two matrices are equal if they have the same order and every pair of corresponding entries is equal; this gives equations for unknowns.

Worked example

Find x and y if $\left[\begin{array}{cc}x+y& 2\\ 5& x-y\end{array}\right]=\left[\begin{array}{cc}6& 2\\ 5& 2\end{array}\right]$.

Try it first, then show the solutionSolution

1. Equate entries: x + y = 6 and x − y = 2.
2. Adding: 2x = 8, x = 4; then y = 2.

Answer: x = 4, y = 2

Addition and scalar multiplication

For matrices of the same order, $A+B=[a_{ij}+b_{ij}]$ and $kA=[k{a}_{ij}]$. Properties: A + B = B + A (commutative), (A + B) + C = A + (B + C), A + O = A, A + (−A) = O, k(A + B) = kA + kB and (k + l)A = kA + lA. Matrix equations such as 2A + 3X = 5B are solved like ordinary linear equations: $X=\frac{1}{3}(5B-2A)$.

Worked example

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

Try it first, then show the solutionSolution

1. $5B-2A=\left[\begin{array}{cc}10& 5\\ 15& 0\end{array}\right]-\left[\begin{array}{cc}2& 4\\ 0& 6\end{array}\right]=\left[\begin{array}{cc}8& 1\\ 15& -6\end{array}\right]$
2. $X=\frac{1}{3}\left[\begin{array}{cc}8& 1\\ 15& -6\end{array}\right]=\left[\begin{array}{cc}8/3& 1/3\\ 5& -2\end{array}\right]$

Answer: $X=\left[\begin{array}{cc}8/3& 1/3\\ 5& -2\end{array}\right]$

Matrix multiplication: compatibility and non-commutativity

If A is m × n and B is n × p, then AB is m × p with $(AB{)}_{ik}={\sum }_j{a}_{ij}{b}_{jk}$ (row of A times column of B). If the inner orders differ, the product is not defined. Multiplication is associative, A(BC) = (AB)C, and distributive, A(B + C) = AB + AC, and AI = IA = A. But in general AB ≠ BA; and AB = O is possible with A ≠ O, B ≠ O. Hence $(A+B{)}^2=A^2+AB+BA+B^2$ and $(A+B)(A-B)=A^2-AB+BA-B^2$, which equal the familiar forms only when AB = BA.

Worked example

For $A=\left[\begin{array}{cc}1& 2\\ 3& 4\end{array}\right]$ and $B=\left[\begin{array}{cc}0& 1\\ 1& 0\end{array}\right]$, find AB and BA. Also give two non-zero 2 × 2 matrices whose product is O.

Try it first, then show the solutionSolution

1. $AB=\left[\begin{array}{cc}2& 1\\ 4& 3\end{array}\right]$ (columns of A swapped); $BA=\left[\begin{array}{cc}3& 4\\ 1& 2\end{array}\right]$ (rows swapped). AB ≠ BA.
2. $P=\left[\begin{array}{cc}0& 1\\ 0& 0\end{array}\right]$, $Q=\left[\begin{array}{cc}1& 0\\ 0& 0\end{array}\right]$: $PQ=\left[\begin{array}{cc}0& 0\\ 0& 0\end{array}\right]$ although P, Q ≠ O.

Answer: AB ≠ BA; PQ = O with P, Q non-zero

Transpose and its properties

The transpose A′ (or $A^T$) of an m × n matrix is the n × m matrix with rows and columns interchanged: $(A^{\prime }{)}_{ij}=a_{ji}$. Properties: (A′)′ = A; (A + B)′ = A′ + B′; (kA)′ = kA′; and the reversal rule (AB)′ = B′A′. The reversal rule matters because AB and BA are usually different.

Worked example

Verify (AB)′ = B′A′ for $A=\left[\begin{array}{cc}1& 2\\ 3& 4\end{array}\right]$, $B=\left[\begin{array}{cc}0& 1\\ 1& 0\end{array}\right]$.

Try it first, then show the solutionSolution

1. $AB=\left[\begin{array}{cc}2& 1\\ 4& 3\end{array}\right]$, so $(AB{)}^{\prime }=\left[\begin{array}{cc}2& 4\\ 1& 3\end{array}\right]$.
2. B′ = B; $A^{\prime }=\left[\begin{array}{cc}1& 3\\ 2& 4\end{array}\right]$; $B^{\prime }{A}^{\prime }=\left[\begin{array}{cc}2& 4\\ 1& 3\end{array}\right]$. Equal.

Answer: Verified

Symmetric and skew-symmetric matrices

A square matrix is symmetric if A′ = A ($a_{ij}=a_{ji}$) and skew-symmetric if A′ = −A ($a_{ij}=-a_{ji}$, so every diagonal entry satisfies $a_{ii}=-a_{ii}$, i.e. is 0). Properties: A + A′ is symmetric and A − A′ is skew-symmetric; for any matrix B, BB′ is symmetric; if A and B are symmetric, AB − BA is skew-symmetric and AB + BA is symmetric. Every square matrix is uniquely the sum $A=\frac{1}{2}(A+A^{\prime })+\frac{1}{2}(A-A^{\prime })$ of a symmetric and a skew-symmetric matrix.

Worked example

Express $A=\left[\begin{array}{cc}3& 5\\ 1& -1\end{array}\right]$ as the sum of a symmetric and a skew-symmetric matrix.

Try it first, then show the solutionSolution

1. $A^{\prime }=\left[\begin{array}{cc}3& 1\\ 5& -1\end{array}\right]$; $P=\frac{1}{2}(A+A^{\prime })=\left[\begin{array}{cc}3& 3\\ 3& -1\end{array}\right]$ (symmetric).
2. $Q=\frac{1}{2}(A-A^{\prime })=\left[\begin{array}{cc}0& 2\\ -2& 0\end{array}\right]$ (skew-symmetric). P + Q = A.

Answer: $A=\left[\begin{array}{cc}3& 3\\ 3& -1\end{array}\right]+\left[\begin{array}{cc}0& 2\\ -2& 0\end{array}\right]$

Invertible matrices and uniqueness of the inverse

A square matrix A is invertible if there is a matrix B of the same order with AB = BA = I; B is written $A^{-1}$. Uniqueness: if B and C are both inverses, B = BI = B(AC) = (BA)C = IC = C. A is non-singular if |A| ≠ 0 (invertible) and singular if |A| = 0 (no inverse). Properties: $(A^{-1}{)}^{-1}=A$, $(AB{)}^{-1}=B^{-1}{A}^{-1}$, $(A^{\prime }{)}^{-1}=(A^{-1}{)}^{\prime }$, $(A^2{)}^{-1}=(A^{-1}{)}^2$. If A satisfies a polynomial equation such as $A^2-pA+qI=O$ with q ≠ 0, multiplying by $A^{-1}$ gives $A^{-1}=\frac{1}{q}(pI-A)$.

Worked example

For $A=\left[\begin{array}{cc}2& 3\\ 1& 2\end{array}\right]$, show that $A^2-4A+I=O$ and hence find $A^{-1}$.

Try it first, then show the solutionSolution

1. $A^2=\left[\begin{array}{cc}7& 12\\ 4& 7\end{array}\right]$; $A^2-4A+I=\left[\begin{array}{cc}7-8+1& 12-12\\ 4-4& 7-8+1\end{array}\right]=O$.
2. Multiply by $A^{-1}$: $A-4I+A^{-1}=O\Rightarrow A^{-1}=4I-A=\left[\begin{array}{cc}2& -3\\ -1& 2\end{array}\right]$.
3. Check: $A\cdot A^{-1}=\left[\begin{array}{cc}4-3& -6+6\\ 2-2& -3+4\end{array}\right]=I$.

Answer: $A^{-1}=\left[\begin{array}{cc}2& -3\\ -1& 2\end{array}\right]$

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

• Symmetric and Skew-Symmetric Matrices
• Matrix Multiplication and Powers
• Invertible Matrices

Matrices formula sheet

Product order

$(m\times n)(n\times p)=m\times p,(AB{)}_{ik}={\sum }_j{a}_{ij}{b}_{jk}$

Defined only if inner orders match.

Non-commutativity

$AB\ne BA\text{in general};(A+B{)}^2=A^2+AB+BA+B^2$

Equals $A^2+2AB+B^2$ only if AB = BA.

Transpose rules

$(A^{\prime }{)}^{\prime }=A,(A+B{)}^{\prime }=A^{\prime }+B^{\prime },(kA{)}^{\prime }=k{A}^{\prime },(AB{)}^{\prime }=B^{\prime }{A}^{\prime }$

Reversal in the product.

Symmetric and skew-symmetric

$A^{\prime }=A;A^{\prime }=-A$

Skew-symmetric diagonal entries are 0.

Decomposition

$A=\frac{1}{2}(A+A^{\prime })+\frac{1}{2}(A-A^{\prime })$

Symmetric part + skew-symmetric part.

Inverse

$A{A}^{-1}=A^{-1}A=I$

Exists only if |A| ≠ 0; unique.

Inverse rules

$(AB{)}^{-1}=B^{-1}{A}^{-1},(A^{\prime }{)}^{-1}=(A^{-1}{)}^{\prime },(A^{-1}{)}^{-1}=A$

Inverse from a matrix equation

$A^2-pA+qI=O\Rightarrow A^{-1}=\frac{1}{q}(pI-A)$

q ≠ 0.

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.

• Inner match, outer order

  (m × n)(n × p): inner numbers must match; outer numbers give the order.

  Use it when: Checking whether a product exists.

• Socks and shoes again

  (AB)′ = B′A′ and (AB)⁻¹ = B⁻¹A⁻¹: reverse the order.

  Use it when: Transpose and inverse of products.

• Skew means zero diagonal

  Skew-symmetric: mirror entries are negatives and the diagonal must be zero.

  Use it when: Find-x skew-symmetric questions.

• Half plus half

  A = ½(A + A′) + ½(A − A′): plus gives symmetric, minus gives skew.

  Use it when: Decomposition questions.

• Multiply by A⁻¹ and tidy

  From a matrix polynomial = O, multiply through by $A^{-1}$; each power drops by one.

  Use it when: Finding inverses without adjoints.

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 D style)

Read the situation, then answer every part. In the board paper this is a 5-mark question split into 1 + 1 + 2 + 1 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: Not recognising that AB − BA is skew-symmetric

  CISCE examiners' comments, 2024: most candidates could not identify AB − BA as skew-symmetric when A and B are symmetric.

  Wrong working

  AB − BA = O, so it is the zero (or identity) matrix.

  Right working

  (AB − BA)′ = B′A′ − A′B′ = BA − AB = −(AB − BA): skew-symmetric.

  ✓Do this: Take the transpose of the expression and use the reversal rule.

  Usually costs: 1 mark.

• ✗Mistake: Mechanical multiplication instead of spotting structure

  CISCE examiners' comments, 2025: for $A^{16}$ with $A=\left[\begin{array}{cc}0& a\\ 0& 0\end{array}\right]$, candidates made multiplication mistakes or guessed, instead of noticing $A^2=O$; in 2024 many marked an assertion-reason item on $A^{100}B=B{A}^{100}$ wrongly.

  Wrong working

  Tries to multiply A by itself 16 times.

  Right working

  $A^2=O$, so every higher power is O. If AB = BA, then $A^{n}B=B{A}^n$.

  ✓Do this: Compute $A^2$ (or AB and BA) first and look for a pattern.

  Usually costs: 1 mark.

• ✗Mistake: Confusing symmetric and skew-symmetric conditions

  CISCE examiners' comments, 2025: asked for x making a matrix skew-symmetric, candidates confused symmetric with skew-symmetric, and some found the inverse instead.

  Wrong working

  Sets $a_{31}=a_{13}$.

  Right working

  Skew-symmetric: $a_{31}=-a_{13}$ and all diagonal entries are 0.

  ✓Do this: Write A′ = −A and compare entries.

  Usually costs: 1 mark.

• ✗Mistake: Not turning data into matrices

  CISCE examiners' comments, 2025: in a question on shopkeepers' bag costs, candidates could not represent the data in matrix form or did not see it as a matrix-multiplication application; in 2024 others could not form the system of equations for a furniture problem.

  Wrong working

  Adds the numbers without arranging them.

  Right working

  Quantities as a matrix with one row per person, prices as a column; their product gives each person's total.

  ✓Do this: Decide what the rows and columns stand for before writing numbers.

  Usually costs: 1-2 marks.

• ✗Mistake: Using (A + B)² = A² + 2AB + B² for matrices

  A frequent slip, and the 2027 specimen asks candidates to explain why such identities fail for matrices.

  Wrong working

  $(A+B)(A-B)=A^2-B^2$ for all matrices.

  Right working

  $(A+B)(A-B)=A^2-AB+BA-B^2$; equal to $A^2-B^2$ only if AB = BA.

  ✓Do this: Expand matrix products term by term, keeping the order.

  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

   For matrices A and B for which the products exist, (AB)′ equals:
   (a) A′B′
   (b) B′A′
   (c) AB
   (d) BA

   Hint

   Reversal rule.

   Answer

   (b)

2. Q2 · 1 mark · MCQBoard level

   For any square matrix A, the matrix A − A′ is:
   (a) symmetric
   (b) skew-symmetric
   (c) diagonal
   (d) the identity

   Hint

   Take its transpose.

   Answer

   (b)

3. Q3 · 1 mark · MCQBoard level

   Which of these is a scalar matrix?
   (a) $\left[\begin{array}{cc}3& 0\\ 0& 2\end{array}\right]$
   (b) $\left[\begin{array}{cc}4& 0\\ 0& 4\end{array}\right]$
   (c) $\left[\begin{array}{cc}0& 4\\ 4& 0\end{array}\right]$
   (d) $\left[\begin{array}{cc}1& 1\\ 0& 1\end{array}\right]$

   Hint

   Diagonal with equal diagonal entries.

   Answer

   (b)

4. Q4 · 1 mark · Very short answerBoard level

   A matrix has 6 elements. Write all its possible orders.

   Hint

   Factor pairs of 6.

   Answer

   1 × 6, 6 × 1, 2 × 3, 3 × 2

5. Q5 · 2 marks · Short answerBoard level

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

   Hint

   Row by column.

   Answer

   $AB=\left[\begin{array}{cc}2& 1\\ 4& 3\end{array}\right]$, $BA=\left[\begin{array}{cc}3& 4\\ 1& 2\end{array}\right]$; not equal.

6. Q6 · 2 marks · Short answerHarder board level

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

   Hint

   X = (5B − 2A)/3.

   Answer

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

7. Q7 · 2 marks · Short answerHarder board level

   Show that $A=\left[\begin{array}{cc}2& -2\\ -1& 1\end{array}\right]$ satisfies $A^2=3A$. Does A have an inverse?

   Hint

   If A⁻¹ existed, A = 3I would follow.

   Answer

   $A^2=\left[\begin{array}{cc}6& -6\\ -3& 3\end{array}\right]=3A$. If $A^{-1}$ existed, multiplying gives A = 3I, which is false; so A is singular (|A| = 2 − 2 = 0).

8. Q8 · 2 marks · Short answerHOTS / competency

   Give an example of two non-zero 2 × 2 matrices A and B with AB = O but BA ≠ O.

   Hint

   Use 0s and 1s.

   Answer

   $A=\left[\begin{array}{cc}0& 1\\ 0& 0\end{array}\right]$, $B=\left[\begin{array}{cc}1& 0\\ 0& 0\end{array}\right]$: AB = O, but $BA=\left[\begin{array}{cc}0& 1\\ 0& 0\end{array}\right]\ne O$.

9. Q9 · 3 marks · Short answerHarder board level

   Express $\left[\begin{array}{cc}3& 5\\ 1& -1\end{array}\right]$ as the sum of a symmetric and a skew-symmetric matrix.

   Hint

   ½(A ± A′).

   Answer

   $\left[\begin{array}{cc}3& 3\\ 3& -1\end{array}\right]+\left[\begin{array}{cc}0& 2\\ -2& 0\end{array}\right]$

10. Q10 · 3 marks · Short answerHarder board level

    If $A=\left[\begin{array}{cc}3& 1\\ -1& 2\end{array}\right]$, show that $A^2-5A+7I=O$ and hence find $A^{-1}$.

    Hint

    Multiply by A⁻¹.

    Answer

    $A^2=\left[\begin{array}{cc}8& 5\\ -5& 3\end{array}\right]$; $A^2-5A+7I=O$. $A^{-1}=\frac{1}{7}(5I-A)=\frac{1}{7}\left[\begin{array}{cc}2& -1\\ 1& 3\end{array}\right]$.

11. Q11 · 3 marks · Short answerHarder board level

    For $A=\left[\begin{array}{cc}1& 2\\ 3& 4\end{array}\right]$ and $B=\left[\begin{array}{cc}0& 1\\ 1& 0\end{array}\right]$, find $(A+B)(A-B)$ and $A^2-B^2$. Show that their difference is $BA-AB$, and state the condition under which the two are equal.

    Hint

    Expand without commuting.

    Answer

    $(A+B)(A-B)=\left[\begin{array}{cc}1& 3\\ 4& 4\end{array}\right]\left[\begin{array}{cc}1& 1\\ 2& 4\end{array}\right]=\left[\begin{array}{cc}7& 13\\ 12& 20\end{array}\right]$; $A^2-B^2=\left[\begin{array}{cc}7& 10\\ 15& 22\end{array}\right]-I=\left[\begin{array}{cc}6& 10\\ 15& 21\end{array}\right]$. Difference $=\left[\begin{array}{cc}1& 3\\ -3& -1\end{array}\right]=BA-AB$, since $(A+B)(A-B)=A^2-AB+BA-B^2$. They are equal only when AB = BA.

12. Q12 · 5 marks · Long answerHarder board level

    (a) If $A=\left[\begin{array}{ccc}1& 2& 2\\ 2& 1& 2\\ 2& 2& 1\end{array}\right]$, show that $A^2-4A-5I=O$. [3]
    (b) Hence find $A^{-1}$ and verify that $A{A}^{-1}=I$. [2]

    Hint

    Each row of A² is 9, 8, 8 in some order.

    Answer

    (a) $A^2=\left[\begin{array}{ccc}9& 8& 8\\ 8& 9& 8\\ 8& 8& 9\end{array}\right]$, and $A^2-4A-5I=O$. (b) $A^{-1}=\frac{1}{5}(A-4I)=\frac{1}{5}\left[\begin{array}{ccc}-3& 2& 2\\ 2& -3& 2\\ 2& 2& -3\end{array}\right]$; $A\times$ this gives $\frac{1}{5}(5I)=I$.

13. Q13 · 2 marks · Short answerHarder board level

    Find a, b, c if $\left[\begin{array}{ccc}0& a& 3\\ 2& b& -1\\ c& 1& 0\end{array}\right]$ is skew-symmetric.

    Hint

    A′ = −A.

    Answer

    a = −2, b = 0, c = −3

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.When is the product AB defined?Show answer

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

2. 2.Define a scalar matrix.Show answer

   A diagonal matrix whose diagonal entries are all equal.

3. 3.(AB)′ = ?Show answer

   B′A′.

4. 4.Define a skew-symmetric matrix.Show answer

   A square matrix with A′ = −A; its diagonal entries are zero.

5. 5.How do you split A into symmetric and skew-symmetric parts?Show answer

   ½(A + A′) and ½(A − A′).

6. 6.When is A invertible?Show answer

   When |A| ≠ 0 (non-singular).

7. 7.(AB)⁻¹ = ?Show answer

   B⁻¹A⁻¹.

8. 8.Can AB = O with A ≠ O and B ≠ O?Show answer

   Yes, e.g. $\left[\begin{array}{cc}0& 1\\ 0& 0\end{array}\right]\left[\begin{array}{cc}1& 0\\ 0& 0\end{array}\right]=O$.

Quick revision checklist

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

• ☐Can you state the order of a product and when it exists?
• ☐Can you show AB ≠ BA with an example?
• ☐Can you give two non-zero matrices with product O?
• ☐Can you prove (AB)′ = B′A′ for a given pair?
• ☐Can you test for symmetric and skew-symmetric matrices?
• ☐Can you split a 3 × 3 matrix into symmetric and skew parts?
• ☐Can you prove the inverse is unique?
• ☐Can you find $A^{-1}$ from $A^2-pA+qI=O$?

Matrices: frequently asked questions