MathsCBSE Class 12 · Matrices

Invertible Matrices: Definition, Uniqueness and Inverse from a Matrix Equation (Class 12)

Updated 15 October 2026 · Fact-checked

A square matrix A is invertible if there is a matrix B of the same order with AB = BA = I; then B is called A⁻¹. The inverse, when it exists, is unique, and (AB)⁻¹ = B⁻¹A⁻¹. If A satisfies an equation like A² − 6A + 5I = O, you can read off A⁻¹ without the adjoint.

To undo AB you reverse the order: first undo B, then undo A, so (AB)⁻¹ = B⁻¹A⁻¹, and A·A⁻¹ gives I.

Illustration drawn for QuestionBuddy.

Understand Invertible Matrices

For numbers, the inverse of 5 is $\frac{1}{5}$ because $5 \times \frac{1}{5} = 1$ . For matrices, the identity matrix $I$ plays the role of 1. A square matrix $A$ of order $n$ is invertible (or non-singular) if there is a square matrix $B$ of the same order such that
$A B = B A = I$ .
Then $B$ is called the inverse of $A$ , written $A^{- 1}$ , and $A$ is also the inverse of $B$ . Both products must give $I$ ; for square matrices of the same order one implies the other, but the definition states both, and board questions on verifying an inverse expect you to show at least $A B = I$ (better, both).

Not every square matrix has an inverse. If $A B = O$ for some non-zero $B$ , then $A$ cannot be invertible: multiplying $A B = O$ on the left by $A^{- 1}$ would give $B = O$ . For example $[1224]$ is not invertible. (In the next chapter you will see the test: $A$ is invertible exactly when $∣ A ∣ \neq = 0$ .)

Uniqueness of the inverse (NCERT Theorem, Section 3.7). Suppose $B$ and $C$ are both inverses of $A$ . Then $A B = B A = I$ and $A C = C A = I$ . Now
$B = B I = B (A C) = (B A) C = I C = C$ .
The proof uses only the identity and associativity; learn it, as it is a short proof question.

Inverse of a product. If $A$ and $B$ are invertible matrices of the same order, then $A B$ is invertible and
$(A B)^{- 1} = B^{- 1} A^{- 1}$ .
Check: $(A B) (B^{- 1} A^{- 1}) = A (B B^{- 1}) A^{- 1} = A I A^{- 1} = A A^{- 1} = I$ , and similarly in the other order. As with the transpose, the order reverses. Two related facts: $(A^{- 1})^{- 1} = A$ , and $(A^{'})^{- 1} = (A^{- 1})^{'}$ , so the inverse of a symmetric invertible matrix is symmetric.

**Finding $A^{- 1}$ from a matrix equation.** The 2026-27 syllabus does not include elementary row operations, so inverses come either from the adjoint (Determinants) or from a polynomial relation. If $A^{2} - 6 A + 5 I = O$ , rewrite it as $A^{2} - 6 A = - 5 I$ , that is $A (A - 6 I) = - 5 I$ , so
$A \cdot \frac{1}{5} (6 I - A) = I$ and $A^{- 1} = \frac{1}{5} (6 I - A)$ .
The trick is always the same: move the $I$ term to one side, take $A$ out as a common factor, and divide by the constant. It only works when the constant term is non-zero. A $3 \times 3$ version uses a cubic, such as $A^{3} - 6 A^{2} + 7 A + 2 I = O$ , giving $A^{- 1} = - \frac{1}{2} (A^{2} - 6 A + 7 I)$ .

Inverses undo actions: if a matrix $K$ scrambles data, $K^{- 1}$ unscrambles it, and if $A X = C$ then $X = A^{- 1} C$ . That second idea is the matrix method for solving linear equations.

Invertible Matrices: formulas to remember

Definition: $A B = B A = I \Rightarrow B = A^{- 1}$; A and B are square matrices of the same order.
Uniqueness: $B = B I = B (A C) = (B A) C = I C = C$; Any two inverses of A are equal.
Inverse of a product: $(A B)^{- 1} = B^{- 1} A^{- 1}$; The order reverses; extends to $(A B C)^{- 1} = C^{- 1} B^{- 1} A^{- 1}$ .
Inverse of an inverse and of a transpose: $(A^{- 1})^{- 1} = A, (A^{'})^{- 1} = (A^{- 1})^{'}$; So a symmetric invertible matrix has a symmetric inverse.
Inverse from a quadratic relation: $A^{2} + p A + q I = O, q \neq = 0 \Rightarrow A^{- 1} = - \frac{1}{q} (A + p I)$; From $A (A + p I) = - q I$ .
Not invertible: $A B = O with B \neq = O \Rightarrow A has no inverse$; Equivalent to $∣ A ∣ = 0$ (Determinants).

How to solve Invertible Matrices questions

Use these steps to verify an inverse, find one from a matrix equation, or prove a property.

1To verify that B is the inverse of A, compute $A B$ and $B A$ and show both equal $I$ .
2To find $A^{- 1}$ from a relation, first verify the relation if asked: compute $A^{2}$ (and $A^{3}$ if needed), then substitute.
3Rearrange so that the $I$ term is alone: for example $A^{2} - 6 A = - 5 I$ .
4Factor out $A$ on the left: $A (A - 6 I) = - 5 I$ .
5Divide by the constant: $A \cdot (- \frac{1}{5}) (A - 6 I) = I$ , so $A^{- 1} = \frac{1}{5} (6 I - A)$ .
6Substitute the matrix and simplify; check one entry of $A A^{- 1}$ equals the matching entry of $I$ .
7For proofs, use only the definition, the identity property, associativity and the reverse-order rules.

Invertible Matrices in real life

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.

An inverse undoes a transformation

Suppose a photo-editing app slants a picture by moving every point $(x, y)$ to $(x + y, y)$ , which is multiplication by $S = [1011]$ . A user wants to undo the slant.

Show that $T = [10 - 1 1]$ is the inverse of $S$ . Where does the point $(3, 2)$ go under $S$ , and which original point became $(7, 3)$ ?

Try it first, then show the solutionSolution

$S T = [10 - 1 + 1 1] = I$ and $T S = [10 1 - 1 1] = I$ , so $T = S^{- 1}$ .
$S [32] = [52]$ : the point moves to $(5, 2)$ .
Undo: $S^{- 1} [73] = [7 - 3 3] = [43]$ .
Check: $(4 + 3, 3) = (7, 3)$ .

Answer: $S^{- 1} = [10 - 1 1]$ ; $(3, 2) \to (5, 2)$ ; the point $(7, 3)$ came from $(4, 3)$ .

Memory hook: The inverse is the undo button.

Using an inverse to find unknown amounts

Imagine a canteen mixing two drinks. A glass of drink X gives 2 units of sugar and 1 unit of vitamin C; a glass of drink Y gives 1 unit of sugar and 3 units of vitamin C. A student's tray supplied 9 units of sugar and 12 units of vitamin C in total.

With $M = [2113]$ , show that $N = \frac{1}{5} [3 - 1 - 1 2]$ is $M^{- 1}$ , and use it to find how many glasses of each drink were on the tray.

Try it first, then show the solutionSolution

If $x$ and $y$ are the numbers of glasses, $M [x y] = [912]$ .
$M N = \frac{1}{5} [6 - 1 3 - 3 - 2 + 2 - 1 + 6] = I$ , so $N = M^{- 1}$ .
$[x y] = N [912] = \frac{1}{5} [27 - 12 - 9 + 24] = [33]$ .
Check: sugar $2 (3) + 3 = 9$ , vitamin $3 + 3 (3) = 12$ .

Answer: 3 glasses of X and 3 glasses of Y.

Memory hook: Totals = M × amounts, so amounts = $M^{- 1}$ × totals.

Tricks to remember Invertible Matrices

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

Both sides must give I
An inverse works from the left and from the right: $A B = I$ and $B A = I$ .
Use it when: Verification questions.
Shoes and socks again
$(A B)^{- 1} = B^{- 1} A^{- 1}$ : undo the last action first, just like $(A B)^{'} = B^{'} A^{'}$ .
Use it when: Reverse-order MCQs and proofs.
Isolate I, pull out A, divide
From $A^{2} - k A + m I = O$ : $A (k I - A) = m I$ , so $A^{- 1} = \frac{1}{m} (k I - A)$ .
Use it when: Finding the inverse from a given matrix equation.
A zero product kills the inverse
If some non-zero $B$ gives $A B = O$ , then $A$ has no inverse.
Use it when: Assertion-reason items on singular matrices.

Invertible Matrices: solved examples

Ordered from board level to HOTS. Try each one before you open the solution.

Example 1: Verify an inverse · 2 marks · Very short answerBoard level

Show that $B = [3 - 1 - 5 2]$ is the inverse of $A = [2153]$ .

Try it first, then show the solutionSolution

$A B = [6 - 5 3 - 3 - 10 + 10 - 5 + 6] = [1001]$ . (1 mark)
$B A = [6 - 5 - 2 + 2 15 - 15 - 5 + 6] = [1001]$ . Since $A B = B A = I$ , $B = A^{- 1}$ . (1 mark)

Answer: $A B = B A = I$ , so $B = A^{- 1}$ .

Example 2: Inverse from a quadratic matrix equation · 3 marks · Short answerBoard level

For $A = [2134]$ , show that $A^{2} - 6 A + 5 I = O$ . Hence find $A^{- 1}$ .

Try it first, then show the solutionSolution

$A^{2} = [761819]$ , $6 A = [1261824]$ , $5 I = [5005]$ . (1 mark)
$A^{2} - 6 A + 5 I = [7 - 12 + 5 0 0 19 - 24 + 5] = O$ . (½ mark)
So $A^{2} - 6 A = - 5 I$ , i.e. $A (6 I - A) = 5 I$ , giving $A^{- 1} = \frac{1}{5} (6 I - A)$ . (1 mark)
$A^{- 1} = \frac{1}{5} [4 - 1 - 3 2]$ . (½ mark)

Answer: $A^{- 1} = \frac{1}{5} [4 - 1 - 3 2]$

Example 3: Uniqueness of the inverse · 2 marks · Very short answerHarder board level

Prove that a square matrix cannot have two different inverses.

Try it first, then show the solutionSolution

Let $B$ and $C$ both be inverses of $A$ : $A B = B A = I$ and $A C = C A = I$ . (½ mark)
Then $B = B I = B (A C)$ . (½ mark)
By associativity, $B (A C) = (B A) C = I C = C$ . (½ mark)
So $B = C$ : the inverse is unique. (½ mark)

Answer: $B = B I = B (A C) = (B A) C = I C = C$ .

Example 4: Inverse of a product from given inverses · 3 marks · Short answerHarder board level

If $A^{- 1} = [1021]$ and $B^{- 1} = [1301]$ , find $(A B)^{- 1}$ . Prove the rule you use.

Try it first, then show the solutionSolution

Rule: $(A B) (B^{- 1} A^{- 1}) = A (B B^{- 1}) A^{- 1} = A A^{- 1} = I$ and $(B^{- 1} A^{- 1}) (A B) = B^{- 1} (A^{- 1} A) B = B^{- 1} B = I$ , so $(A B)^{- 1} = B^{- 1} A^{- 1}$ . (1½ marks)
$(A B)^{- 1} = B^{- 1} A^{- 1} = [1301] [1021]$ . (½ mark)
$= [13 2 6 + 1] = [1327]$ . (1 mark)

Answer: $(A B)^{- 1} = [1327]$

Example 5: A 3 × 3 inverse from a cubic relation · 5 marks · Long answerHOTS / competency

Let $A = 102020213$ . Show that $A^{3} - 6 A^{2} + 7 A + 2 I = O$ , and hence find $A^{- 1}$ .

Try it first, then show the solutionSolution

$A^{2} = 5280408513$ . (1 mark)
$A^{3} = A^{2} A = 211234080342355$ . (1 mark)
$A^{3} - 6 A^{2} + 7 A + 2 I$ : entry (1, 1) $21 - 30 + 7 + 2 = 0$ ; (1, 3) $34 - 48 + 14 = 0$ ; (2, 1) $12 - 12 + 0 = 0$ ; (2, 2) $8 - 24 + 14 + 2 = 0$ ; (2, 3) $23 - 30 + 7 = 0$ ; (3, 1) $34 - 48 + 14 = 0$ ; (3, 3) $55 - 78 + 21 + 2 = 0$ ; the rest are $0$ . So the relation holds. (1 mark)
Multiply by $A^{- 1}$ (it exists, since the constant term 2 is non-zero): $A^{2} - 6 A + 7 I + 2 A^{- 1} = O$ , so $A^{- 1} = - \frac{1}{2} (A^{2} - 6 A + 7 I)$ . (1 mark)
$A^{2} - 6 A + 7 I = 62 - 4 0 - 1 0 - 4 - 1 2$ , so $A^{- 1} = - 3 - 1 2 0 \frac{1}{2} 0 2 \frac{1}{2} - 1$ . Check row 1 of $A$ times column 1: $- 3 + 0 + 4 = 1$ . (1 mark)

Answer: $A^{- 1} = - 3 - 1 2 0 \frac{1}{2} 0 2 \frac{1}{2} - 1$

Example 6: Case study: a secret code · 4 marks · Case-basedHOTS / competency

Suppose two friends send coded messages. Each pair of letters is turned into numbers (A = 1, B = 2, ..., Z = 26) and multiplied by the key $K = [2111]$ .
(i) Show that $L = [1 - 1 - 1 2]$ is $K^{- 1}$ . (1 mark)
(ii) Encode the pair 'HI' (8, 9). (1 mark)
(iii) A received pair is $[3119]$ . Decode it. (1 mark)
(iv) Why must the key matrix be invertible? (1 mark)

Try it first, then show the solutionSolution

(i) $K L = [2 - 1 1 - 1 - 2 + 2 - 1 + 2] = I$ and $L K = [2 - 1 - 2 + 2 1 - 1 - 1 + 2] = I$ . (1 mark)
(ii) $K [89] = [16 + 9 8 + 9] = [2517]$ . (1 mark)
(iii) $L [3119] = [31 - 19 - 31 + 38] = [127]$ , that is 'LG'. (1 mark)
(iv) Decoding multiplies by $K^{- 1}$ . If $K$ had no inverse, two different messages could give the same code and the receiver could not recover the message. (1 mark)

Answer: (i) $K L = L K = I$ . (ii) $(25, 17)$ . (iii) $(12, 7)$ , 'LG'. (iv) Decoding needs $K^{- 1}$ .

Where students lose marks in Invertible Matrices

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

Mistake: Checking only one product
The definition needs $A B = I$ and $B A = I$ .
Wrong working
$A B = I$ , hence $B = A^{- 1}$ (BA not shown).
Right working
Show $A B = I$ and $B A = I$ , then conclude $B = A^{- 1}$ .
Do this: Write both products; it costs one extra line.
Usually costs: Half to one mark.
Mistake: Writing (AB)⁻¹ = A⁻¹B⁻¹
Inverses of products reverse the order.
Wrong working
$(A B)^{- 1} = A^{- 1} B^{- 1} = [1021] [1301] = [7321]$ .
Right working
$(A B)^{- 1} = B^{- 1} A^{- 1} = [1327]$ .
Do this: Remember 'undo the last action first'.
Usually costs: The whole answer.
Mistake: Dividing by a matrix
Matrix division is not defined; you multiply by an inverse.
Wrong working
$A^{2} - 6 A = - 5 I \Rightarrow A - 6 = \frac{- 5 I}{A}$ .
Right working
$A (A - 6 I) = - 5 I \Rightarrow A^{- 1} = - \frac{1}{5} (A - 6 I)$ .
Do this: Factor out A and divide only by numbers.
Usually costs: The method marks.
Mistake: Forgetting the I after factoring
$A^{2} - 6 A = A (A - 6)$ subtracts a number from a matrix.
Wrong working
$A^{2} - 6 A = A (A - 6)$ .
Right working
$A^{2} - 6 A = A (A - 6 I)$ .
Do this: Every constant inside a matrix bracket needs an I.
Usually costs: Half a mark, and errors in the final matrix.

Exam tips

Finding $A^{- 1}$ from a given polynomial relation appeared as an MCQ in the 2023 board paper; in longer questions, show the relation is satisfied before using it.
Write the line ' $A (A - 6 I) = - 5 I$ ' explicitly; it is the step examiners look for.
Fractions inside an inverse may be kept outside the matrix, as in $\frac{1}{5} [4 - 1 - 3 2]$ .
For verification questions, compute $A B$ and $B A$ in full; do not say 'similarly'.
Remember that only square matrices can have inverses; an MCQ may test this with a $2 \times 3$ matrix.

Practice questions on Invertible Matrices

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

Q1 · 1 mark · MCQBoard level
If A is an invertible matrix, then $(A^{- 1})^{- 1}$ equals:
(a) $A$
(b) $A^{- 1}$
(c) $I$
(d) $A^{'}$
Hint
$A^{- 1} A = A A^{- 1} = I$ : whose inverse is A?
Answer
(a) $A$ .
Q2 · 1 mark · MCQBoard level
If A and B are invertible matrices of the same order, then $(A B)^{- 1}$ is:
(a) $A^{- 1} B^{- 1}$
(b) $B^{- 1} A^{- 1}$
(c) $A B$
(d) $B A$
Hint
Undo the last action first.
Answer
(b) $B^{- 1} A^{- 1}$ .
Q3 · 2 marks · Very short answerBoard level
Show that $A = [34 - 2 - 3]$ satisfies $A^{2} = I$ . Hence write $A^{- 1}$ .
Hint
If $AA = I$ , then A is its own inverse.
Answer
$A^{2} = [9 - 8 12 - 12 - 6 + 6 - 8 + 9] = I$ . So $A \cdot A = I$ and $A^{- 1} = A = [34 - 2 - 3]$ .
Q4 · 3 marks · Short answerHarder board level
Show that $A = [3212]$ satisfies $A^{2} - 5 A + 4 I = O$ and hence find $A^{- 1}$ .
Hint
$A (5 I - A) = 4 I$ .
Answer
$A^{2} = [111056]$ , and $11 - 15 + 4 = 0$ , $5 - 5 = 0$ , $10 - 10 = 0$ , $6 - 10 + 4 = 0$ . So $A (5 I - A) = 4 I$ and $A^{- 1} = \frac{1}{4} (5 I - A) = \frac{1}{4} [2 - 2 - 1 3] = [\frac{1}{2} - \frac{1}{2} - \frac{1}{4} \frac{3}{4}]$ .
Q5 · 1 mark · Assertion-reasonHarder board level
Assertion (A): The matrix $A = [1224]$ is not invertible.
Reason (R): There is a non-zero matrix B such that $A B = O$ .
(a) Both A and R are true and R is the correct explanation of A.
(b) Both A and R are true but R is not the correct explanation of A.
(c) A is true but R is false.
(d) A is false but R is true.
Hint
Try $B = [2 - 1 2 - 1]$ ; then imagine multiplying $A B = O$ by $A^{- 1}$ .
Answer
(a). With $B = [2 - 1 2 - 1] \neq = O$ , $A B = O$ . If $A^{- 1}$ existed, $B = A^{- 1} A B = O$ , a contradiction; so R explains A.
Q6 · 3 marks · Short answerHarder board level
If A is an invertible symmetric matrix, prove that $A^{- 1}$ is symmetric.
Hint
Transpose both sides of $A A^{- 1} = I$ .
Answer
Transposing $A A^{- 1} = I$ gives $(A^{- 1})^{'} A^{'} = I$ , and transposing $A^{- 1} A = I$ gives $A^{'} (A^{- 1})^{'} = I$ . So $(A^{- 1})^{'}$ is the inverse of $A^{'}$ . Since $A^{'} = A$ , $(A^{- 1})^{'}$ is the inverse of $A$ , and by uniqueness $(A^{- 1})^{'} = A^{- 1}$ . Hence $A^{- 1}$ is symmetric.
Q7 · 3 marks · Short answerHOTS / competency
A square matrix A satisfies $A^{2} = A$ . (i) Show that $(I - A)^{2} = I - A$ . (ii) If A is also invertible, show that $A = I$ .
Hint
In (ii), multiply $A^{2} = A$ by $A^{- 1}$ .
Answer
(i) $(I - A)^{2} = I - I A - A I + A^{2} = I - 2 A + A = I - A$ . (ii) $A^{- 1} (AA) = A^{- 1} A$ gives $(A^{- 1} A) A = I$ , so $I A = I$ , i.e. $A = I$ .

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.When is a square matrix A invertible?Show answer
When there is B of the same order with $A B = B A = I$ .
2.State the reverse-order law for inverses.Show answer
$(A B)^{- 1} = B^{- 1} A^{- 1}$ .
3.If $A^{2} - 3 A + 2 I = O$ , what is $A^{- 1}$ ?Show answer
$A (3 I - A) = 2 I$ , so $A^{- 1} = \frac{1}{2} (3 I - A)$ .
4.If $A^{2} = I$ , what is $A^{- 1}$ ?Show answer
$A$ itself.
5.Can a $2 \times 3$ matrix have an inverse?Show answer
No; only square matrices can be invertible.
6.Write the key line of the uniqueness proof.Show answer
$B = B I = B (A C) = (B A) C = I C = C$ .

Invertible Matrices: frequently asked questions

Is finding the inverse by elementary row operations in the 2026-27 syllabus?

No. Elementary operations and the inverse by elementary operations were dropped. Find inverses with the adjoint formula from Determinants, or from a given matrix equation.

How do I find the inverse from A² − 5A + 7I = O?

Write A² − 5A = −7I, factor A(A − 5I) = −7I, and divide by −7: A⁻¹ = (1/7)(5I − A). Then substitute the matrix and simplify.

Why is the inverse unique?

If B and C are both inverses of A, then B = BI = B(AC) = (BA)C = IC = C, using associativity. So two inverses must be equal.

Which NCERT section covers invertible matrices?

Section 3.7 (Invertible Matrices) of Chapter 3 in NCERT Mathematics Part I for Class 12, including the theorems on uniqueness and the inverse of a product.

How do I know a matrix has no inverse?

If its determinant is zero (Chapter 4), or if a non-zero matrix B gives AB = O. For example, a 2 × 2 matrix whose second row is a multiple of the first has no inverse.