MathsISC Class 12 · Matrices

Invertible Matrices: Uniqueness of the Inverse and Its Properties

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 the inverse A⁻¹, and it is unique. A is invertible exactly when |A| ≠ 0 (non-singular). Key rules: (AB)⁻¹ = B⁻¹A⁻¹, (A′)⁻¹ = (A⁻¹)′ and |A⁻¹| = 1/|A|.

Swap a and d, negate b and c, divide by ad − bc; this works only when the determinant is not zero.

Illustration drawn for QuestionBuddy.

Understand Invertible Matrices

For numbers, $a \times \frac{1}{a} = 1$ whenever $a \neq = 0$ . For matrices, the role of 1 is played by the identity matrix $I$ , and the role of "non-zero" is played by a non-zero determinant.

Definition. A square matrix $A$ of order $n$ is invertible if there is a square matrix $B$ of order $n$ such that $A B = B A = I_{n}$ . Then $B$ is called the inverse of $A$ and written $A^{- 1}$ . Both products must give $I$ ; for square matrices it is enough to check one, but the definition asks for both. A non-square matrix never has an inverse in this sense.

Uniqueness (proof in the syllabus). Suppose $B$ and $C$ are both inverses of $A$ . Then
$B = B I = B (A C) = (B A) C = I C = C .$
So an invertible matrix has exactly one inverse. Notice that the proof uses only associativity and the definition; this is the proof the CISCE syllabus names.

Singular and non-singular. $A$ is singular if $∣ A ∣ = 0$ and non-singular if $∣ A ∣ \neq = 0$ . A square matrix is invertible if and only if it is non-singular. One direction is quick: if $A A^{- 1} = I$ , then $∣ A ∣∣ A^{- 1} ∣ = ∣ I ∣ = 1$ , so $∣ A ∣ \neq = 0$ , and also $∣ A^{- 1} ∣ = \frac{1}{∣ A ∣}$ . The other direction uses the adjoint, $A^{- 1} = \frac{1}{∣ A ∣} adj A$ , which is worked out on the adjoint and inverse topic page.

The 2 × 2 inverse. For $A = [a c b d]$ with $a d - b c \neq = 0$ :
$A^{- 1} = \frac{1}{a d - b c} [d - c - b a] .$
Swap the diagonal entries, change the signs of the other two, and divide by the determinant. Always check by multiplying: $A A^{- 1}$ should give $I$ .

Properties. For invertible matrices $A$ , $B$ of the same order and a non-zero scalar $k$ :

$(A^{- 1})^{- 1} = A$
$(A B)^{- 1} = B^{- 1} A^{- 1}$ (order reversed, as with the transpose), because $(A B) (B^{- 1} A^{- 1}) = A (B B^{- 1}) A^{- 1} = A A^{- 1} = I$
$(A^{'})^{- 1} = (A^{- 1})^{'}$ , because $A^{'} (A^{- 1})^{'} = (A^{- 1} A)^{'} = I^{'} = I$
$(k A)^{- 1} = \frac{1}{k} A^{- 1}$
$(A^{n})^{- 1} = (A^{- 1})^{n}$ , so $(A^{2})^{- 1} = (A^{- 1})^{2}$ , the statement tested in ISC 2025
$∣ A^{- 1} ∣ = ∣ A ∣^{- 1}$

Using the inverse. The inverse lets you "divide" in matrix equations, but on the correct side. If $A X = B$ , multiply on the left by $A^{- 1}$ : $X = A^{- 1} B$ . If $X A = B$ , multiply on the right: $X = B A^{- 1}$ . If $A X B = C$ , then $X = A^{- 1} C B^{- 1}$ . It also explains cancellation: $A B = A C$ gives $B = C$ only when $A$ is invertible. A matrix equation such as $A^{2} - 6 A + 5 I = O$ gives the inverse directly: multiply by $A^{- 1}$ to get $A^{- 1} = \frac{1}{5} (6 I - A)$ .

Why singular matrices cannot be inverted. A singular matrix squashes the plane (or space) onto a line or a point, so different inputs can give the same output. Information is lost, and nothing can undo it.

Invertible Matrices: formulas to remember

Definition: $A A^{- 1} = A^{- 1} A = I$; Only square matrices can have inverses.
Condition: $A^{- 1} exists ⟺ ∣ A ∣ \neq = 0$; Singular: |A| = 0; non-singular: |A| ≠ 0.
2 × 2 inverse: $[a c b d]^{- 1} = \frac{1}{a d - b c} [d - c - b a]$; Swap the diagonal, negate the rest, divide by the determinant.
Inverse of a product: $(A B)^{- 1} = B^{- 1} A^{- 1}$; Reverse the order.
Inverse of a transpose: $(A^{'})^{- 1} = (A^{- 1})^{'}$; Transpose and inverse can be taken in either order.
Scalar and powers: $(k A)^{- 1} = \frac{1}{k} A^{- 1}, (A^{n})^{- 1} = (A^{- 1})^{n}$; k ≠ 0.
Determinant of the inverse: $∣ A^{- 1} ∣ = \frac{1}{∣ A ∣}$; From |A||A⁻¹| = |I| = 1.
Solving matrix equations: $A X = B \Rightarrow X = A^{- 1} B; X A = B \Rightarrow X = B A^{- 1}$; Multiply by the inverse on the same side as A.

How to solve Invertible Matrices questions

Most questions here ask you to test invertibility, find a 2 × 2 inverse, use the properties, or solve a matrix equation.

1Find the determinant first. If it is zero, the matrix is singular and has no inverse; say so.
2For a $2 \times 2$ matrix, swap the leading diagonal entries, change the signs of the other two entries, and divide by the determinant.
3Check your inverse by multiplying it by the original matrix; you should get $I$ .
4For products and transposes, use $(A B)^{- 1} = B^{- 1} A^{- 1}$ and $(A^{'})^{- 1} = (A^{- 1})^{'}$ ; keep the order reversed.
5For a matrix equation, note which side the known matrix is on, and multiply both sides by its inverse on that side.
6For an equation such as $A^{2} + p A + q I = O$ with $q \neq = 0$ , multiply by $A^{- 1}$ and rearrange to get $A^{- 1}$ in terms of $A$ and $I$ .
7State the final answer as a matrix, with the scalar factor outside if the entries are fractions.

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.

Solving a system with an inverse matrix

Suppose two customers at a stationery shop buy pens and notebooks. The first pays Rs 130 for 3 pens and 2 notebooks; the second pays Rs 240 for 5 pens and 4 notebooks.

Write this as $A X = B$ and use $A^{- 1}$ to find the price of a pen and of a notebook.

Try it first, then show the solutionSolution

$A = [3524]$ , $X = [p n]$ , $B = [130240]$ .
$∣ A ∣ = 12 - 10 = 2 \neq = 0$ , so $A$ is invertible.
$A^{- 1} = \frac{1}{2} [4 - 5 - 2 3]$ .
$X = A^{- 1} B = \frac{1}{2} [520 - 480 - 650 + 720] = \frac{1}{2} [4070] = [2035]$ .
Check: $3 (20) + 2 (35) = 130$ and $5 (20) + 4 (35) = 240$ .

Answer: A pen costs Rs 20 and a notebook Rs 35.

Memory hook: Two bills, two unknowns: undo the price matrix with its inverse.

Inverse as an undo operation

Imagine an image editor that slants a picture with the matrix $S = [10 0.5 1]$ , which sends a point $(x, y)$ to $(x + 0.5 y, y)$ . It also has a "flatten" tool $P = [1000]$ that sends $(x, y)$ to $(x, 0)$ .

After slanting, a point is at $(5, 2)$ . Where was it before? Can the flatten tool be undone?

Try it first, then show the solutionSolution

$∣ S ∣ = 1 \neq = 0$ , so $S^{- 1} = [10 - 0.5 1]$ .
Original point $= S^{- 1} [52] = [5 - 1 2] = [42]$ . Check: $S$ sends $(4, 2)$ to $(4 + 1, 2) = (5, 2)$ .
$∣ P ∣ = 0$ , so $P$ is singular. It sends $(3, 1)$ and $(3, 7)$ to the same point $(3, 0)$ , so no matrix can tell them apart afterwards.

Answer: The point was at $(4, 2)$ . The flatten tool cannot be undone because its matrix is singular.

Memory hook: Undo exists only if nothing was squashed: non-zero determinant means an inverse.

Tricks to remember Invertible Matrices

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

Swap, negate, divide
For a 2 × 2 inverse: swap a and d, negate b and c, divide by ad − bc.
Use it when: Every 2 × 2 inverse in under a minute.
Socks and shoes
(AB)⁻¹ = B⁻¹A⁻¹, just like (AB)′ = B′A′.
Use it when: Product rules in MCQs and proofs.
Zero determinant, no inverse
Singular = |A| = 0 = no inverse. Check the determinant before doing anything else.
Use it when: Avoiding long work on a matrix that has no inverse.
Same side
Undo A on the side where it sits: AX = B gives X = A⁻¹B, never BA⁻¹.
Use it when: Solving matrix equations.

Invertible Matrices: solved examples

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

Example 1: A 2 × 2 inverse with determinant 1 · 1 mark · MCQBoard level

The inverse of $[3512]$ is:
(a) $[2 - 5 - 1 3]$
(b) $[2513]$
(c) $[- 2 5 1 - 3]$
(d) $[3 - 5 - 1 2]$

Try it first, then show the solutionSolution

$∣ A ∣ = 6 - 5 = 1$ . Swap 3 and 2, negate 1 and 5, divide by 1.

Answer: (a) $[2 - 5 - 1 3]$ .

Example 2: Singular value and an inverse · 2 marks · Short answerBoard level

For $A = [x 2 42]$ : (i) find $x$ for which $A$ is singular; (ii) find $A^{- 1}$ when $x = 5$ .

Try it first, then show the solutionSolution

(i) $∣ A ∣ = 2 x - 8 = 0 \Rightarrow x = 4$ . (1 mark)
(ii) For $x = 5$ , $∣ A ∣ = 2$ , so $A^{- 1} = \frac{1}{2} [2 - 2 - 4 5]$ . (1 mark)

Answer: (i) $x = 4$ ; (ii) $A^{- 1} = \frac{1}{2} [2 - 2 - 4 5]$ .

Example 3: Proof of uniqueness of the inverse · 2 marks · Short answerHarder board level

Prove that an invertible square matrix has only one inverse.

Try it first, then show the solutionSolution

Let $B$ and $C$ both be inverses of $A$ , so $A B = B A = I$ and $A C = C A = I$ . (½ mark)
$B = B I = B (A C)$ . (½ mark)
$= (B A) C$ by associativity $= I C = C$ . (1 mark)

Answer: Hence $B = C$ : the inverse is unique.

Example 4: Verifying (AB)⁻¹ = B⁻¹A⁻¹ · 3 marks · Short answerHarder board level

For $A = [2111]$ and $B = [1021]$ , verify that $(A B)^{- 1} = B^{- 1} A^{- 1}$ and show that $(A B)^{- 1} \neq = A^{- 1} B^{- 1}$ .

Try it first, then show the solutionSolution

$∣ A ∣ = 1$ , $A^{- 1} = [1 - 1 - 1 2]$ ; $∣ B ∣ = 1$ , $B^{- 1} = [10 - 2 1]$ . (1 mark)
$A B = [2153]$ , $∣ A B ∣ = 1$ , so $(A B)^{- 1} = [3 - 1 - 5 2]$ . (1 mark)
$B^{- 1} A^{- 1} = [1 + 2 - 1 - 1 - 4 2] = [3 - 1 - 5 2]$ , which matches. But $A^{- 1} B^{- 1} = [1 - 1 - 3 4]$ , which does not. (1 mark)

Answer: Verified: $(A B)^{- 1} = B^{- 1} A^{- 1}$ , and the unreversed product is different.

Example 5: Two statements on inverses · 1 mark · MCQHarder board level

Let $A$ and $B$ be invertible matrices of the same order.
Statement 1: $(A B)^{- 1} = A^{- 1} B^{- 1}$ .
Statement 2: $(A^{'})^{- 1} = (A^{- 1})^{'}$ .
(a) Both true (b) Both false (c) 1 true, 2 false (d) 1 false, 2 true

Try it first, then show the solutionSolution

Statement 1 is false: the order must reverse.
Statement 2 is true: $A^{'} (A^{- 1})^{'} = (A^{- 1} A)^{'} = I^{'} = I$ .

Answer: (d) Statement 1 false, Statement 2 true.

Example 6: A matrix sandwiched between two others · 5 marks · Long answerHOTS / competency

Find the matrix $X$ such that $[2312] X [- 3 5 2 - 3] = [1001]$ .

Try it first, then show the solutionSolution

Write $A X B = I$ with $A$ and $B$ as given. Then $X = A^{- 1} I B^{- 1} = A^{- 1} B^{- 1}$ , multiplying by $A^{- 1}$ on the left and $B^{- 1}$ on the right. (1 mark)
$∣ A ∣ = 4 - 3 = 1$ , so $A^{- 1} = [2 - 3 - 1 2]$ . (1 mark)
$∣ B ∣ = 9 - 10 = - 1$ , so $B^{- 1} = \frac{1}{- 1} [- 3 - 5 - 2 - 3] = [3523]$ . (1 mark)
$X = [2 - 3 - 1 2] [3523] = [6 - 5 - 9 + 10 4 - 3 - 6 + 6] = [1110]$ . (1 mark)
Check: $A X = [3523]$ and $(A X) B = [- 9 + 10 - 15 + 15 6 - 6 10 - 9] = I$ . (1 mark)

Answer: $X = [1110]$ . Writing $X = B^{- 1} A^{- 1}$ instead would be wrong, because the inverses must sit on the same sides as $A$ and $B$ .

Where students lose marks in Invertible Matrices

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

Mistake: Writing (AB)⁻¹ = A⁻¹B⁻¹
The order must reverse, because matrix multiplication is not commutative.
Wrong working
$(A B)^{- 1} = A^{- 1} B^{- 1}$ .
Right working
$(A B)^{- 1} = B^{- 1} A^{- 1}$ , since $(A B) (B^{- 1} A^{- 1}) = I$ .
Do this: Check any claimed inverse by multiplying it with the original.
Usually costs: 1 mark in two-statement MCQs; more in equations.
Mistake: Inverting a singular matrix
Without checking the determinant, students divide by zero or produce a meaningless answer.
Do this: Compute |A| first; if it is 0, stop and write "A is singular, so A⁻¹ does not exist".
Usually costs: The whole question.
Mistake: Multiplying by the inverse on the wrong side
From AX = B, writing X = BA⁻¹ assumes commutativity.
Wrong working
$A X = B \Rightarrow X = B A^{- 1}$ .
Right working
$A X = B \Rightarrow A^{- 1} A X = A^{- 1} B \Rightarrow X = A^{- 1} B$ .
Do this: Pre-multiply both sides explicitly, and write the step.
Usually costs: 2 to 3 marks.
Mistake: Sign errors in the 2 × 2 inverse
Students negate the diagonal instead of the off-diagonal entries.
Do this: Diagonal entries swap places but keep their signs; off-diagonal entries stay in place but change sign.
Usually costs: 1 mark, and every later step.

Exam tips

Show the determinant as a separate line (" $∣ A ∣ = 1 \neq = 0$ , so $A^{- 1}$ exists"); examiners expect the existence check.
Keep the scalar outside the matrix: $\frac{1}{10} [6 - 2 - 7 4]$ is clearer than fractions in every entry.
In properties MCQs, test each statement by a one-line proof rather than numbers; it is faster and leaves no doubt.
For the uniqueness proof, write each equality with its reason (definition, associativity); it is a short proof but every link is marked.
After solving a matrix equation, multiply back to check; it takes one line and catches side errors.

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 a $2 \times 2$ matrix with $∣ A ∣ = 4$ , then $∣ A^{- 1} ∣$ is:
(a) 4
(b) $\frac{1}{4}$
(c) 16
(d) $\frac{1}{16}$
Hint
$∣ A ∣∣ A^{- 1} ∣ = 1$ .
Answer
(b) $\frac{1}{4}$ .
Q2 · 2 marks · Short answerBoard level
Find the inverse of $[4276]$ .
Hint
Determinant first.
Answer
$∣ A ∣ = 24 - 14 = 10$ ; $A^{- 1} = \frac{1}{10} [6 - 2 - 7 4]$ .
Q3 · 2 marks · Short answerHarder board level
If $A$ is a square matrix with $A^{2} = I$ , show that $A$ is invertible and $A^{- 1} = A$ . Give a non-identity example.
Hint
Read $A \cdot A = I$ as the definition of an inverse.
Answer
$AA = I$ , so $A$ satisfies the definition with inverse $A$ itself. Example: $[0110]$ , whose square is $I$ .
Q4 · 3 marks · Short answerHarder board level
For $A = [3725]$ and $B = [6879]$ , find $(A B)^{- 1}$ .
Hint
Either invert $A B$ directly or compute $B^{- 1} A^{- 1}$ .
Answer
$A B = [34823994]$ , $∣ A B ∣ = 3196 - 3198 = - 2$ (equal to $∣ A ∣∣ B ∣ = 1 \times (- 2)$ ). $(A B)^{- 1} = - \frac{1}{2} [94 - 82 - 39 34] = [- 47 41 19.5 - 17]$ .
Q5 · 2 marks · Short answerHarder board level
Find the values of $k$ for which $[k 3 2 k - 1]$ has no inverse.
Hint
Set the determinant to zero.
Answer
$k (k - 1) - 6 = 0 \Rightarrow k^{2} - k - 6 = 0 \Rightarrow (k - 3) (k + 2) = 0$ , so $k = 3$ or $k = - 2$ .
Q6 · 3 marks · Short answerHOTS / competency
(i) If $A$ is invertible and $A B = A C$ , prove that $B = C$ . (ii) Show that this fails for $A = [1111]$ , $B = I$ and $C = [0110]$ .
Hint
Multiply on the left by $A^{- 1}$ .
Answer
(i) $A^{- 1} (A B) = A^{- 1} (A C) \Rightarrow (A^{- 1} A) B = (A^{- 1} A) C \Rightarrow B = C$ . (ii) $A B = A = [1111]$ and $A C = [1111]$ , so $A B = A C$ but $B \neq = C$ ; this is possible because $∣ A ∣ = 0$ .
Q7 · 4 marks · Case-basedHOTS / competency
Case study: Suppose two friends send each other pairs of numbers (letters as positions in the alphabet) using the coding matrix $E = [3121]$ : a pair $X$ is sent as $E X$ .
(i) Show that $E$ is invertible. (1 mark)
(ii) Find the decoding matrix $E^{- 1}$ . (1 mark)
(iii) Decode the received pair $(31, 12)$ and write the letters. (1 mark)
(iv) Explain why $[2142]$ would be a bad coding matrix. (1 mark)
Hint
Decoding is multiplication by the inverse.
Answer
(i) $∣ E ∣ = 3 - 2 = 1 \neq = 0$ . (ii) $E^{- 1} = [1 - 1 - 2 3]$ . (iii) $E^{- 1} [3112] = [75]$ : the letters G and E. (iv) Its determinant is $4 - 4 = 0$ , so it has no inverse: different messages can be sent as the same pair and cannot be decoded.

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 invertible?Show answer
When its determinant is non-zero (it is non-singular).
2.Write the inverse of $[a c b d]$ .Show answer
$\frac{1}{a d - b c} [d - c - b a]$ , if $a d - b c \neq = 0$ .
3.State $(A B)^{- 1}$ in terms of $A^{- 1}$ and $B^{- 1}$ .Show answer
$B^{- 1} A^{- 1}$ .
4.If $∣ A ∣ = - 3$ , what is $∣ A^{- 1} ∣$ ?Show answer
$- \frac{1}{3}$ .
5.Solve $X A = B$ for $X$ .Show answer
$X = B A^{- 1}$ .
6.Is $(A^{2})^{- 1} = (A^{- 1})^{2}$ for an invertible $A$ ?Show answer
Yes, since $A^{2} (A^{- 1})^{2} = A (A A^{- 1}) A^{- 1} = I$ .

Invertible Matrices: frequently asked questions

What is the difference between a singular and a non-singular matrix?

A singular matrix has determinant zero and no inverse. A non-singular matrix has a non-zero determinant and exactly one inverse.

Why does a non-square matrix have no inverse?

An inverse must satisfy both AB = I and BA = I with the same identity. If A is m × n with m ≠ n, then AB and BA would be identities of different orders, so the definition cannot be met. Determinants are also defined only for square matrices.

How do I prove that the inverse of a matrix is unique?

Assume two inverses B and C. Then B = BI = B(AC) = (BA)C = IC = C. The proof uses only the definition of an inverse and the associative law.

How do I find the inverse of a 3 × 3 matrix?

Use A⁻¹ = (adj A)/|A|: find the cofactors, transpose them to get the adjoint, and divide by the determinant. This is covered step by step on the adjoint and inverse topic page under Determinants.

Is the proof of uniqueness of the inverse in the ISC 2027 syllabus?

Yes. The CISCE 2027 syllabus lists invertible matrices and the proof of the uniqueness of the inverse, if it exists, for matrices with real entries.