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CLASS 12 · MATHEMATICS · CHAPTER 4

Inverse of
a Matrix

Using determinants and the adjoint to find the inverse of a matrix.

1

Concept of Inverse

📘 Definition — Inverse of Matrix A

For a square matrix A of order n, another matrix B is called the inverse of A if:

A · B = B · A = I

where I is the identity matrix of order n. We write B = A⁻¹.

⚠️ Condition: The inverse of A exists if and only if |A| ≠ 0. A matrix with |A| = 0 is called singular and has no inverse.

💡

Always check |A| ≠ 0 before attempting to find the inverse. This is Step 0 in every inverse problem.

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The Inverse Formula

⭐ The Master Formula for Inverse
A⁻¹ = (1 / |A|) × adj(A)
valid only when |A| ≠ 0

This formula is derived from the fundamental result: A · adj(A) = |A| · I

Dividing both sides by |A|: A · (adj(A)/|A|) = I, which means A⁻¹ = adj(A)/|A|

🔑

For a 2×2 matrix A = [[a,b],[c,d]], the inverse has a shortcut:
A⁻¹ = (1/|A|) × [[d, −b],[−c, a]]
Just swap the diagonal elements and negate the off-diagonal elements, then divide by |A|.

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Step-by-Step Method

0

Check |A| ≠ 0

Compute the determinant first. If |A| = 0, inverse does not exist — stop.

1

Find all Cofactors Aᵢⱼ

Compute the cofactor for every element using (−1)^(i+j) × Mᵢⱼ.

2

Form the Cofactor Matrix

Arrange all cofactors into a matrix at their corresponding positions.

3

Transpose → get adj(A)

Take the transpose of the cofactor matrix to get adj(A).

4

Divide by |A|

A⁻¹ = (1/|A|) × adj(A). Multiply each element of adj(A) by 1/|A|.

Verify: A · A⁻¹ = I

Multiply A by A⁻¹ and confirm you get the identity matrix.

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Inverse of a 2×2 Matrix

✏️ Solved Example
Find A⁻¹ for A = [[3, 1], [5, 2]]
1

|A| = (3)(2) − (1)(5) = 6 − 5 = 1 ≠ 0 ✓ Inverse exists

2

adj(A) = [[2, −1],[−5, 3]] (swap diagonal, negate off-diagonal)

3

A⁻¹ = (1/1) × [[2,−1],[−5,3]] = [[2,−1],[−5,3]]

✅ Verify: A × A⁻¹ = [[3,1],[5,2]] × [[2,−1],[−5,3]] = [[6−5, −3+3],[10−10, −5+6]] = [[1,0],[0,1]] = I ✓

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Inverse of a 3×3 Matrix

✏️ Full Solved Example
Find A⁻¹ for A = [[1, 0, 0], [3, 3, 0], [5, 2, −1]]
|A|

Expand along R1: 1·|3,0;2,−1| − 0 + 0 = 1·(−3−0) = −3 ≠ 0

C₁₁

(+1)|3,0;2,−1| = −3

C₁₂

(−1)|3,0;5,−1| = −(−3−0) = 3

C₁₃

(+1)|3,3;5,2| = 6−15 = −9

C₂₁

(−1)|0,0;2,−1| = −(0−0) = 0

C₂₂

(+1)|1,0;5,−1| = −1−0 = −1

C₂₃

(−1)|1,0;5,2| = −(2−0) = −2

C₃₁

(+1)|0,0;3,0| = 0−0 = 0

C₃₂

(−1)|1,0;3,0| = −(0−0) = 0

C₃₃

(+1)|1,0;3,3| = 3−0 = 3

adj

Cofactor matrix transposed:
adj(A) = [[−3, 0, 0],[3, −1, 0],[−9, −2, 3]]

A⁻¹

A⁻¹ = (1/−3) × adj(A) = [[1, 0, 0],[−1, 1/3, 0],[3, 2/3, −1]]

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Important Results & Properties

📌 Key Results

(A⁻¹)⁻¹ = A — Inverse of inverse is the original matrix
(AB)⁻¹ = B⁻¹A⁻¹ — Reversal rule for products
(Aᵀ)⁻¹ = (A⁻¹)ᵀ — Inverse of transpose
|A⁻¹| = 1/|A| — Determinant of inverse
A·A⁻¹ = A⁻¹·A = I — Definition check

Matrix TypeConditionInverse Exists?
Non-Singular|A| ≠ 0Yes ✓
Singular|A| = 0No ✗
Identity I|I| = 1Yes (I⁻¹ = I)
Scalar Matrix kIk ≠ 0Yes ((kI)⁻¹ = (1/k)I)