9 min readConcept sheet1-mark MCQ on |adj A|; 2 and 3-mark questions; the main step of every inverse question; also a JEE favourite

Adjoint of a square matrix

The adjoint of A is the transpose of its matrix of cofactors. For every square matrix, A(adj A) = (adj A)A = |A| I.

Definitionadj A = transpose of [Aᵢⱼ], the matrix of cofactors of A

2 × 2 shortcutFor A = [a, b; c, d], adj A = [d, −b; −c, a]: swap the diagonal, change the signs of the other two

Key theoremA(adj A) = (adj A)A = |A| I, where I is the identity matrix of the same order

Determinant of the adjoint|adj A| = |A|ⁿ⁻¹ for a matrix of order n

What the adjoint is

Replace every element of A by its cofactor. This gives the cofactor matrix [Aᵢⱼ]. Its transpose (rows written as columns) is the adjoint of A, written adj A. Every square matrix has an adjoint, even a singular one. We write matrices with rows separated by a semicolon.

How to find adj A for a 3 × 3 matrix

1
Find all nine minors Mᵢⱼ.
2
Attach signs to get the cofactors: Aᵢⱼ = (−1)ⁱ⁺ʲ Mᵢⱼ.
3
Write the cofactor matrix: row 1 is A₁₁, A₁₂, A₁₃, and so on.
4★
Transpose it: row 1 of the cofactor matrix becomes column 1 of adj A.
5
Check: A(adj A) must equal |A| I.

★ key step

Example 1: a 2 × 2 matrix

Problem

Find adj A for A = [3, −4; 2, 5] and verify that A(adj A) = |A| I.

Working

For [a, b; c, d], adj A = [d, −b; −c, a].
adj A = [5, 4; −2, 3]
A(adj A) = [3 × 5 + (−4)(−2), 3 × 4 + (−4) × 3; 2 × 5 + 5 × (−2), 2 × 4 + 5 × 3]
= [15 + 8, 12 − 12; 10 − 10, 8 + 15] = [23, 0; 0, 23]
|A| = 3 × 5 − (−4) × 2 = 15 + 8 = 23, so |A| I = [23, 0; 0, 23].

Answer

adj A = [5, 4; −2, 3] and A(adj A) = 23 I = |A| I

Example 2: a 3 × 3 matrix

Problem

Find adj A for A = [1, 0, 2; −1, 3, 1; 2, 1, 0].

Working

Row 1 cofactors: A₁₁ = +(0 − 1) = −1, A₁₂ = −(0 − 2) = 2, A₁₃ = +(−1 − 6) = −7
Row 2 cofactors: A₂₁ = −(0 − 2) = 2, A₂₂ = +(0 − 4) = −4, A₂₃ = −(1 − 0) = −1
Row 3 cofactors: A₃₁ = +(0 − 6) = −6, A₃₂ = −(1 + 2) = −3, A₃₃ = +(3 − 0) = 3
Cofactor matrix = [−1, 2, −7; 2, −4, −1; −6, −3, 3]
Transpose it: adj A = [−1, 2, −6; 2, −4, −3; −7, −1, 3]
Check: |A| = 1(−1) + 0(2) + 2(−7) = −15, and row 1 of A times column 1 of adj A = −1 + 0 − 14 = −15.

Answer

adj A = [−1, 2, −6; 2, −4, −3; −7, −1, 3], with |A| = −15

Results to remember

A(adj A) = (adj A)A = |A| I for every square matrix A.
|adj A| = |A|ⁿ⁻¹ for order n: |adj A| = |A| for a 2 × 2 matrix and |A|² for a 3 × 3 matrix.
If |A| = 0, then A(adj A) = O, the zero matrix.
For a 2 × 2 matrix: swap the diagonal elements and change the signs of the other two.
Do not forget to transpose. The cofactor matrix itself is not adj A.

In real life

The adjoint is the hand method behind matrix inversion. Software that designs power grids and bridges solves the same kind of matrix problems, only with thousands of rows.

Exam angle

1 marks: “If A is a 3 × 3 matrix with |A| = 4, find |adj A|.” |adj A| = |A|ⁿ⁻¹ = 4² = 16.

2 marks: “Find adj A for A = [2, −1; 3, 4] and verify that A(adj A) = |A| I.” adj A = [4, 1; −3, 2]. A(adj A) = [8 + 3, 2 − 2; 12 − 12, 3 + 8] = [11, 0; 0, 11]. |A| = 8 + 3 = 11, so A(adj A) = 11 I = |A| I.

2 marks: “If A is a square matrix of order 3 with |A| = −5, find |A(adj A)|.” A(adj A) = |A| I = −5I. So |A(adj A)| = (−5)³ × |I| = −125.

Remember it

Cofactors, then transpose. For a 2 × 2 matrix: swap the diagonal, change the signs of the rest.