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Adjoint questions in JEE look like a list of formulas to memorise: adj(kA), adj(AB), adj(adj A), (adj A)⁻¹. Students who memorise them mix up the powers under time pressure. This post shows that every one of them comes from a single line in NCERT, works through a complete 3×3 inverse two ways, and lists the slips that cost marks.
What does the syllabus ask for?
The JEE (Main) 2026 syllabus lists "Adjoint and inverse of a square matrix" in Unit 3, next to the test of consistency. The JEE (Advanced) 2026 brochure asks for the adjoint, the inverse of a square matrix of order up to three and "properties of these matrix operations". The current NCERT Class 12 chapter on determinants (reprint 2026-27) covers the core in Section 4.5. Check the documents for your exam year.
Which one identity gives all the adjoint properties?
NCERT defines adj A as the transpose of the matrix of cofactors, then states Theorem 1: A(adj A) = (adj A)A = |A|I. From this it proves that A is invertible exactly when |A| ≠ 0, with A⁻¹ = (1/|A|) adj A, and derives |adj A| = |A|ⁿ⁻¹ for a matrix of order n.
Turn the inverse formula around and you get the line to remember:
adj A = |A| A⁻¹ (for |A| ≠ 0)
Now every property is two steps of algebra. For an n×n matrix:
adj(kA). |kA| = kⁿ|A| and (kA)⁻¹ = A⁻¹/k, so adj(kA) = kⁿ⁻¹ adj A.
adj(AB). |AB| = |A||B| and (AB)⁻¹ = B⁻¹A⁻¹, so adj(AB) = (|B|B⁻¹)(|A|A⁻¹) = (adj B)(adj A). The order reverses, as it does for the inverse.
adj(adj A). |adj A| = |A|ⁿ⁻¹ and (adj A)⁻¹ = A/|A|, so adj(adj A) = |A|ⁿ⁻² A.
(adj A)⁻¹ and adj(A⁻¹). Both equal A/|A|.
adj(Aᵀ) = (adj A)ᵀ, because |Aᵀ| = |A| and (Aᵀ)⁻¹ = (A⁻¹)ᵀ.
These derivations assume |A| ≠ 0. The results also hold for singular matrices, but that needs a different argument, and JEE questions almost always give you an invertible matrix.
Worked example: a 3×3 inverse two ways
Let A have rows (2, 1, 1), (1, 2, 1) and (1, 1, 2).
Method 1: cofactors. Expand along the first row: |A| = 2(4 − 1) − 1(2 − 1) + 1(1 − 2) = 6 − 1 − 1 = 4.
The cofactors are A₁₁ = 3, A₁₂ = −1, A₁₃ = −1, A₂₂ = 3, A₂₃ = −1 and A₃₃ = 3. Because A is symmetric, A₂₁ = A₁₂ and so on, so adj A has rows (3, −1, −1), (−1, 3, −1) and (−1, −1, 3). Then A⁻¹ = (1/4) adj A.
Check one entry of A(adj A): row 1 of A times column 1 of adj A is 6 − 1 − 1 = 4, and row 1 times column 2 is −2 + 3 − 1 = 0. That matches |A|I = 4I.
Method 2: a matrix equation. NCERT's Example 15 finds an inverse from an equation the matrix satisfies. Here A = I + J, where J is the matrix of all ones, and J² = 3J. So A² = I + 2J + 3J = I + 5J, and A² − 5A + 4I = I + 5J − 5I − 5J + 4I = O. Multiply by A⁻¹: A − 5I + 4A⁻¹ = O, so A⁻¹ = (5I − A)/4. The diagonal of 5I − A is 3 and the off-diagonal entries are −1, the same matrix as before.
Worked example: properties without entries
Let A be 3×3 with |A| = 3. Then:
adj(2A) = 2² adj A = 4 adj A, not 2 adj A.
adj(adj A) = 3¹ A = 3A.
|adj(adj A)| = |3A| = 3³ × 3 = 81. As a check, |adj A| = 3² = 9, and |adj(adj A)| = 9² = 81.
(adj A)⁻¹ = A/3.
None of these needs a single entry of A. Our post on matrices and determinants for JEE covers the determinant versions of these shortcuts.
Which mistakes cost the most marks?
Forgetting the transpose. adj A is the transpose of the cofactor matrix; for a non-symmetric matrix, skipping it gives the wrong answer with no warning.
Sign errors in cofactors. The cofactor of aᵢⱼ is (−1)^(i+j) times the minor.
Writing adj(kA) = k adj A, or using kⁿ instead of kⁿ⁻¹.
Writing adj(AB) = (adj A)(adj B). Matrix products do not commute, so the order matters.
Using A⁻¹ = adj A/|A| when |A| = 0. A singular matrix has no inverse.
Mixing up the 2×2 shortcut. For rows (a, b) and (c, d), adj A has rows (d, −b) and (−c, a): swap the diagonal, change the sign of the other two.
A practice routine
Studying solved examples is a sound way to start: Sweller and Cooper (1985) found that algebra students who studied worked examples later solved similar problems faster and with fewer errors. They also found the gain was specific to problems with the same structure, so do not stop there.
Derive the list (once a week, 10 minutes). From a blank page, derive the five properties above from adj A = |A|A⁻¹.
Entry-free set (three times a week). Ten questions of the "|A| = 3, find adj(adj 2A)" type, no matrix entries allowed.
Two-method inverse. Once a week, invert one 3×3 matrix by cofactors and check it with A(adj A) = |A|I.
Mix it in. Put these questions into mixed sets with other chapters. In a study of 126 seventh-grade students, interleaved practice beat blocked practice on tests one day and 30 days later.
If your cofactor answers keep disagreeing with the answer key, the Fermi Problem Library has worked problems with a step-by-step solver that explains the reasoning, so you can find the exact step where a sign went wrong.
Today, take the 3×3 matrix above, change its last row to (1, 1, 3), and find its inverse both ways.
