Matrix Inverse Techniques
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Direct answer
A square matrix admits an inverse exactly when its determinant is nonzero; then A^(−1) = (adj A)/|A|, where adj A is the transpose of the cofactor matrix. For 2 × 2 matrices the adjoint is mechanical — swap the diagonal entries, negate the off-diagonal — and for 3 × 3, row reduction from [A | I] to [I | A^(−1)] beats nine cofactors on time. Inverses reverse order, (AB)^(−1) = B^(−1)A^(−1), and commute with transposes, (A^T)^(−1) = (A^(−1))^T. They settle systems too: AX = B has the unique solution X = A^(−1)B precisely when |A| ≠ 0; when |A| = 0, testing (adj A)B delivers the verdict — zero means infinitely many solutions, nonzero means none.
What you must remember
- Existence: invertible ⇔ |A| ≠ 0 ⇔ rows linearly independent; singular means |A| = 0 and no inverse exists in any form.
- Master identity: A(adj A) = (adj A)A = |A| I — from it both the inverse formula and most adjoint questions fall out.
- 2 × 2 shortcut: for rows (a, b) and (c, d), the adjoint swaps to (d, −b) and (−c, a), so A^(−1) = (1/(ad − bc)) times that matrix.
- Order reversal: (AB)^(−1) = B^(−1)A^(−1) and adj(AB) = (adj B)(adj A) — like undoing socks then shoes.
- Special matrices: involutory (A² = I) is its own inverse; orthogonal (A^T A = I) has A^(−1) = A^T; idempotent (A² = A) is singular unless it equals I.
- Systems by rank of verdict: |A| ≠ 0 → unique solution; |A| = 0 with (adj A)B = O → infinitely many; (adj A)B ≠ O → inconsistent.
- Determinant links: |A^(−1)| = 1/|A| and |adj A| = |A|^(n−1) — quick sanity checks after any inversion.
A worked inversion
Invert A with rows (2, 3) and (1, 4), then solve 2x + 3y = 8 and x + 4y = 6 in one stroke. The determinant is 2 × 4 − 3 × 1 = 5, nonzero, so the inverse is (1/5) times the matrix with rows (4, −3) and (−1, 2). Multiply by the right-hand side: x = (4 × 8 − 3 × 6)/5 = 14/5 and y = (−8 + 2 × 6)/5 = 4/5. Verify in the original equations: 2(14/5) + 3(4/5) = 40/5 = 8 and 14/5 + 4(4/5) = 30/5 = 6. The point of writing X = A^(−1)B rather than eliminating by hand is that the same inverse answers every right-hand side you are later handed — one matrix division replaces repeated substitution, which is precisely how multi-part Main numericals are constructed.
How the exam frames it
JEE Main stays close to the machinery: compute a 2 × 2 or 3 × 3 inverse, use adjoint identities, or decide the nature of a system — and the favourite distractor is order: for (AB)^(−1), the options always include (A^(−1)B^(−1)). JEE Advanced prefers the functional style: given A² − A + I = O, rearrange to A(I − A) = I, so A^(−1) = I − A with no determinant ever computed. The same trick handles A² = A − I, where multiplying through shows A³ = −I and A^(−1) = −A². The systematic loss is pre-multiplication discipline: X = A^(−1)B pre-multiplies, and writing BA^(−1) is dimensionally legal in square problems but logically wrong whenever compositions appear. When a system question reports |A| = 0, do not stop — compute (adj A)B before concluding anything; the half-finished "singular, so no solution" is exactly the trap the answer options reward wrongly.
Frequently asked questions
When does a square matrix fail to have an inverse?
Exactly when |A| = 0 — the rows are then linearly dependent and no matrix can undo the collapse.
What is the inverse of a 2 × 2 matrix?
(1/(ad − bc)) times the adjoint with rows (d, −b) and (−c, a) — one determinant, four sign-conscious entries.
Why does (AB)^(−1) reverse the order?
Because undoing AB means undoing B first: (AB)(B^(−1)A^(−1)) = AIA^(−1) = I.
If A² − A + I = O, what is A^(−1)?
I − A, read off by rearranging to A(I − A) = I — no adjoint or determinant needed.
How do you classify AX = B when |A| = 0?
Compute (adj A)B: zero gives infinitely many solutions, nonzero means the system is inconsistent.