Read through concept notes with worked explanations and diagrams, chapter by chapter.
Tip: use←→to switch concepts
Concept 08 / 09Matrices and Determinants
Adjoint and Inverse of a Matrix
Cofactor matrix and adjoint, the identity A adj A = |A| I, the inverse formula and the invertibility criterion, the properties of adj A (|adj A| = |A|^(n-1), adj adj A = |A|^(n-2) A), reversal rules for inverses, and the 2 x 2 shortcut.
The adjoint (classical adjugate) of a square matrix A is the transpose of its matrix of cofactors: adjA=[Cij]T. So the (i,j) entry of adjA is the cofactor Cji.
The key identity. A⋅adjA=adjA⋅A=∣A∣I. The (i,i) entry of A⋅adjA is row i of A times its own cofactors, which is ∣A∣. The (i,j) entry for i=j is row i of A times the cofactors of row j, which is 0 (it is the determinant of a matrix with two equal rows). Hence the diagonal of ∣A∣ values and zeros elsewhere.
Inverse. A square matrix A is invertible (non-singular) if there is B with AB=BA=I; then B=A−1 is unique. From the key identity, if ∣A∣=0,
A−1=∣A∣1adjA.
Conversely, if A is invertible then ∣A∣∣A−1∣=∣I∣=1, so ∣A∣=0. Therefore A is invertible if and only if ∣A∣=0. For square matrices, AB=I alone already implies BA=I.
The 2×2 shortcut: for A=[acbd], adjA=[d−c−ba] (swap the diagonal entries and negate the off-diagonal ones). So A−1=ad−bc1[d−c−ba].
Rules for inverses. (AB)−1=B−1A−1 (reverse order, like the transpose rule). (AT)−1=(A−1)T. (kA)−1=k1A−1 for k=0. (A−1)−1=A. ∣A−1∣=∣A∣1. The inverse of a diagonal matrix is the diagonal matrix of reciprocals. If A is symmetric, so is A−1; if A is skew-symmetric and invertible (even order), so is A−1.
Properties of the adjoint for an n×n matrix A. These are asked very often, usually with n=3.
∣adjA∣=∣A∣n−1. Proof: take determinants in AadjA=∣A∣I to get ∣A∣∣adjA∣=∣A∣n, then divide by ∣A∣ when it is non-zero (the result also holds when ∣A∣=0).
adj(adjA)=∣A∣n−2A.
∣adj(adjA)∣=∣A∣(n−1)2.
adj(AB)=adjB⋅adjA.
adj(AT)=(adjA)T.
adj(kA)=kn−1adjA.
adj(A−1)=(adjA)−1=∣A∣A.
For n=3: ∣adjA∣=∣A∣2 and adj(adjA)=∣A∣A, and ∣adj(adjA)∣=∣A∣4.
Elementary row operations give a second method for finding the inverse: write [A∣I] and apply row operations to turn A into I; the same operations turn I into A−1. Use only row operations (or only column operations) throughout, never a mixture.
Key Takeaways:
- adjA=(cofactor matrix)T; AadjA=adjAA=∣A∣I.
- A−1=∣A∣adjA exists iff ∣A∣=0.
- 2×2: swap the diagonal, negate the off-diagonal, divide by ad−bc.
- (AB)−1=B−1A−1; (AT)−1=(A−1)T.
- ∣adjA∣=∣A∣n−1; adjadjA=∣A∣n−2A; adj(kA)=kn−1adjA.
- ∣A−1∣=∣A∣1 and ∣adj(A−1)∣=∣A∣1−n.
- If AB=AC and ∣A∣=0, then B=C (multiply on the left by A−1).
- Pitfall: forgetting the transpose when forming the adjoint from cofactors.
Brain-Teaser 1: A is 3×3 with ∣A∣=5. Find ∣adjA∣, ∣adj(adjA)∣, ∣2adjA∣ and ∣A−1adjA∣.
Solution: ∣adjA∣=52=25. ∣adj(adjA)∣=∣A∣(n−1)2=54=625. ∣2adjA∣=23×25=200. And A−1adjA=∣A∣adjAadjA=51(adjA)2, so its determinant is 531×252=125625=5. Every one of these is a direct application of ∣kM∣=kn∣M∣ and the product rule, with adjA treated as just another matrix whose determinant is known.
Brain-Teaser 2: If A2−5A+7I=O for a 2×2 matrix A, find A−1 in terms of A.
Solution: Rearrange: A2−5A=−7I, so A(A−5I)=−7I and A⋅75I−A=I. Hence A−1=71(5I−A). The existence of the inverse is guaranteed because the equation exhibits a matrix that multiplies A to give I. Such polynomial equations usually come from the Cayley-Hamilton theorem; here, for A=[3−112], the trace is 5 and the determinant is 7, and A−1=71[21−13]=71(5I−A) indeed.
Brain-Teaser 3: Is it true that adj(A+B)=adjA+adjB? Test with 2×2 matrices, then explain what happens for 3×3 matrices.
Solution: For 2×2 matrices, adj is linear: it swaps the diagonal and negates the off-diagonal entries, each operation being linear in the entries. So the identity holds for every pair of 2×2 matrices. For 3×3 matrices, each entry of adjA is a 2×2 minor, a degree-2 expression in the entries, so adj is not linear. For example, adj(2I)=4I, whereas adjI+adjI=2I. This matches the rule adj(kA)=kn−1adjA, which is linear only when n−1=1.
