Try
All numerical examples below use real scalars. The general identities will be proved over any field. Consider the matrix A whose rows are (1,2,0), (0,1,3), (2,0,1). We want to solve Ax=b, but first construct an exact certificate that can be checked by multiplication. For an entry in row i and column j, delete that row and column. The determinant of the remaining matrix is its minor. Multiply it by (−1)^(i+j) to obtain the cofactor C(i,j). The signs follow a checkerboard starting with plus in the top-left corner.
For example, deleting row 1 and column 2 leaves rows (0,3),(2,1), whose determinant is −6. The checkerboard sign is negative, so C(1,2)=6. Predict C(2,1) before checking the table. Its minor has rows (2,0),(0,1), determinant 2, and its negative sign gives −2. These two cofactors differ: their positions must not be silently exchanged.
| Row | Entries of A | Cofactors in the same row | Row expansion |
|---|
| 1 | (1,2,0) | (1,6,−2) | 1·1+2·6+0·(−2)=13 |
| 2 | (0,1,3) | (−2,1,4) | 0·(−2)+1·1+3·4=13 |
| 3 | (2,0,1) | (6,−3,1) | 2·6+0·(−3)+1·1=13 |
Each row gives the same determinant. Recompute one complete row of minors, including every sign. Agreement is a useful arithmetic check; the general proof below explains why the row choice cannot change the answer.
Predict
The cofactor matrix C keeps each cofactor in its original position. The adjugate, written adj A, is Cᵀ: transpose that matrix. Our adjugate therefore has rows (1,−2,6), (6,1,−3), (−2,4,1).
Predict the entry in row 1, column 2 of A adj A. It is (1,2,0) dotted with (−2,1,4), giving zero. The diagonal entry in row 1 is (1,2,0) dotted with (1,6,−2), giving 13. The transpose aligns a matrix row with an entire row of cofactors. Using C directly instead would give the wrong pattern.
Try b=(−1,5,4). Multiplying adj A by b gives (13,−13,26). If A adj A=13I, dividing by 13 should give the solution x=(1,−1,2). Substitute this vector into all three original equations now. The output is indeed (−1,5,4). We still owe proofs of the identity and of the division condition; a successful example is not a proof for every matrix.
Experiment
Change to the 2×2 matrix B with rows (3,1),(2,1). Build its signed-minor table on paper, transpose it, and multiply on both sides. Then replace its second row by (6,2). Predict which statements survive when the determinant becomes zero.
Check the changed matrices
For the original B, det B=1 and adj B has rows (1,−1),(−2,3). Both products equal I. For the changed matrix D with rows (3,1),(6,2), det D=0 and adj D has rows (2,−1),(−6,3). Both products are zero. The adjugate still exists and is nonzero, but it is not an inverse. The system Dx=b is consistent exactly when b₂=2b₁. Each column of adj D is a solution of Dx=0.
Now test a matrix with all entries zero and size 3×3. Its adjugate is zero because every 2×2 minor is zero. The equation adj A b=0 now holds for every b, including unattainable requests. A certificate useful at rank n−1 can become uninformative at lower rank.
Understand
Work over a field, with an n×n matrix and n≥1. Use the row-indexed permutation formula: σ(i) is the column chosen in row i. It agrees with the preceding column-indexed convention by replacing each permutation with its inverse. Fix row i and group permutations according to the column j selected in that row. Delete position i and value j; the remaining permutation acts on the ordered remaining rows and columns. This deletion is a bijection with permutations of n−1 indices. Move row i to the first position and column j to the first position: there are i−1 and j−1 adjacent moves. The combined sign is (−1)^(i+j−2)=(−1)^(i+j). The first selected entry is then a(i,j), and the remaining signed products sum to the deleted minor determinant. Summing all groups proves the Laplace expansion
det A = Σ over j of a(i,j) C(i,j).
No minor is assumed invertible. For a 1×1 matrix [a], deleting its entry leaves the empty matrix. Its determinant is the empty product 1, so the cofactor is 1 and the formula returns a. This convention also gives adj[a]=[1], even if a=0.
Column expansion follows from transpose invariance. In the permutation sum for Aᵀ, replace a permutation by its inverse. Its sign is unchanged, and reindexing the product gives precisely the sum for A. Thus det Aᵀ=det A, and row expansion of Aᵀ is column expansion of A.
For the adjugate identity, the (i,j) entry of A adj A is Σ over k of a(i,k) C(j,k). If i=j, this is row expansion and equals det A. If i≠j, it is the determinant of the matrix obtained by replacing row j with row i: its row-j cofactors are unchanged because that row is deleted when they are formed. The new matrix has two equal rows, so its determinant is zero. Consequently A adj A=(det A)I. The same argument with column i replaced by column j proves adj A A=(det A)I: its (i,j) entry is Σ C(k,i)a(k,j). Both identities are necessary to describe both kernel directions.
If det A≠0, multiplication by its reciprocal yields an explicit two-sided inverse, A⁻¹=adj A/det A. Conversely, invertibility forces a nonzero determinant by the previously established determinant criterion. These are exact algebraic identities. Cofactor enumeration and division near a small determinant are not being recommended as a stable numerical solution algorithm.
To derive Cramer’s rule, write the columns of A as a₁,…,aₙ and b=Ax=Σ x(k)aₖ. Replace column j by b. Multilinearity expands its determinant into a sum. Each term using aₖ with k≠j has repeated columns and vanishes. The remaining term is x(j)det A. Therefore x(j)=det Aⱼ(b)/det A when det A≠0. Existence and uniqueness come from the inverse; the ratio identity alone would only be a necessary statement about a solution assumed to exist.
Finally suppose det A=0 and n≥2. Both adjugate products vanish. Thus its columns lie in ker A and its rows lie in the left kernel. Use the fact that rank equals the largest size of a nonzero square minor, justified below. At rank n−1, some cofactor is nonzero, and the kernel is one-dimensional. Hence adj A is nonzero of rank one. At rank at most n−2, every cofactor vanishes and adj A=0. This is why a singular adjugate can either identify a direction or carry no information.
Why rank is the largest nonzero minor size
A nonzero k×k minor is invertible by the preceding determinant criterion. Its selected columns are independent even after restricting to the selected rows, so the corresponding full columns are independent too. Therefore the original matrix has rank at least k.
Conversely, if its rank is r, choose r independent columns and put them into a matrix C. The equality of row rank and column rank, proved by row reduction, gives r independent rows of C. Those rows form an invertible r×r matrix, so its determinant is nonzero. They select an r×r minor of the original matrix. Together the two arguments give the claimed largest size. At rank zero all entries vanish; the empty minor has determinant 1 and size zero, consistently with the statement. This argument uses row-rank equality and the already proved determinant criterion, not the adjugate rank conclusion.
Apply
For E with rows (1,2),(2,4), compute adj E and use its rows to test b=(3,6) and b=(3,5). You should obtain adj E with rows (4,−2),(−2,1). The first request is consistent, with all solutions (3−2t,t); the second is not. Explain why these statements require checking the image condition, rather than dividing by the zero determinant.
Generalize that check. At rank n−1, choose nonzero vectors z in the kernel and y in the left kernel, so yᵀA=0. All adjugate columns are multiples of z, so adj A=zvᵀ for a nonzero v. The other product identity gives vᵀA=0, hence v=λy for nonzero λ. Therefore adj A b=0 is equivalent to yᵀb=0. The image of A lies in this hyperplane and has the same dimension n−1, so the two spaces are equal. At this rank the adjugate condition is sufficient for consistency. Revisit the zero 3×3 matrix to explain exactly why the proof fails when the rank is smaller.
Continue
A complete cofactor argument records the deleted row and column, the sign, the transpose, and the determinant condition before dividing. A singular matrix still has cofactors and an adjugate; it does not thereby acquire an inverse. Keep exact existence certificates separate from numerical accuracy and from efficient algorithms. The written tasks ask for permutation-level proofs, changed computations and rank-dependent consistency certificates so that the method extends beyond the worked table.
