From Minors to the Adjugate
The adjugate is a matrix built from the determinants of smaller matrices obtained by deleting rows and columns. Its construction requires careful attention to two details: the sign attached to each smaller determinant, and the transposition that places the resulting cofactors in the adjugate. These steps apply to every square matrix, whether or not its determinant is zero.
Throughout, let \(\mathbf{A}=[a_{ij}]\) be an \(n\times n\) matrix over \(\mathbb{R}\). The minor at position \((i,j)\) is found by deleting row \(i\) and column \(j\). The determinant of what remains is denoted \(M_{ij}\). The cofactor at that position adds a sign to the minor. For \(n=1\), use the convention that the determinant of the empty matrix is \(1\), so the single cofactor is \(1\).
Thus, the cofactor \(C_{ij}\) is not placed in position \((i,j)\) in the adjugate. It appears in position \((j,i)\): \((\operatorname{adj}\mathbf{A})_{ij}=C_{ji}\). This distinction is essential when assembling the final matrix.
The Sign Pattern and a Small Formula
The factor \((-1)^{i+j}\) produces an alternating checkerboard of signs, beginning with a plus sign in the upper-left corner. The signs depend only on the row and column indices, not on the values in the matrix.
Proof. By definition, the sign is \((-1)^{i+j}\). If \(i\) and \(j\) have the same parity, their sum is even, so \((-1)^{i+j}=1\). If they have different parity, their sum is odd, so \((-1)^{i+j}=-1\). Applying this rule to \(i,j\in\{1,2,3\}\) gives the displayed array. \(\square\)
For a two-by-two matrix, each minor is a single remaining entry. This gives a compact formula that is also a useful check on the indexing.
Proof. Deleting row 1 and column 1 leaves \(d\), so \(M_{11}=d\) and \(C_{11}=d\). Deleting row 1 and column 2 leaves \(c\), and the sign is negative, so \(C_{12}=-c\). Deleting row 2 and column 1 leaves \(b\), with negative sign, so \(C_{21}=-b\). Deleting row 2 and column 2 leaves \(a\), with positive sign, so \(C_{22}=a\). Therefore \(\mathbf{C}=\begin{bmatrix}d&-c\\-b&a\end{bmatrix}\), and transposing gives the stated adjugate. \(\square\)
Constructing the Matrix in Practice
A reliable construction separates the arithmetic into three stages: compute the unsigned minors, apply the checkerboard signs to get the cofactors, and transpose the cofactor matrix. Keep the row and column labels attached to each result until the final stage; otherwise it is easy to transpose too soon or not at all.
For each position \((i,j)\), delete row \(i\) and column \(j\), then evaluate the determinant of the remaining matrix.
Multiply \(M_{ij}\) by \((-1)^{i+j}\). Record the result as \(C_{ij}\) in position \((i,j)\) of the cofactor matrix.
Interchange rows and columns of the cofactor matrix. The resulting matrix is \(\operatorname{adj}\mathbf{A}\).
Worked Example: A Two-by-Two Adjugate
Construct the adjugate of \(\mathbf{A}=\begin{bmatrix}2&1\\3&4\end{bmatrix}\). For each position, delete the indicated row and column and apply its sign:
The cofactor matrix retains these cofactors in their original positions. Transpose it to obtain the adjugate:
Check: the entry in adjugate position \((1,2)\) is \(C_{21}=-1\), and the entry in position \((2,1)\) is \(C_{12}=-3\). The diagonal entries \(C_{11}=4\) and \(C_{22}=2\) stay in place under transposition, as required.
Worked Example: Computing All Cofactors of a Three-by-Three Matrix
Construct the adjugate of \(\mathbf{A}=\begin{bmatrix}1&2&0\\3&1&4\\2&-1&1\end{bmatrix}\). The sign array is \(\begin{bmatrix}+&-&+\\-&+&-\\+&-&+\end{bmatrix}\). Compute each minor from its remaining two-by-two matrix:
Place these nine signed values into the cofactor matrix by their original indices, then transpose:
Check: transposition changes the first row of \(\mathbf{C}\), \((5,5,-5)\), into the first column of the adjugate. It changes the second row \((-2,1,5)\) into the second column, and the third row \((8,-4,-5)\) into the third column. Each displayed adjugate entry is therefore \(C_{ji}\), not \(C_{ij}\).
A Transpose Relationship
The construction is consistent with transposing the original matrix. The determinant fact used below is that a square matrix and its transpose have the same determinant. The same fact applies to every square submatrix formed by deleting a row and a column.
Proof. Let \(C_{ij}(\mathbf{A})\) denote the cofactor at position \((i,j)\). Deleting row \(i\) and column \(j\) from \(\mathbf{A}^{\top}\) leaves the transpose of the matrix obtained by deleting row \(j\) and column \(i\) from \(\mathbf{A}\). Since transposition preserves determinants, the corresponding minors satisfy \(M_{ij}(\mathbf{A}^{\top})=M_{ji}(\mathbf{A})\). Their signs agree because \((-1)^{i+j}=(-1)^{j+i}\). Hence \(C_{ij}(\mathbf{A}^{\top})=C_{ji}(\mathbf{A})\), so \(\mathbf{C}(\mathbf{A}^{\top})=\mathbf{C}(\mathbf{A})^{\top}\). By the definition of adjugate, \(\operatorname{adj}(\mathbf{A}^{\top})=\mathbf{C}(\mathbf{A}^{\top})^{\top}=\mathbf{C}(\mathbf{A})=(\operatorname{adj}\mathbf{A})^{\top}\). \(\square\)
Worked Example: A Three-by-Three Matrix with a Parameter
Construct the adjugate of \(\mathbf{B}=\begin{bmatrix}1&0&2\\0&1&3\\1&1&t\end{bmatrix}\), where \(t\in\mathbb{R}\). Keeping \(t\) symbolic, calculate the minors and apply the signs:
Therefore the cofactor matrix and its transpose are
Check: the first column of the displayed adjugate is the first row of \(\mathbf{C}\), \((t-3,3,-1)^{\top}\). The second and third columns are respectively \((2,t-2,-3)^{\top}\) and \((-1,-1,1)^{\top}\), the second and third rows of \(\mathbf{C}\). The parameter changes the numerical value of a cofactor, but not its sign rule or its transposed location.
Common Mistakes and a Reliable Check
Copying the cofactor matrix without transposing it is a common error. In the second worked example, copying \(\mathbf{C}\) directly would give the wrong candidate \(\begin{bmatrix}5&5&-5\\-2&1&5\\8&-4&-5\end{bmatrix}\) for the adjugate. For instance, that candidate has \(5\) in position \((1,2)\), but the definition requires \((\operatorname{adj}\mathbf{A})_{12}=C_{21}=-2\). The correction is to transpose the whole cofactor matrix, giving \(\begin{bmatrix}5&-2&8\\5&1&-4\\-5&5&-5\end{bmatrix}\).
Forgetting the sign is a different error. The minor \(M_{12}\) in that same example is \(-5\), but the cofactor is \(C_{12}=(-1)^3(-5)=5\). The sign factor multiplies the minor; it does not replace the minor's own sign. A good final check is to label the rows and columns of the cofactor matrix, verify the checkerboard signs, and then check that adjugate entry \((i,j)\) equals cofactor \(C_{ji}\).
The adjugate is a structured way to collect determinant calculations: every entry comes from a particular deleted-row, deleted-column submatrix. Its construction is useful in formulas involving determinants and matrices, but the construction itself needs no invertibility test and no division. Even when some minors vanish, or the original matrix is singular, the cofactors and adjugate are still defined.
Check Your Understanding
For each question, keep the minor indices, cofactor signs, and final transposition distinct.
- For \(\mathbf{A}=\begin{bmatrix}3&2\\1&5\end{bmatrix}\), find all four cofactors and construct \(\operatorname{adj}\mathbf{A}\).
- For a three-by-three matrix, what sign belongs to the cofactor at position \((2,3)\), and why?
- Explain the difference between the cofactor matrix and the adjugate in terms of the entry at position \((1,2)\).
- Compute \(M_{31}\) and \(C_{31}\) for \(\begin{bmatrix}1&2&0\\0&1&3\\2&-1&1\end{bmatrix}\).
- State the relationship between \(\operatorname{adj}(\mathbf{A}^{\top})\) and \(\operatorname{adj}\mathbf{A}\).
