The Recursive Definition of Determinants

Why a Recursive Definition Is Needed

The determinant of a \(2\times2\) matrix has a direct formula, but larger matrices involve more entries and more terms. A recursive definition builds each larger determinant from determinants of smaller matrices. It starts with a single entry, then repeatedly removes one row and one column until only \(1\times1\) determinants remain. The definition specifies both how to continue and where to stop.

Throughout, matrices are square and have real entries. If \(\mathbf{A}\) is an \(n\times n\) matrix, let \(\mathbf{A}_{1j}\) denote the \((n-1)\times(n-1)\) matrix formed by deleting row 1 and column \(j\). This is the minor matrix associated with the entry \(a_{1j}\). Its determinant is the determinant of a smaller square matrix, so it can be defined before the determinant of \(\mathbf{A}\) is defined. The sign attached to position \((1,j)\) is \((-1)^{1+j}\), giving the alternating pattern \(+,-,+,-,\ldots\) along the first row.

Each term uses one entry from the first row, its sign, and the determinant of the corresponding minor matrix. The minor matrix is smaller by one row and one column, and the same definition applies to it. The recurrence uses the first row specifically; it does not, by itself, authorize switching to another row or column. The alternating signs are essential: the term from column 1 is positive, the term from column 2 is negative, and the signs continue to alternate.

Starting from the Base Case

For an \(n\times n\) matrix with \(n\ge2\), every term in the recurrence refers to an \((n-1)\times(n-1)\) determinant. For a \(2\times2\) matrix, those minor matrices are \(1\times1\), so a single expansion reaches the base case.

Following the Recurrence Through Another Level

For a \(3\times3\) matrix, expanding along the first row produces three \(2\times2\) determinants. Each of those can be evaluated by expanding its own first row, which reaches \(1\times1\) determinants. The original determinant is therefore reduced to the base case in two rounds of expansion.

Why the Recursive Rule Is Well-Defined

Proof. We use induction on the order \(n\). If \(n=1\), the definition assigns the matrix's single entry as its determinant, so there is exactly one value. Assume that every \((n-1)\times(n-1)\) real matrix has exactly one determinant. Let \(\mathbf{A}\) be any \(n\times n\) real matrix. Each minor matrix \(\mathbf{A}_{1j}\) has order \(n-1\), so the induction assumption assigns exactly one real number to \(\det(\mathbf{A}_{1j})\) for each \(j\). Every entry \(a_{1j}\) and sign \((-1)^{1+j}\) is fixed. The recurrence is a finite sum of these fixed real numbers, so it assigns exactly one real number to \(\det\mathbf{A}\). This proves the claim for order \(n\), and induction proves it for every positive integer \(n\). \(\square\)

The induction also explains why the definition is not circular. When defining the determinant of an \(n\times n\) matrix, it calls only on determinants of matrices with smaller order, which have already been assigned values at the previous induction step.

Proof. For \(n=1\), no expansion is needed, and the determinant is already at the base case; the number of expansions is \(0=1-1\). Suppose the statement holds for matrices of order \(n-1\), and start with an \(n\times n\) matrix where \(n\ge2\). One expansion replaces it with minor matrices of order \(n-1\). By the induction assumption, each such determinant reaches \(1\times1\) after \((n-1)-1=n-2\) further expansions. Including the first expansion gives \(1+(n-2)=n-1\) expansions in total. Thus every branch reaches the base case in the stated number of steps, and induction proves the proposition. \(\square\)

A calculation can be viewed as a branching process: each expansion produces one term for each first-row entry, and each non-base determinant in a term can be expanded again. The proposition guarantees that no branch continues indefinitely. A zero entry makes its term zero, but the rule and the size of its minor are still determined in the same way.

Keeping Signs and Minor Matrices Straight

There are two separate choices in each recursive term: which minor matrix belongs to the entry, and which sign belongs to that position. For the first row, the minor for column \(j\) is formed by deleting column \(j\), not by deleting row \(j\). Its sign depends on the column position through \((-1)^{1+j}\). Keeping those choices separate helps prevent errors.

Common mistake: using the same sign for every term. A tempting but incorrect calculation for \(\begin{bmatrix}3&-2\\5&4\end{bmatrix}\) is \(3(4)+(-2)(5)=2\), because it uses a plus sign for both first-row terms. The second cofactor sign is negative, so the correct calculation is \(3(4)-(-2)(5)=12+10=22\). The correction is not to change the entry \(-2\); it is to apply the required minus sign to the entire second term.

Another error is to stop at the minor matrices without evaluating their determinants. The recurrence is an instruction to continue until the base case applies. For example, in the \(3\times3\) calculation, the three \(2\times2\) determinants must each be expanded or evaluated by the established \(2\times2\) formula before their values can be combined.

The recursive definition is useful because it turns the determinant of any positive-order square matrix into a finite calculation using smaller square matrices. Its stopping rule is exact, and its signs are fixed by position. The next time a determinant is evaluated, keep track of the matrix order at every stage: each recursive step lowers the order by one, and order one is where the calculation ends.