Expansion Trees for Four-by-Four Determinants

From One Expansion to a Tree

A four-by-four determinant can look like a long calculation because its first-row expansion creates four three-by-three determinants, and each of those can be expanded again. The useful idea is to keep the calculation organized: record each signed contribution as a branch, label every smaller determinant by the rows and columns that remain, and reuse a subcalculation when the same minor appears twice.

For a square matrix, a minor of an entry is the determinant of the matrix obtained by deleting that entry’s row and column. The cofactor of an entry in row \(i\), column \(j\), is its minor multiplied by the sign \((-1)^{i+j}\). The first-row cofactor expansion, established in the earlier material on determinants, says that the determinant is the sum of the first-row entries multiplied by their cofactors. In a four-by-four matrix, the first-row signs are \(+,-,+,-\).

The local-position rule matters. After deleting a column, the remaining columns are renumbered within the smaller matrix for the purpose of its cofactor signs. Thus every three-by-three expansion has the local sign pattern \(+,-,+\), regardless of which original columns its entries came from. At the final stage, a two-by-two determinant is evaluated by \(ad-bc\).

The Tree as a Calculation Rule

Let \(\mathbf{A}=(a_{ij})\) be a four-by-four matrix. For each \(j\), let \(\mathbf{B}_j\) be the three-by-three matrix formed by deleting row 1 and column \(j\) of \(\mathbf{A}\). The first-row expansion gives

Now expand each \(\det\mathbf{B}_j\) along the first row of that three-by-three matrix. Write \(b_{jk}\) for its entry in local column \(k\), and let \(\mathbf{C}_{jk}\) be the two-by-two matrix left after deleting the first row and local column \(k\) of \(\mathbf{B}_j\). Then each branch has the form

This gives a mechanical route through the calculation: first record the four root signs, then record \(+,-,+\) separately at every three-by-three node, then evaluate each resulting two-by-two determinant. A zero entry makes its branch contribute zero, but it does not change the signs on the other branches.

Proof. Let \(\mathbf{A}\) be any four-by-four real matrix. By first-row cofactor expansion, its determinant is the alternating sum of the four first-row entries multiplied by the determinants of the corresponding matrices \(\mathbf{B}_j\). Each \(\mathbf{B}_j\) is three-by-three, so applying first-row cofactor expansion to it gives the displayed three-term expression with signs \(+,-,+\). Each matrix \(\mathbf{C}_{jk}\) is two-by-two, whose determinant is \(ad-bc\) by the two-by-two determinant formula. Substituting these three-by-three expansions into the four terms of the original expansion, and then evaluating the two-by-two determinants, is exactly the process recorded by the tree. Each substitution preserves the value of the expression, so the final tree calculation equals \(\det\mathbf{A}\). \(\square\)

A Tree with Two Nonzero Root Branches

When Every Root Branch Is Present

Reusing a Repeated Minor

Different branches can lead to exactly the same smaller matrix. When that happens, its determinant needs to be calculated only once. This is not an extra determinant rule; it is simply reusing the value of the same expression wherever it occurs.

Proof. A two-by-two determinant is determined by its four entries: for \(\begin{bmatrix}a&b\\c&d\end{bmatrix}\), its value is \(ad-bc\). If two leaves contain the same matrix, their corresponding four entries are equal. Their products \(ad\) and \(bc\), and hence their differences \(ad-bc\), are equal. Replacing either occurrence by this common value therefore leaves the tree’s sum unchanged. \(\square\)

Common Sign Error

Do not use an original column number for a local cofactor sign. In the four-branch example, \(\mathbf{B}_2\) comes from deleting original column 2. Its first row is \((0,2,0)\), and its middle entry is in local column 2. The local sign for that entry is negative, so \(\det\mathbf{B}_2=-2(-3)=6\). A wrong calculation might notice that the entry came from original column 3 and use the positive sign for column 3 instead, obtaining \(2(-3)=-6\). The correction is to use the positions in \(\mathbf{B}_2\), not the original matrix: the entry is in local column 2, so its sign is negative. The root’s negative sign for \(\mathbf{B}_2\) is applied separately afterward.

Keeping those two levels separate prevents a second common error: changing the sign inside a minor and then forgetting to apply the root sign. A reliable tree records the signed minor value first, then combines it with its parent branch.