Why Zeros in the First Row Save Work
A cofactor expansion expresses a determinant as a sum of terms, one for each entry in the chosen row. Each term combines that entry with the determinant of a smaller matrix. When the first row contains zeros, some of those terms vanish immediately. The useful consequence is specific: you need to calculate the smaller determinants only for the nonzero entries in that row.
This tutorial focuses on expanding along the first row of a \(3\times3\) or \(4\times4\) matrix. The determinants of the smaller matrices can be calculated using the methods for \(2\times2\) and \(3\times3\) determinants. The signs still matter: across the first row, they alternate \(+,-,+,-\), beginning with a plus sign.
Minors, Cofactors, and the First-Row Formula
Let \(\mathbf{A}=[a_{ij}]\) be a square matrix. For an entry \(a_{1j}\) in the first row, its minor \(M_{1j}\) is the matrix formed by deleting row 1 and column \(j\) from \(\mathbf{A}\). Its cofactor is the signed number \(C_{1j}=(-1)^{1+j}\det(M_{1j})\). Thus a cofactor includes a sign as well as a minor determinant; the minor itself has no cofactor sign attached to it.
The cofactor expansion along the first row, established in the earlier material on determinants, is \(\det(\mathbf{A})=\sum_j a_{1j}C_{1j}\). Substituting the definition of \(C_{1j}\) gives the following form. For a \(3\times3\) matrix, each minor is \(2\times2\); for a \(4\times4\) matrix, each minor is \(3\times3\).
Proof. By the cofactor expansion along the first row, \(\det(\mathbf{A})=\sum_{j=1}^{n}a_{1j}C_{1j}\). By definition, \(C_{1j}=(-1)^{1+j}\det(M_{1j})\). Substituting this expression for each cofactor gives the displayed sum. For any index \(j\) with \(a_{1j}=0\), the corresponding summand is \((-1)^{1+j}\cdot 0\cdot\det(M_{1j})=0\), regardless of the value of the minor determinant. Therefore that minor is unnecessary for evaluating the expansion. \(\square\)
The signs in the formula follow the first-row pattern \(+,-,+,-,\dots\). For a \(3\times3\) matrix with first row \((a_{11},a_{12},a_{13})\), the terms have signs \(+,-,+\). For a \(4\times4\) matrix, they have signs \(+,-,+,-\). A zero entry removes a term; it does not change the signs of the terms that remain.
Proof. The expansion has one summand for each first-row entry. If an entry is zero, the theorem shows that its summand is zero without regard to its minor. If an entry is nonzero, its summand is included in the expansion and may require the corresponding minor determinant. Hence only the \(k\) minors paired with nonzero entries need to be evaluated. A contributing term may still turn out to be zero if its minor determinant is zero, but that possibility does not reduce the number of minors one must check in advance. \(\square\)
Using the Shortcut for a 3×3 Determinant
Worked Example: One Nonzero Entry in the First Row
Evaluate the determinant of \(\mathbf{A}=\begin{bmatrix}0&2&0\\1&3&4\\4&5&6\end{bmatrix}\). Only the entry in column 2 of the first row is nonzero, so only \(M_{12}\) contributes. Delete row 1 and column 2:
The sign for column 2 is negative, because \((-1)^{1+2}=-1\). Therefore
Check: The only nonzero first-row term is the column-2 term, and its value is \(20\). In particular, neither \(M_{11}\) nor \(M_{13}\) is needed: its coefficient in the expansion is zero.
This example illustrates why a sparse row helps even if some omitted minors would be easy to compute. The determinant formula already determines that their terms contribute nothing. Calculating those minors adds work without changing the result.
Worked Example: Two Nonzero Entries in the First Row
Evaluate \(\det(\mathbf{B})\) for \(\mathbf{B}=\begin{bmatrix}3&0&2\\1&4&-1\\2&5&3\end{bmatrix}\). The first-row entries in columns 1 and 3 are nonzero, so calculate \(M_{11}\) and \(M_{13}\), but not \(M_{12}\).
Deleting row 1 and column 1 gives \(M_{11}=\begin{bmatrix}4&-1\\5&3\end{bmatrix}\), so \(\det(M_{11})=4\cdot3-(-1)\cdot5=12+5=17\). Deleting row 1 and column 3 gives \(M_{13}=\begin{bmatrix}1&4\\2&5\end{bmatrix}\), so \(\det(M_{13})=1\cdot5-4\cdot2=5-8=-3\).
Column 1 has a positive sign and column 3 also has a positive sign. The column-2 term is zero because its first-row entry is zero. Thus
Check: The two contributing terms are \(51\) and \(-6\); their sum is \(45\). The sign of the column-3 term is positive, so its value is \(2(-3)=-6\), not \(+6\).
Using 3×3 Minors in a 4×4 Expansion
For a \(4\times4\) matrix, each first-row minor is \(3\times3\). When the first row is sparse, the expansion can avoid calculating some of these \(3\times3\) determinants entirely. For a minor that does contribute, use the \(3\times3\) determinant calculation carefully, keeping its entries in the same row and column order after deletion.
Worked Example: Two Terms in a 4×4 Expansion
Evaluate \(\det(\mathbf{C})\) for \(\mathbf{C}=\begin{bmatrix}0&2&0&-1\\1&1&0&2\\2&0&1&1\\0&3&2&1\end{bmatrix}\). The nonzero first-row entries are \(c_{12}=2\) and \(c_{14}=-1\), so only \(M_{12}\) and \(M_{14}\) are needed.
For \(M_{12}\), delete row 1 and column 2. The remaining columns are 1, 3, and 4, in that order:
Expand this \(3\times3\) determinant along its first row, using the \(3\times3\) formula. Its middle first-row entry is zero, so only two terms remain:
For \(M_{14}\), delete row 1 and column 4, leaving columns 1, 2, and 3:
Its determinant is
In the original \(4\times4\) expansion, the sign in column 2 is negative and the sign in column 4 is negative. Therefore
Check: The column-2 contribution is \(-14\). The column-4 contribution is also negative: its sign is \(-1\), its entry is \(-1\), and its minor determinant is \(-7\), giving \((-1)(-1)(-7)=-7\). The total is \(-14-7=-21\).
Worked Example: A Single Nonzero Entry in a 4×4 First Row
Evaluate the determinant of \(\mathbf{D}=\begin{bmatrix}0&0&-2&0\\1&2&0&1\\0&1&3&2\\2&0&1&1\end{bmatrix}\). The only nonzero first-row entry is in column 3. Delete row 1 and column 3, keeping the remaining columns in order, to obtain
Calculate this minor using the \(3\times3\) formula:
The sign for column 3 is positive. It follows that
Check: There is just one nonzero term in the first-row expansion, and that term is \((-2)(7)=-14\). The other three minors do not need to be calculated.
Common Pitfall: A Zero Entry Does Not Make Its Minor Zero
Do not confuse an entry with the minor obtained by deleting its row and column. A zero entry makes the product in the expansion zero; it does not imply that the associated minor determinant is zero. For instance, consider \(\mathbf{E}=\begin{bmatrix}0&1&0\\1&2&3\\4&5&6\end{bmatrix}\). The minor for \(e_{11}=0\) is \(M_{11}=\begin{bmatrix}2&3\\5&6\end{bmatrix}\), with determinant \(2\cdot6-3\cdot5=12-15=-3\), which is not zero.
A wrong calculation would keep only the first term and write \(\det(\mathbf{E})=e_{11}\det(M_{11})=0(-3)=0\). That omits the nonzero entry \(e_{12}=1\). The correct expansion is
The correction is to inspect every entry in the first row: keep each nonzero entry's signed term and omit only terms whose first-row entry is zero. The minor is not what determines whether a term can be skipped in advance; the first-row entry does.
A Reliable Calculation Routine
For these \(3\times3\) and \(4\times4\) problems, a short routine helps prevent both wasted calculations and sign errors:
These, and only these, determine which terms can contribute to the expansion.
Delete row 1 and the column containing that entry; preserve the order of the remaining rows and columns.
The first-row signs alternate \(+,-,+,-,\dots\); multiply the signed minor determinant by its first-row entry.
Check the sign of each term separately, especially when the entry or minor determinant is negative.
The main benefit is not a new determinant rule but a more economical use of the first-row expansion. With one nonzero entry, only one minor is required; with two, only two are required. Still, every contributing minor must be formed from the correct rows and columns, and every term must retain its alternating sign.
Check Your Understanding
For each question, identify which minors are needed before calculating any determinants.
- In a \(3\times3\) matrix with first row \((0,-3,0)\), which minor contributes, and what sign does its term have?
- For \(\begin{bmatrix}2&0&-1\\1&3&4\\0&2&5\end{bmatrix}\), calculate the determinant using only the contributing first-row terms.
- A \(4\times4\) matrix has first row \((0,0,0,5)\). How many minors must be evaluated, and what is the sign of the surviving term?
- Explain why a zero first-row entry allows its minor to be skipped even if that minor has a nonzero determinant.
- For a \(4\times4\) matrix whose first row has three nonzero entries, how many terms can contribute to its first-row expansion?
