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Determinants by Cofactor Expansion · Tutorial 228 of 1000

Common Sign and Minor Errors

You will learn to calculate each cofactor contribution carefully and use the structure of an expansion to locate common errors.

Beginner 9 min read

What You'll Learn

  • Define the minor and signed cofactor associated with a matrix entry.
  • Apply the checkerboard sign pattern to a chosen row or column.
  • Identify exactly which row and column to delete when forming a minor.
  • Distinguish a minor from its signed cofactor in an expansion.
  • Calculate how a single sign or minor substitution changes an answer.
  • Check a cofactor expansion using a different row or column.

Three Places a Cofactor Calculation Can Go Wrong

A cofactor expansion is a sum of contributions, one for each entry in a chosen row or column. Each contribution depends on three pieces of information: the entry, the determinant left after deleting its row and column, and a checkerboard sign. A correct calculation keeps all three pieces attached to the same position. Many errors arise when one piece is copied from a neighbouring position or when the unsigned minor is mistaken for the signed cofactor.

Throughout, let \(\mathbf{A}=[a_{ij}]\) be an \(n\times n\) real matrix with \(n\ge 2\). For an entry in row \(i\), column \(j\), deleting a row means removing all its entries, and deleting a column means removing all its entries. The remaining square matrix has \(n-1\) rows and \(n-1\) columns. Its determinant is the minor associated with that entry.

Definition: The minor \(M_{ij}\) is the determinant of the matrix obtained from \(\mathbf{A}\) by deleting row \(i\) and column \(j\). The cofactor \(C_{ij}\) is the signed minor \(C_{ij}=(-1)^{i+j}M_{ij}\). Thus a minor is a determinant, while a cofactor includes the sign determined by the entry’s position.

The factor \((-1)^{i+j}\) gives the checkerboard pattern. Starting with a positive sign in position \((1,1)\), signs alternate across each row and down each column:

$$ \begin{bmatrix} +&-&+&-\\ -&+&-&+\\ +&-&+&-\\ -&+&-&+ \end{bmatrix}. $$

For instance, the sign in position \((2,3)\) is positive because \(2+3=5\) is odd? No: the sign is negative when the sum is odd, so position \((2,3)\) has a negative sign. In general, \(i+j\) even gives a positive sign, and \(i+j\) odd gives a negative sign. The checkerboard is attached to positions in the original matrix, not to the entries’ values.

Keeping the Expansion Indices Together

The cofactor expansion along row \(r\) is \(\det\mathbf{A}=\sum_{j=1}^{n}a_{rj}C_{rj}\). Along column \(s\), it is \(\det\mathbf{A}=\sum_{i=1}^{n}a_{is}C_{is}\). These are the cofactor expansion formulas from the earlier material on determinants by cofactor expansion. The bookkeeping is the same in either direction: retain the entry’s original position, find the minor by deleting the corresponding row and column, and multiply that minor by its position’s sign.

When expanding along row \(r\), the row deleted from each minor is always row \(r\). The column deleted changes with the term: the term involving \(a_{rj}\) uses the minor \(M_{rj}\), made by deleting column \(j\). For a column expansion, the column deleted is fixed and the deleted row changes. This distinction is easy to lose when writing several small determinants in sequence.

1
Record the position.
For the term \(a_{ij}\), keep both indices \(i\) and \(j\) visible.
2
Delete the matching row and column.
Form \(M_{ij}\) by deleting row \(i\) and column \(j\) from the original matrix.
3
Apply the positional sign.
Compute \(C_{ij}=(-1)^{i+j}M_{ij}\), then multiply by \(a_{ij}\).
4
Add the contributions.
Include every term in the chosen row or column, including zero terms if they help keep the indexing clear.

Worked Examples: Signs, Deletions, and Cofactors

Worked Example: Finding a Sign Error in a Row Expansion

Expand the determinant along row 2 of \(\mathbf{A}=\begin{bmatrix}1&2&0\\3&1&4\\2&-1&5\end{bmatrix}\), and find the result of reversing just the sign of the third cofactor contribution.

The sign pattern in row 2 is \(-,+,-\). For the first entry, delete row 2 and column 1:

$$ M_{21}= \begin{vmatrix}2&0\\-1&5\end{vmatrix} =2\cdot5-0\cdot(-1)=10, \qquad C_{21}=-10. $$

For the second entry, delete row 2 and column 2. For the third, delete row 2 and column 3:

$$ M_{22}= \begin{vmatrix}1&0\\2&5\end{vmatrix} =1\cdot5-0\cdot2=5, \qquad C_{22}=5, $$
$$ M_{23}= \begin{vmatrix}1&2\\2&-1\end{vmatrix} =1\cdot(-1)-2\cdot2=-5, \qquad C_{23}=-(-5)=5. $$

The expansion is \(3(-10)+1(5)+4(5)=-30+5+20=-5\). If the sign of only the third cofactor contribution is reversed, that term becomes \(-20\), so the incorrect total is \(-30+5-20=-45\).

Check: Expanding along row 1 instead gives \(1(1\cdot5-4(-1))-2(3\cdot5-4\cdot2)+0=9-14=-5\), agreeing with the correct row 2 expansion.

Worked Example: Detecting the Wrong Deleted Row

Use the same matrix and expand along row 2. Suppose the first minor is formed by deleting row 1 and column 1 instead of deleting row 2 and column 1. Find the resulting error.

The first entry in row 2 is \(a_{21}=3\). Its correct minor comes from deleting row 2 and column 1, leaving \(\begin{bmatrix}2&0\\-1&5\end{bmatrix}\), so \(M_{21}=10\). The sign at \((2,1)\) is negative and its correct contribution is \(3(-10)=-30\).

Deleting row 1 and column 1 by mistake leaves \(\begin{bmatrix}1&4\\-1&5\end{bmatrix}\), whose determinant is \(1\cdot5-4(-1)=9\). Using this incorrect minor gives a first contribution of \(3(-9)=-27\). Leaving the other two contributions unchanged, the incorrect total is \(-27+5+20=-2\), rather than \(-5\).

Check: The independent row 1 expansion gives \(9-14=-5\), as in the previous check. The discrepancy is not a new determinant value: it comes from using a matrix that does not correspond to the entry \(a_{21}\). To form its minor, both the row and column named by that entry must be deleted.

Worked Example: Using the Minor Instead of the Cofactor

Expand along row 1 of \(\mathbf{B}=\begin{bmatrix}2&1&3\\0&4&-1\\5&2&1\end{bmatrix}\). Then calculate the result of using the unsigned minor for the middle term.

For position \((1,1)\), the sign is positive. Deleting row 1 and column 1 gives a minor \(4\cdot1-(-1)\cdot2=6\), so the cofactor is \(C_{11}=6\). At position \((1,2)\), the sign is negative. The minor is \(0\cdot1-(-1)\cdot5=5\), and the cofactor is \(C_{12}=-5\). At position \((1,3)\), the sign is positive, and the minor is \(0\cdot2-4\cdot5=-20\), so \(C_{13}=-20\).

The correct expansion is \(2(6)+1(-5)+3(-20)=12-5-60=-53\). If the unsigned minor \(M_{12}=5\) is used in place of \(C_{12}=-5\), the result becomes \(12+5-60=-43\).

Check: Expanding along row 2, the middle minor is \(M_{22}=2\cdot1-3\cdot5=-13\), with positive sign, giving contribution \(4(-13)=-52\). The last minor is \(M_{23}=2\cdot2-1\cdot5=-1\), with negative sign, so its cofactor is \(1\) and its contribution is \((-1)(1)=-1\). The first contribution is zero. Thus the row 2 expansion is \(0-52-1=-53\).

What a Single Mistake Does to the Answer

The size of a discrepancy can help identify its source. The following two results assume a single error in one term and that every other contribution is kept unchanged. They do not say that every incorrect calculation must produce a different number: a zero entry or a coincidental equality can conceal an error.

Proposition: Suppose a cofactor expansion contains the term \(a_{rk}C_{rk}\), and only the sign of this cofactor is reversed. If \(D\) is the correct expansion and \(D_{\mathrm{wrong}}\) is the result after the reversal, then \(D_{\mathrm{wrong}}-D=-2a_{rk}C_{rk}\).

Proof. The correct term is \(a_{rk}C_{rk}\). Reversing only its cofactor sign replaces it with \(a_{rk}(-C_{rk})=-a_{rk}C_{rk}\). All other terms cancel when we subtract the correct sum from the incorrect sum. Therefore \(D_{\mathrm{wrong}}-D=-a_{rk}C_{rk}-a_{rk}C_{rk}=-2a_{rk}C_{rk}\). If \(a_{rk}=0\) or \(C_{rk}=0\), the difference is zero, so this particular sign error may not change the numerical answer. \(\square\)

In the first worked example, \(a_{23}=4\) and \(C_{23}=5\). The proposition predicts a change of \(-2(4)(5)=-40\). Indeed, \(-45-(-5)=-40\). This gives a useful arithmetic check when the incorrect and correct totals are both available.

Proposition: Suppose only the minor \(M_{rk}\) is replaced by \(M_{st}\) in the term at position \((r,k)\), while its position’s sign and all other terms remain unchanged. Then \(D_{\mathrm{wrong}}-D=a_{rk}(-1)^{r+k}(M_{st}-M_{rk})\).

Proof. The correct term is \(a_{rk}(-1)^{r+k}M_{rk}\). The substituted term is \(a_{rk}(-1)^{r+k}M_{st}\), because the sign still belongs to the original position \((r,k)\). Subtracting the correct term from the substituted term gives \(a_{rk}(-1)^{r+k}(M_{st}-M_{rk})\). Every other term is unchanged and cancels in the difference of the two sums. \(\square\)

In the second worked example, \(a_{21}=3\), the position sign is negative, the correct minor is \(10\), and the mistakenly used minor is \(9\). The predicted change is \(3(-1)(9-10)=3\). The incorrect result exceeds the correct one by \(-2-(-5)=3\), as predicted.

A Practical Check Before Adding

Before summing an expansion, check each term in isolation. This is more reliable than trying to recover a lost sign after combining several positive and negative numbers. In particular, do not infer a cofactor’s sign from the minor’s value: a negative minor remains negative if its position sign is positive, but becomes positive if its position sign is negative.

  • Sign: Calculate \((-1)^{i+j}\) from the entry’s position in the original matrix.
  • Deletion: Cross out row \(i\) and column \(j\) for the term \(a_{ij}\).
  • Minor: Evaluate the determinant of what remains before applying the cofactor sign.
  • Contribution: Multiply the original entry by the signed cofactor, not by the unsigned minor.
  • Check: If practical, expand along a different row or column and compare the result.

A different expansion is a check on the calculation, not permission to change the determinant. Each choice of row or column must yield the same value when its entries, minors, and positional signs are all handled correctly. If two expansions disagree, compare their terms one at a time; the error is often visible in a single deleted row, deleted column, or checkerboard sign.

Common Pitfall: The Sign Belongs to the Position

A tempting shortcut is to put a minus sign on a term whenever its minor is negative. That confuses the minor’s numerical sign with the cofactor’s positional sign. For the middle term in the third example, \(M_{12}=5\) is positive, but the position \((1,2)\) has a negative checkerboard sign, so \(C_{12}=-5\). The contribution is \(1(-5)\), not \(1(5)\). Conversely, a negative minor in a positive-sign position remains a negative cofactor, as \(M_{13}=-20\) showed.

Another risky shortcut is to write several \(2\times2\) determinants without labeling the positions that produced them. A determinant may be evaluated correctly and still be the wrong minor for the term. Keeping an explicit label such as \(M_{21}\) beside each small determinant makes that error easier to catch. The row and column indices are part of the definition, not optional bookkeeping.

ObjectWhat it recordsQuestion to ask
Entry \(a_{ij}\)Value at row \(i\), column \(j\)Which original position is this?
Minor \(M_{ij}\)Determinant after deleting row \(i\), column \(j\)Did I delete both matching indices?
Cofactor \(C_{ij}\)\((-1)^{i+j}M_{ij}\)Did I apply the checkerboard sign?
Expansion term\(a_{ij}C_{ij}\)Did I multiply the entry by its signed cofactor?

Check Your Understanding

For each question, keep the entry’s row and column indices attached to its minor and cofactor.

  1. For an entry in position \((3,2)\), what sign is used, and which row and column are deleted to form its minor?
  2. In a row 1 expansion, a term has entry \(a_{12}=2\) and minor \(M_{12}=-3\). What is its cofactor and its contribution?
  3. A student replaces a single cofactor by its negative. State the formula for the change in the expansion, using the entry and the original cofactor.
  4. For the matrix \(\begin{bmatrix}1&2\\3&4\end{bmatrix}\), calculate the contribution from position \((1,2)\), naming its minor and cofactor.
  5. Explain why deleting row 1 and column 2 cannot produce the minor associated with an entry in position \((2,1)\).

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