LINEAR-ALGEBRA · Unit 18 · Video 2 · Interactive Practice

Determinants Inside Determinants — Cofactor Expansion

IKey Formulas

Formula	Name	Description
$\det\begin{pmatrix} a & b \\ c & d \end{pmatrix} = ad - bc$	2×2 Determinant	Base case for cofactor expansion
$M_{ij}$ = det(submatrix after deleting row $i$ , col $j$ )	Minor	The smaller determinant
$C_{ij} = (-1)^{i+j} \cdot M_{ij}$	Cofactor	Minor with checkerboard sign
$\det(A) = \displaystyle\sum_{j=1}^{n} a_{ij} \, C_{ij}$	Cofactor Expansion	Expand along any row $i$ (or any column $j$ )

IIVisualization 1 — The Checkerboard Sign

The cofactor's sign $(-1)^{i+j}$ depends only on the position $(i,j)$ — it alternates like a chessboard.

Matrix size 4×4

💡 Positions $(i,j)$ and $(j,i)$ always share a sign, since $i+j$ ignores order.

IIIVisualization 2 — Expanding a 3×3 Determinant

Expanding $\det A$ along any row or column rewrites it as a signed sum of entries times their $2\times2$ minors.

Expand along

💡 Challenge: find a row or column whose expansion needs only two $2\times2$ minors.

IVVisualization 3 — Zero Hunting

Every row and column yields the same determinant — but each zero you land on erases a whole cofactor.

Matrix

💡 Before expanding, scan for the line with the most zeros — it needs the fewest minors.

VQuiz Questions

Question 1 · Computing a Cofactor

Consider a matrix where the entry at position $(2, 3)$ has value $a_{23} = 5$ . After deleting row 2 and column 3, the remaining submatrix has determinant (minor) $M_{23} = 8$ .

What is the cofactor $C_{23}$ ?

✅ Correct! Since $i+j = 2+3 = 5$ is odd, the checkerboard sign is $-1$ , giving $C_{23} = (-1)(8) = -8$ .

❌ Not quite. You may have forgotten the checkerboard sign. Since $i+j = 5$ is odd, the sign is $-1$ , not $+1$ .

❌ Not quite. You multiplied the entry value (5) into the result. The cofactor uses only the sign and the minor — not the entry itself.

❌ Not quite. Two issues: the cofactor doesn't include the entry value, and the sign at position (2,3) is negative, not positive.

❌ Not quite. Try again — the hints above can help.

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Solution:

The cofactor formula is $C_{ij} = (-1)^{i+j} \cdot M_{ij}$ .

For position $(2, 3)$ : $C_{23} = (-1)^{2+3} \cdot M_{23} = (-1)^5 \cdot 8 = (-1)(8) = -8$

Key distinction: The cofactor combines only the checkerboard sign and the minor. The entry value $a_{23} = 5$ is used separately when computing the determinant contribution: $a_{23} \cdot C_{23} = 5 \times (-8) = -40$ .

Question 2 · Strategic Row/Column Choice

Consider the 4×4 matrix: $B = \begin{pmatrix} 3 & 0 & 0 & 2 \\ 1 & 5 & 0 & 4 \\ 0 & 7 & 1 & 0 \\ 6 & 0 & 0 & 3 \end{pmatrix}$

Which expansion requires computing the fewest 3×3 cofactors?

✅ Correct! Column 3 has three zeros, so only 1 cofactor needs computing — maximum efficiency!

❌ Not quite. Look for the row or column with the most zeros. Each zero eliminates an entire cofactor computation.

Show solution

Solution:

Count the nonzero entries (= cofactors you must compute) for each option:

Option	Entries	Zeros	Cofactors needed
Row 1	3, 0, 0, 2	2	2
Row 3	0, 7, 1, 0	2	2
Column 3	0, 0, 1, 0	3	1 ✅
Column 4	2, 4, 0, 3	1	3

Column 3 has three zeros, leaving only one nonzero entry (the 1 in row 3). Only one 3×3 cofactor must be computed — the minimum possible!

Question 3 · Entry Sign vs. Checkerboard Sign

An entry $a_{21} = -4$ sits at position $(2, 1)$ in a matrix. The minor is $M_{21} = 11$ .

What is the determinant contribution of this entry, i.e., $a_{21} \cdot C_{21}$ ?

✅ Correct! The cofactor is $C_{21} = (-1)(11) = -11$ , and the contribution is $(-4)(-11) = +44$ . Two negatives make a positive!

❌ Not quite. Watch the signs carefully: the entry is $-4$ and the cofactor is $-11$ . Negative times negative equals positive 44.

❌ Not quite. The contribution is entry $\times$ cofactor = $a_{21} \times C_{21}$ . Make sure you multiply the entry value by the full cofactor (sign $\times$ minor).

Show solution

Solution:

Three separate components:

Entry value: $a_{21} = -4$ (from the matrix)
Checkerboard sign: $(-1)^{2+1} = (-1)^3 = -1$
Minor: $M_{21} = 11$

Step 1 — Cofactor: $C_{21} = (-1)^{2+1} \cdot M_{21} = (-1)(11) = -11$

Step 2 — Contribution: $a_{21} \cdot C_{21} = (-4)(-11) = +44$

The entry's negative sign ( $-4$ ) and the checkerboard's negative sign ( $-1$ ) are independent. Negative times negative = positive!

Question 4 · True or False

Statement: Expanding a determinant along different rows of the same matrix can produce different determinant values.

✅ Correct! The determinant is unique — every expansion (any row or column) gives the same value. Only the workload differs.

❌ Incorrect. The determinant is a fixed property of the matrix. Different expansion choices change the computation path, but always arrive at the same number.

❌ Not quite. Try again — the hints above can help.

Show solution

Solution: False

The determinant is a unique number associated with each square matrix. Cofactor expansion along any row or any column always produces the same value:

$\det(A) = \sum_{j=1}^{n} a_{ij}\, C_{ij} \quad \text{(same for every row } i\text{)}$

$\det(A) = \sum_{i=1}^{n} a_{ij}\, C_{ij} \quad \text{(same for every column } j\text{)}$

The choice only affects the amount of computation — which is precisely why we can "hunt for zeros" to minimize work without changing the answer.

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