Determinant of 3×3 matrix — cofactor expansion

Think of it like this: you're measuring the "signed volume" of a parallelepiped in 3D space. Instead of computing it all at once, you slice it along one direction and add up contributions from each slice.

What is cofactor expansion?

The cofactor $C_{ij}$ of element $a_{ij}$ is: $C_{ij} = (-1)^{i+j} M_{ij}$ where $M_{ij}$ is the minor (determinant of the 2×2 matrix obtained by deleting row $i$ and column $j$).

The determinant is: $\det(A) = \sum_{j=1}^{3} a_{ij} C_{ij} \quad \text{(expanding along row } i\text{)}$ or $\det(A) = \sum_{i=1}^{3} a_{ij} C_{ij} \quad \text{(expanding along column } j\text{)}$

WHY does this work?

The determinant measures how much a linear transformation scales volumes. When we expand along a row, we're expressing the transformation as a linear combination of simpler transformations (the 2×2 minors), weighted by the elements of that row. The alternating signs $(-1)^{i+j}$ ensure the correct orientation (positive/negative volume).

Derivation from first principles:

Start with the permutation definition of determinant: $\det(A) = \sum_{\sigma \in S_3} \text{sgn}(\sigma) \, a_{1\sigma(1)} a_{2\sigma(2)} a_{3\sigma(3)}$

Group terms by the first row element they contain:

• Terms with $a_{11}$: all permutations starting with column 1
• Terms with $a_{12}$: all permutations starting with column 2
• Terms with $a_{13}$: all permutations starting with column 3

For terms with $a_{11}$, after factoring it out, the remaining product involves only elements from rows 2,3 and columns 2,3—exactly the minor $M_{11}$. The sign from the permutation gives $(-1)^{1+1} = +1$.

Similarly for $a_{12}$ and $a_{13}$, yielding: $\det(A) = a_{11}(-1)^{1+1}M_{11} + a_{12}(-1)^{1+2}M_{12} + a_{13}(-1)^{1+3}M_{13}$

This is cofactor expansion along row 1.

Sign pattern (checkerboard): $\begin{bmatrix} + & - & + \\ - & + & - \\ + & - & + \end{bmatrix}$

The sign for position $(i,j)$ is $(-1)^{i+j}$.

HOW to compute: Step-by-step

Step 1: Choose a row or column (pick one with zeros to simplify).

Step 2: For each element in that row/column:

• Determine the sign using $(-1)^{i+j}$ (or use the checkerboard pattern)
• Cross out its row and column to get the 2×2 minor
• Compute the 2×2 determinant
• Multiply: element × sign × minor

Step 3: Sum all three contributions.

Why row 1? We could pick any row/column—let's demonstrate the standard approach first.

Expanding along row 1: $\det(A) = 2 \cdot C_{11} + (-3) \cdot C_{12} + 1 \cdot C_{13}$

Compute $C_{11}$:

• Sign: $(-1)^{1+1} = +1$
• Minor $M_{11}$: delete row 1, column 1: $M_{11} = \begin{vmatrix} 0 & 6 \\ 4 & 2 \end{vmatrix} = 0 \cdot 2 - 6 \cdot 4 = -24$
• $C_{11} = (+1)(-24) = -24$

Why this step? We're isolating the contribution of element $a_{11} = 2$ to the overall volume scaling. The minor captures how the remaining2D "slice" behaves.

Compute $C_{12}$:

• Sign: $(-1)^{1+2} = -1$
• Minor $M_{12}$: delete row 1, column 2: $M_{12} = \begin{vmatrix} 5 & 6 \\ -1 & 2 \end{vmatrix} = 5 \cdot 2 - 6 \cdot (-1) = 10 + 6 = 16$
• $C_{12} = (-1)(16) = -16$

Compute $C_{13}$:

• Sign: $(-1)^{1+3} = +1$
• Minor $M_{13}$: delete row 1, column 3: $M_{13} = \begin{vmatrix} 5 & 0 \\ -1 & 4 \end{vmatrix} = 5 \cdot 4 - 0 \cdot (-1) = 20$
• $C_{13} = (+1)(20) = 20$

Final sum: $\det(A) = 2(-24) + (-3)(-16) + 1(20) = -48 + 48 + 20 = 20$

Why column 2? Notice column 2 has TWO zeros! This means we only compute ONE minor instead of three.

Expanding along column 2: $\det(B) = 0 \cdot C_{12} + 5 \cdot C_{22} + 0 \cdot C_{32} = 5 \cdot C_{22}$

Compute $C_{22}$:

• Sign: $(-1)^{2+2} = +1$
• Minor $M_{22}$: delete row 2, column 2: $M_{22} = \begin{vmatrix} 3 & 2 \\ 4 & 7 \end{vmatrix} = 3 \cdot 7 - 2 \cdot 4 = 21 - 8 = 13$
• $C_{22} = (+1)(13) = 13$

Final: $\det(B) = 5 \cdot 13 = 65$

Why this strategy? Zeros eliminate entire terms, drastically reducing computation. Always scan for the row/column with the most zeros.

Expanding along column 3: $\det(C) = 3 \cdot C_{13} + 6 \cdot C_{23} + 9 \cdot C_{33}$

Compute $C_{13}$:

• Sign: $(-1)^{1+3} = +1$
• $M_{13} = \begin{vmatrix} 4 & 5 \\ 7 & 8 \end{vmatrix} = 32 - 35 = -3$
• $C_{13} = +1 \cdot (-3) = -3$

Compute $C_{23}$:

• Sign: $(-1)^{2+3} = -1$
• $M_{23} = \begin{vmatrix} 1 & 2 \\ 7 & 8 \end{vmatrix} = 8 - 14 = -6$
• $C_{23} = -1 \cdot (-6) = 6$

Compute $C_{33}$:

• Sign: $(-1)^{3+3} = +1$
• $M_{33} = \begin{vmatrix} 1 & 2 \\ 4 & 5 \end{vmatrix} = 5 - 8 = -3$
• $C_{33} = +1 \cdot (-3) = -3$

Final: $\det(C) = 3(-3) + 6(6) + 9(-3) = -9 + 36 - 27 = 0$

Why zero? This matrix has linearly dependent rows: notice that $\text{row 3} = 2 \cdot \text{row 2} - \text{row 1}$, since $2(4,5,6) - (1,2,3) = (7,8,9)$. Because the three rows are linearly dependent, they lie in a 2D plane (a flat sheet), so the parallelepiped they span has zero volume. Zero determinant means the transformation collapses 3D space down to 2D.

Common mistakes

Why it feels right: The formula looks simpler without the signs.

The fix: The signs $(-1)^{i+j}$ are not optional—they encode whether each contribution adds or subtracts from the total signed volume. Use the checkerboard pattern to remember.

Why it feels right: It's easy to confuse the indices, especially when working quickly.

The fix: Double-check: "I'm finding the cofactor of $a_{ij}$, so I cross out row $i$ AND column $j$."

The truth: All expansions give the same determinant—it's a unique property of the matrix. Choosing different rows/columns is just a computational strategy.

Why it feels wrong: Different rows produce different intermediate calculations, making it seem like the result should differ.

The fix: The determinant is invariant. Different paths, same destination.

Correct: $\begin{vmatrix} a & b \\ c & d \end{vmatrix} = ad - bc$

The fix: Remember "down-right minus up-left" or "main diagonal minus anti-diagonal."

Practice problems

1. Compute $\det(A)$ for $A = \begin{bmatrix} 1 & 3 & 2 \\ -1 & 0 & 4 \\ 2 & 5 & 1 \end{bmatrix}$ by expanding along row 2.

2. Find $\det(B)$ where $B = \begin{bmatrix} 2 & 0 & 0 \\ 3 & 1 & 0 \\ 5 & 4 & 6 \end{bmatrix}$ (lower-triangular matrix—expand along row 1 and notice the pattern: the determinant equals the product of diagonal entries $2 \cdot 1 \cdot 6 = 12$).

3. Show that expanding $A = \begin{bmatrix} 1 & 2 & 3 \\ 0 & 4 & 5 \\ 0 & 0 & 6 \end{bmatrix}$ along column 1 gives the same result as expanding along row 1.

For a big3×3 matrix, instead of computing this all at once, we use a clever trick: we pick one row, and for each number in that row, we:

1. Cover up its row and column with your hands
2. Look at the remaining 2×2 grid and compute its determinant (which you already know!)
3. Multiply that 2×2 determinant by the original number
4. Add a + or − sign based on a checkerboard pattern

Then you add up all three pieces. The magic is that no matter which row you pick, you get the same answer! It's like climbing a mountain: different paths, same summit.

The checkerboard pattern of signs is like the "handedness" of the box—sometimes a transformation flips the box inside-out, and the signs keep track of that.

Connections

• 2.6.01-What-is-a-determinant — foundational definition and geometric meaning
• 2.6.05-Determinant-of-2×2-matrix — the base case for cofactor expansion
• 2.6.09-Properties-of-determinants — shortcuts that make cofactor expansion faster
• 2.6.11-Determinant-and-matrix-inverse — $\det(A) \neq 0$ iff $A$ is invertible
• 2.7.03-Cramers-rule — uses cofactor expansion to solve systems
• 3.4.02-Cross-product-and-determinants — 3D cross product as a3×3 determinant

#flashcards/maths

What is the cofactor $C_{ij}$ of element $a_{ij}$ in a matrix? :: $C_{ij} = (-1)^{i+j} M_{ij}$ where $M_{ij}$ is the minor (determinant of the matrix with row $i$ and column $j$ deleted)

What is the sign pattern (checkerboard) for cofactors in a 3×3 matrix? :: $\begin{bmatrix} + & - & + \\ - & + & - \\ + & - & + \end{bmatrix}$, following $(-1)^{i+j}$

What does $M_{23}$ mean for a 3×3 matrix? :: The minor $M_{23}$ is the determinant of the 2×2 matrix obtained by deleting row 2 and column 3

Concept Map

Hinglish (regional understanding)