The 3x3 Determinant

Part 18, Chapter 18: Matrices and Determinants

Learning objectives

Compute a 3x3 determinant by cofactor expansion
Apply Sarrus' rule for 3x3 determinants
Recognize when a 3x3 determinant is zero

The $2 \times 2$ determinant measured the area of a parallelogram. The $3 \times 3$ determinant measures the volume of a parallelepiped. Same idea, one dimension up. The formula gets longer, six terms instead of two, but the underlying picture is the same, and the same five properties from the previous section still hold. This section gives you two ways to compute the $3 \times 3$ determinant and shows why both deliver the same number.

The picture: signed volume

Take the three columns of a $3 \times 3$ matrix $A$ as vectors $\mathbf{u}, \mathbf{v}, \mathbf{w}$ in space. They span a parallelepiped, a slanted box. The signed volume of that box equals $\det(A)$ . If the three vectors form a right-handed triple, the determinant is positive; left-handed gives negative; coplanar (the box flattens) gives zero.

Where this shows up

Physics: The cross product $\vec{a} \times \vec{b}$ is computed as a 3x3 determinant with unit vectors in the top row; angular momentum, magnetic force, and torque all use this formula.
Computer Graphics: Computing the surface normal of a triangle in 3D space (for lighting) uses a 3x3 cross-product determinant on two edge vectors; this runs once per triangle per frame in every 3D engine.
Geometry Processing: The volume of a parallelepiped with edge vectors $\vec{a}, \vec{b}, \vec{c}$ is $|\det[\vec{a}, \vec{b}, \vec{c}]|$ ; meshing software uses this to detect inverted tetrahedra (negative determinant) and re-orient them.

(Edit the entries of $\mathbf{u}, \mathbf{v}, \mathbf{w}$ in the three columns. The widget projects the resulting parallelepiped obliquely so you can see all three column vectors at once. Start with the standard basis (the identity matrix) and you get a unit cube of volume $1$ . Make any two columns equal and the parallelepiped collapses to a flat parallelogram, volume zero, determinant zero.)

Cofactor expansion along the first row

The general formula. Let $A = (a_{ij})$ ij). Then

$\det(A) = a_{11}(a_{22} a_{33} - a_{23} a_{32}) - a_{12}(a_{21} a_{33} - a_{23} a_{31}) + a_{13}(a_{21} a_{32} - a_{22} a_{31}).$ 11(a22a33−a23a32)−a12(a21a33−a23a31)+a13(a21a32−a22a31).

Read it as: each entry $a_{1j}$ 1j in the first row, times the $2 \times 2$ determinant you get by deleting row 1 and column $j$ , with alternating signs $+, -, +$ . Those signed $2 \times 2$ subdeterminants are the cofactors of the row-1 entries.

You can expand along any row or column, not just the first. The sign of each cofactor is $(-1)^{i+j}$ , a checkerboard pattern $\begin{pmatrix}+ & - & +\\ - & + & -\\ + & - & +\end{pmatrix}$ . Picking a row or column with many zeros is what makes cofactor expansion efficient: zeros kill their own terms before you do any multiplication.

Sarrus rule, a $3 \times 3$ shortcut

For the $3 \times 3$ case only, there is a graphical trick. Write the matrix down and copy the first two columns to the right of it. Add the three products on the down-right diagonals, subtract the three products on the up-right diagonals:

$\det(A) = a_{11} a_{22} a_{33} + a_{12} a_{23} a_{31} + a_{13} a_{21} a_{32} - a_{13} a_{22} a_{31} - a_{11} a_{23} a_{32} - a_{12} a_{21} a_{33}.$ 12a23a31+a13a21a32−a13a22a31−a11a23a32−a12a21a33.

This is just the cofactor expansion written out long-hand. Sarrus is convenient because it has a memorable visual pattern, but it does not generalise to $4 \times 4$ or larger. Cofactor expansion does.

Triangular matrices, the cheapest case

If $A$ is upper triangular (zeros below the diagonal) or lower triangular (zeros above), the determinant is just the product of the diagonal entries:

$\det\begin{pmatrix} a & * & * \\ 0 & b & * \\ 0 & 0 & c \end{pmatrix} = abc.$

This is why Gaussian elimination is the universal algorithm for large determinants: row-reduce to triangular form (tracking sign changes from row swaps and scale factors from row scalings), then multiply the diagonal.

The cross product connection

In multivariable calculus, the cross product $\mathbf{u} \times \mathbf{v}$ of two vectors in $\mathbb{R}^3$ is computed as a symbolic $3 \times 3$ determinant:

$\mathbf{u} \times \mathbf{v} = \det\begin{pmatrix} \mathbf{i} & \mathbf{j} & \mathbf{k} \\ u_1 & u_2 & u_3 \\ v_1 & v_2 & v_3 \end{pmatrix}.$

So determinants are not just an algebra trick, they show up directly in geometry, physics, and any field that uses oriented area or volume.

Try it

Predict first: what is the determinant of the $3 \times 3$ identity matrix? Enter $\mathbf{u} = (1,0,0), \mathbf{v} = (0,1,0), \mathbf{w} = (0,0,1)$ and confirm the volume readout is $1$ .
Predict first: if $\mathbf{w}$ lies in the same plane as $\mathbf{u}$ and $\mathbf{v}$ , what should the determinant be? Change $\mathbf{w}$ to $(1, 1, 0)$ and verify the volume collapses to $0$ , the determinant detects coplanarity.
Compute by cofactor expansion: $\det\begin{pmatrix} 1 & 0 & 2 \\ 3 & 1 & 0 \\ 0 & 2 & 1 \end{pmatrix}$ . Expand along the first row.
Repeat using Sarrus rule. Confirm both yield $13$ .

Pause: an entire column of $0$ s gives $\det = 0$ . Why is this obvious from the volume picture?

A trap to watch for

The most common error: forgetting the alternating signs in cofactor expansion. Students write $a_{11}M_{11} + a_{12}M_{12} + a_{13}M_{13}$ a13M13 when the correct formula is $a_{11}M_{11} - a_{12}M_{12} + a_{13}M_{13}$ a13M13. The minus on the middle term is mandatory; without it your answer is wrong, and the wrongness is hard to spot because the formula looks like a sum. Always write the $(-1)^{i+j}$ sign next to each cofactor before multiplying.

A second trap, lethal for big matrices: Sarrus rule does not extend to $4 \times 4$ . The diagonal trick is a coincidence specific to $3 \times 3$ . Students who try to "draw all the diagonals" on a $4 \times 4$ matrix produce 8 terms; the correct answer has 24 terms. For $n \geq 4$ you must use cofactor expansion or row reduction. There is no diagonal shortcut.

What you now know

You can compute any $3 \times 3$ determinant by cofactor expansion or Sarrus rule, recognise triangular matrices and shortcut their determinants, and interpret the result as signed volume. The next section catalogues the $3 \times 3$ properties, same five as before plus a couple of consequences specific to size three.

Quick check

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References

Lang, S. (1971). Basic Mathematics. Springer. Chapter 17, §4, the $3 \times 3$ cofactor expansion and Sarrus rule.
Strang, G. (2009). Linear Algebra and its Applications (4th ed.). Cengage. Chapter 4: determinants as signed volume; row reduction as the universal computation strategy.
Hoffman, K.; Kunze, R. (1971). Linear Algebra (2nd ed.). Prentice-Hall. §5.3, the proof that all cofactor expansions (along any row or column) give the same value.
Axler, S. (2015). Linear Algebra Done Right (3rd ed.). Springer. Chapter 10: alternating multilinear forms and why the $3 \times 3$ formula has exactly $3! = 6$ terms.