Recipe

Compute a 3×3 determinant by expanding along row 1

Use when: you have a 3\times3 matrix and need its determinant, before the general cofactor-expansion technique is developed

Goal: the value of \det A, plus what its sign and magnitude say about the parallelepiped A's columns span

Taught in The 3x3 determinant and volume

1. 1

   For each entry in row 1, cross out its own row and its own column; the 2\times2 block of numbers left behind is that entry's minor.

   Example

   Take M=\begin{pmatrix}2&1&3\\0&2&1\\4&1&2\end{pmatrix}. Crossing out row 1 and column 1 leaves the minor for the 2 at position (1,1): \begin{pmatrix}2&1\\1&2\end{pmatrix}. Crossing out row 1 and column 2 leaves the minor for the 1 at (1,2): \begin{pmatrix}0&1\\4&2\end{pmatrix}. Crossing out row 1 and column 3 leaves the minor for the 3 at (1,3): \begin{pmatrix}0&2\\4&1\end{pmatrix}.

   Why: deleting an entry's own row and column is what isolates the 2\times2 piece of A that entry's term in the determinant formula multiplies.

2. 2

   Compute the ordinary 2\times2 determinant of each minor from Step 1.

   Example

   \begin{vmatrix}2&1\\1&2\end{vmatrix}=2(2)-1(1)=3, \qquad \begin{vmatrix}0&1\\4&2\end{vmatrix}=0(2)-1(4)=-4, \qquad \begin{vmatrix}0&2\\4&1\end{vmatrix}=0(1)-2(4)=-8

   Why: this is nothing but the 2\times2 determinant from the previous section, applied three times — no new computation, just reused three times.

3. 3

   Multiply the first minor by + its row-1 entry, the second minor by - its row-1 entry, the third minor by + its row-1 entry, then add the three signed terms.

   Example

   Row 1 of M is (2,1,3), so \det M = +2(3) - 1(-4) + 3(-8) = 6 + 4 - 24 = -14

   Why: the alternating +,-,+ pattern (never +,+,+) is baked into the definition of the 3\times3 determinant — dropping the sign on the middle term is the single most common way to get this wrong.

4. 4

   The magnitude |\det A| is the parallelepiped's volume; the sign tells you which way the three columns are twisted.

   Example

   \det M = -14 < 0, so M's three columns form a left-handed triple, and together they span a parallelepiped of volume |{-14}|=14 cubic units.

   • \det A>0 — the columns (\mathbf u,\mathbf v,\mathbf w) form a right-handed triple.
   • \det A<0 — the columns form a left-handed (mirror-image) triple.
   • \det A=0 — the columns are coplanar; the "box" has collapsed to zero volume.

   Why: the sign and the magnitude are exactly what the technical derivation showed \det A measures — orientation and volume in one number, not two separate computations.