Mathematics · Ch 4 — Determinants

Determinant of a Matrix of Order 3 × 3

4.2.3

Determinant of a Matrix of Order 3 × 3

4.2.3 Determinant of a Matrix of Order 3 × 3

The Core Idea: Expanding a 3×3 Determinant

A determinant of order 3 is found by expressing it in terms of second-order ( $2 \times 2$ ) determinants — a process called expansion of a determinant along a row (or a column).

There are exactly six ways to expand a $3 \times 3$ determinant — along each of the three rows ( $R_1, R_2, R_3$ ) and each of the three columns ( $C_1, C_2, C_3$ ). Remarkably, all six expansions give the same numerical value, a fundamental property that makes determinants consistent and powerful.

Expansion Along the First Row ( $R_1$ )

Consider a general matrix $A = [a_{ij}]$ and its determinant:

|A| = \begin{vmatrix} a_{11} & a_{12} & a_{13} \\ a_{21} & a_{22} & a_{23} \\ a_{31} & a_{32} & a_{33} \end{vmatrix}

The expansion along the first row is built element by element:

›Proof

Step-by-step expansion along $R_1$

Step 1: Take $a_{11}$ (row 1, column 1). Multiply it by the sign factor $(-1)^{1+1}$ and by the $2 \times 2$ determinant obtained by deleting row 1 and column 1:

$(-1)^{1+1} a_{11} \begin{vmatrix} a_{22} & a_{23} \\ a_{32} & a_{33} \end{vmatrix}$

Step 2: For $a_{12}$ (row 1, column 2), multiply by $(-1)^{1+2}$ and by the determinant left after deleting row 1 and column 2:

$(-1)^{1+2} a_{12} \begin{vmatrix} a_{21} & a_{23} \\ a_{31} & a_{33} \end{vmatrix}$

Step 3: For $a_{13}$ (row 1, column 3), multiply by $(-1)^{1+3}$ and by the determinant left after deleting row 1 and column 3:

$(-1)^{1+3} a_{13} \begin{vmatrix} a_{21} & a_{22} \\ a_{31} & a_{32} \end{vmatrix}$

Step 4: The determinant is the sum of these three terms:

$|A| = (-1)^{1+1} a_{11} \begin{vmatrix} a_{22} & a_{23} \\ a_{32} & a_{33} \end{vmatrix} + (-1)^{1+2} a_{12} \begin{vmatrix} a_{21} & a_{23} \\ a_{31} & a_{33} \end{vmatrix} + (-1)^{1+3} a_{13} \begin{vmatrix} a_{21} & a_{22} \\ a_{31} & a_{32} \end{vmatrix}$

Evaluating each $2 \times 2$ determinant (recall $\begin{vmatrix} p & q \\ r & s \end{vmatrix} = ps - qr$ ):

|A| = a_{11}(a_{22}a_{33} - a_{32}a_{23}) - a_{12}(a_{21}a_{33} - a_{31}a_{23}) + a_{13}(a_{21}a_{32} - a_{31}a_{22})

which expands to

|A| = a_{11}a_{22}a_{33} - a_{11}a_{32}a_{23} - a_{12}a_{21}a_{33} + a_{12}a_{31}a_{23} + a_{13}a_{21}a_{32} - a_{13}a_{31}a_{22}

This is equation (1) — the standard expansion along the first row.

Tip

In practice you apply all four steps together in one go; the step-by-step breakdown is only for understanding the logic.

Expansion Along the Second Row ( $R_2$ )

Expanding along a different row must give the same result:

›Proof

Expansion along $R_2$

For each element of row 2, multiply by $(-1)^{\text{row}+\text{col}}$ and by the $2 \times 2$ determinant from deleting its row and column:

$|A| = -a_{21}\begin{vmatrix} a_{12} & a_{13} \\ a_{32} & a_{33} \end{vmatrix} + a_{22}\begin{vmatrix} a_{11} & a_{13} \\ a_{31} & a_{33} \end{vmatrix} - a_{23}\begin{vmatrix} a_{11} & a_{12} \\ a_{31} & a_{32} \end{vmatrix}$

$= -a_{21}(a_{12}a_{33} - a_{32}a_{13}) + a_{22}(a_{11}a_{33} - a_{31}a_{13}) - a_{23}(a_{11}a_{32} - a_{31}a_{12})$

Expanding and rearranging the terms gives

$|A| = a_{11}a_{22}a_{33} - a_{11}a_{23}a_{32} - a_{12}a_{21}a_{33} + a_{12}a_{23}a_{31} + a_{13}a_{21}a_{32} - a_{13}a_{31}a_{22}$

This is exactly equation (1) again — the same six products with the same signs, in a different order.

Expansion Along the First Column ( $C_1$ )

›Proof

Expansion along $C_1$

For each element of column 1:

$|A| = a_{11}\begin{vmatrix} a_{22} & a_{23} \\ a_{32} & a_{33} \end{vmatrix} - a_{21}\begin{vmatrix} a_{12} & a_{13} \\ a_{32} & a_{33} \end{vmatrix} + a_{31}\begin{vmatrix} a_{12} & a_{13} \\ a_{22} & a_{23} \end{vmatrix}$

$= a_{11}(a_{22}a_{33} - a_{23}a_{32}) - a_{21}(a_{12}a_{33} - a_{13}a_{32}) + a_{31}(a_{12}a_{23} - a_{13}a_{22})$

Expanding gives

$|A| = a_{11}a_{22}a_{33} - a_{11}a_{23}a_{32} - a_{21}a_{12}a_{33} + a_{21}a_{13}a_{32} + a_{31}a_{12}a_{23} - a_{31}a_{13}a_{22}$

which again matches equation (1).

Important

The value of a $3 \times 3$ determinant is the same regardless of the row or column you expand along, so you can choose the most convenient one. (Expansions along $R_3$ , $C_2$ , $C_3$ are left as an exercise.)

Practical Remarks for Calculation

Tip

For easier calculation, expand along the row or column with the most zeros — each zero eliminates a term.

The sign pattern. Instead of computing $(-1)^{i+j}$ each time, use the checkerboard pattern: …

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Determinant of a Matrix of Order 3 × 3