Determinant of a 3x3 Matrix: From First Row Expansion to Sarrus' Rule

The Story: Two Perspectives on the $3 \times 3$ Determinant

The determinant of a $3 \times 3$ matrix measures the signed volume scaling factor of the parallelepiped spanned by its column vectors in $\mathbb{R}^3$. If $\det(A) = 0$, the transformation collapses 3D space onto a flat plane or line, rendering the matrix non-invertible.

Computing this volume algebraically requires evaluating $3! = 6$ distinct permutation products. In this 2-part masterclass, we explore how two classic techniques—Laplace’s First Row Expansion and Sarrus’ Rule—tackle the exact same computation from two fundamentally different angles.

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Part 1: Laplace’s Cofactor Expansion along the First Row

Given an arbitrary $3 \times 3$ matrix $A$:

$$ A = \begin{pmatrix} a & b & c \\ d & e & f \\ g & h & i \end{pmatrix} $$

Laplace expansion decomposes the determinant along any row or column into smaller $(n-1) \times (n-1)$ sub-determinants (called minors $M_{ij}$), weighted by cofactors $C_{ij} = (-1)^{i+j} M_{ij}$.

Expanding along the first row ($i = 1$):

1. Element $a$ (row 1, col 1 $\to +$): Cross out row 1 and column 1:

   $$+a \cdot \begin{vmatrix} e & f \\ h & i \end{vmatrix} = a(ei - fh)$$

2. Element $b$ (row 1, col 2 $\to -$): Cross out row 1 and column 2:

   $$-b \cdot \begin{vmatrix} d & f \\ g & i \end{vmatrix} = -b(di - fg)$$

3. Element $c$ (row 1, col 3 $\to +$): Cross out row 1 and column 3:

   $$+c \cdot \begin{vmatrix} d & e \\ g & h \end{vmatrix} = c(dh - eg)$$

Combining these three pieces yields the complete expanded polynomial:

$$ \det(A) = aei - afh - bdi + bfg + cdh - ceg $$

Reordering positive terms first and negative terms second:

$$ \det(A) = (aei + bfg + cdh) - (afh + bdi + ceg) $$

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Part 2: Sarrus’ Rule (The Diagonal Shortcut)

In 1846, French mathematician Pierre Frédéric Sarrus discovered that the reordered 6 terms can be read off the matrix visually without calculating sub-matrices.

By augmenting the matrix with copies of its first two columns:

$$ \begin{array}{c|ccc|c} c & a & b & c & a\\ f & d & e & f & d\\ i & g & h & i & g \end{array} $$

The terms naturally group into two sets of three diagonals:

1. Downward Diagonals (Add)

Tracing top-left to bottom-right:

• First downward diagonal : $c \cdot d \cdot h$
• Second downward diagonal : $a \cdot e \cdot i$
• Third downward diagonal : $b \cdot f \cdot g$

$$\Sigma_+ = cdh + aei + bfg$$

2. Upward Diagonals (Subtract)

Tracing bottom-left to top-right:

• First upward diagonal : $b \cdot d \cdot i$
• Second upward diagonal : $c \cdot e \cdot g$
• Third upward diagonal : $a \cdot f \cdot h$

$$\Sigma_- = bdi + ceg + afh$$

The Unified Result:

$$ \det(A) = \Sigma_+ - \Sigma_- = (cdh + aei + bfg) - (bdi + ceg + afh) $$

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Side-by-Side Equivalence: Why Both Methods Agree

Term in Expansion	Laplace Origin	Sarrus Diagonal	Sign
$+aei$	From $a(ei - fh)$	Downward diagonal $(a \cdot e \cdot i)$	$+$
$+bfg$	From $-b(di - fg) \to +bfg$	Downward diagonal $(b \cdot f \cdot g)$	$+$
$+cdh$	From $c(dh - eg)$	Downward diagonal $(c \cdot d \cdot h)$	$+$
$-afh$	From $a(ei - fh) \to -afh$	Upward diagonal $(h \cdot f \cdot a)$	$-$
$-bdi$	From $-b(di - fg) \to -bdi$	Upward diagonal $(i \cdot d \cdot b)$	$-$
$-ceg$	From $c(dh - eg) \to -ceg$	Upward diagonal $(g \cdot e \cdot c)$	$-$

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Practical Takeaway: When to Use Which?

• Use Sarrus’ Rule when calculating a concrete $3 \times 3$ numerical matrix by hand under time pressure (e.g., cross products, calculating eigenvalues of a $3 \times 3$ system).
• Use Laplace Expansion when the matrix contains zeros (expand along the row or column with the most zeros to eliminate sub-matrices instantly).
• Beware the $4 \times 4$ Trap: Sarrus’ Rule only works for $3 \times 3$. A $4 \times 4$ matrix has $4! = 24$ permutation terms, but the diagonal visual trick produces only 8 diagonals. For matrices of order $4 \times 4$ and larger, Laplace expansion or Gaussian row-reduction is mandatory!