Determinants (2×2 and 3×3)

The determinant is a scalar value associated with a square matrix. It encodes important geometric and algebraic information: whether a matrix is invertible, the area/volume scaling factor of a transformation, and more.

---

The 2×2 Determinant

For $A = \begin{pmatrix} a & b \\ c & d \end{pmatrix}$A=(ac​bd​):

$\det(A) = |A| = ad - bc$det(A)=∣A∣=ad−bc

Worked Example 1: $\det\begin{pmatrix} 3 & 2 \\ 1 & 5 \end{pmatrix} = 3(5) - 2(1) = 15 - 2 = 13$det(31​25​)=3(5)−2(1)=15−2=13

Worked Example 2: $\det\begin{pmatrix} 4 & 6 \\ 2 & 3 \end{pmatrix} = 4(3) - 6(2) = 12 - 12 = 0$det(42​63​)=4(3)−6(2)=12−12=0

When the determinant is zero, the matrix is called singular and has no inverse.

---

The 3×3 Determinant

For $A = \begin{pmatrix} a & b & c \\ d & e & f \\ g & h & k \end{pmatrix}$A=​adg​beh​cfk​​:

$\det(A) = a(ek - fh) - b(dk - fg) + c(dh - eg)$det(A)=a(ek−fh)−b(dk−fg)+c(dh−eg)

This is expansion along the first row using cofactors.

Worked Example 3: Find $\det\begin{pmatrix} 2 & 1 & 3 \\ 0 & -1 & 2 \\ 1 & 4 & -1 \end{pmatrix}$det​201​1−14​32−1​​.

$= 2((-1)(-1) - (2)(4)) - 1((0)(-1) - (2)(1)) + 3((0)(4) - (-1)(1))$=2((−1)(−1)−(2)(4))−1((0)(−1)−(2)(1))+3((0)(4)−(−1)(1)) $= 2(1 - 8) - 1(0 - 2) + 3(0 + 1)$=2(1−8)−1(0−2)+3(0+1) $= 2(-7) - 1(-2) + 3(1)$=2(−7)−1(−2)+3(1) $= -14 + 2 + 3 = -9$=−14+2+3=−9

---

Cofactor Expansion

You can expand along any row or column. The signs follow a checkerboard pattern:

$$\begin{pmatrix} + & - & + \\ - & + & - \\ + & - & + \end{pmatrix}$$​+−+​−+−​+−+​​

Tip: Choose the row or column with the most zeros to simplify the calculation.

---

Properties of Determinants

Property	Statement
$\det(I) = 1$det(I)=1	The identity matrix has determinant 1
$\det(AB) = \det(A) \cdot \det(B)$det(AB)=det(A)⋅det(B)	Determinants multiply
$\det(A^T) = \det(A)$det(AT)=det(A)	Transpose does not change the determinant
$\det(kA) = k^n \det(A)$det(kA)=kndet(A)	For an $n \times n$n×n matrix
Row swap	Changes the sign of the determinant
Two identical rows	Determinant is 0
$\det(A) = 0$det(A)=0	$A$A is singular (not invertible)
$\det(A^{-1}) = \frac{1}{\det(A)}$det(A−1)=det(A)1​	When $A$A is invertible

---

Geometric Interpretation