| | 2025.12.06

9: The Determinant

The determinant is a single number that captures something essential about a square matrix: how it scales volume. A determinant of zero means the matrix collapses space,it’s singular. A nonzero determinant means the matrix is invertible. This geometric meaning drives everything else.

Geometric Meaning

(What the Determinant Measures)

For a square matrix $A$ , the determinant $\det(A)$ measures:

Signed volume scaling: How much $A$ scales the volume of any region
Orientation: Whether $A$ preserves or reverses orientation (sign of det)

If $A$ is $n \times n$ and $R$ is any region in $\mathbb{R}^n$ :

\text{Volume}(A(R)) = |\det(A)| \cdot \text{Volume}(R)

Examples:

$\det(A) = 2$ : Doubles volumes, preserves orientation
$\det(A) = -3$ : Triples volumes, reverses orientation (reflection)
$\det(A) = 0$ : Collapses to lower dimension, volume becomes zero

(The Unit Cube Picture)

The columns of $A$ are the images of the standard basis vectors. The determinant equals the signed volume of the parallelepiped spanned by these column vectors.

For a $2 \times 2$ matrix, the columns span a parallelogram. The determinant is its signed area.

For a $3 \times 3$ matrix, the columns span a parallelepiped. The determinant is its signed volume.

The 2×2 Determinant

(Formula)

For $A = \begin{bmatrix} a & b \\ c & d \end{bmatrix}$ :

\det(A) = ad - bc

Derivation from area: The columns $\begin{bmatrix} a \\ c \end{bmatrix}$ and $\begin{bmatrix} b \\ d \end{bmatrix}$ span a parallelogram. Using the cross product formula for area (or direct geometry), we get $|ad - bc|$ . The sign tracks orientation.

(Example)

\det\begin{bmatrix} 3 & 1 \\ 2 & 4 \end{bmatrix} = 3(4) - 1(2) = 10

This transformation scales areas by a factor of $10$ .

(Singular Case)

\det\begin{bmatrix} 2 & 4 \\ 1 & 2 \end{bmatrix} = 2(2) - 4(1) = 0

The columns $\begin{bmatrix} 2 \\ 1 \end{bmatrix}$ and $\begin{bmatrix} 4 \\ 2 \end{bmatrix}$ are parallel,they span a line, not a parallelogram. Zero area means the matrix is singular.

The 3×3 Determinant

(Formula via Sarrus’ Rule)

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

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

This can be remembered by the “rule of Sarrus”: copy the first two columns to the right, then take products along diagonals (down-right positive, up-right negative).

Note: Sarrus’ rule only works for $3 \times 3$ . For larger matrices, use cofactor expansion.

(Example)

\det\begin{bmatrix} 1 & 2 & 3 \\ 4 & 5 & 6 \\ 7 & 8 & 9 \end{bmatrix} = 1(45) + 2(42) + 3(32) - 3(35) - 2(36) - 1(48)

= 45 + 84 + 96 - 105 - 72 - 48 = 0

The determinant is zero,these columns are linearly dependent (the third column is the average of the first two).

Cofactor Expansion

For matrices larger than $3 \times 3$ , we use cofactor expansion (also called Laplace expansion).

(Minor and Cofactor)

For an $n \times n$ matrix $A$ :

The $(i,j)$ minor $M_{ij}$ is the determinant of the $(n-1) \times (n-1)$ matrix obtained by deleting row $i$ and column $j$
The $(i,j)$ cofactor $C_{ij}$ is the signed minor:

C_{ij} = (-1)^{i+j} M_{ij}

(The Checkerboard Sign Pattern)

The factor $(-1)^{i+j}$ creates a checkerboard of signs:

\begin{bmatrix} + & - & + & - & \cdots \\ - & + & - & + & \cdots \\ + & - & + & - & \cdots \\ - & + & - & + & \cdots \\ \vdots & \vdots & \vdots & \vdots & \ddots \end{bmatrix}

Position $(1,1)$ is positive, and signs alternate from there.

(Cofactor Expansion Along a Row)

The determinant can be computed by expanding along any row $i$ :

\det(A) = \sum_{j=1}^{n} a_{ij} C_{ij} = \sum_{j=1}^{n} (-1)^{i+j} a_{ij} M_{ij}

Expanding along row 1:

\det(A) = a_{11}C_{11} - a_{12}C_{12} + a_{13}C_{13} - \cdots

(Cofactor Expansion Along a Column)

Equivalently, expand along any column $j$ :

\det(A) = \sum_{i=1}^{n} a_{ij} C_{ij}

Key insight: Choose the row or column with the most zeros to minimize computation.

(Example: 3×3 via Cofactor Expansion)

A = \begin{bmatrix} 2 & 1 & 3 \\ 0 & 4 & 5 \\ 1 & 0 & 2 \end{bmatrix}

Expand along row 1:

\det(A) = 2 \cdot \det\begin{bmatrix} 4 & 5 \\ 0 & 2 \end{bmatrix} - 1 \cdot \det\begin{bmatrix} 0 & 5 \\ 1 & 2 \end{bmatrix} + 3 \cdot \det\begin{bmatrix} 0 & 4 \\ 1 & 0 \end{bmatrix}

= 2(8 - 0) - 1(0 - 5) + 3(0 - 4) = 16 + 5 - 12 = 9

(Example: 4×4 Determinant)

A = \begin{bmatrix} 1 & 0 & 2 & 0 \\ 3 & 1 & 0 & 1 \\ 0 & 0 & 1 & 0 \\ 2 & 0 & 0 & 1 \end{bmatrix}

Column 2 has three zeros,expand along it:

\det(A) = 0 \cdot C_{12} + 1 \cdot C_{22} + 0 \cdot C_{32} + 0 \cdot C_{42} = C_{22}

C_{22} = (-1)^{2+2} \det\begin{bmatrix} 1 & 2 & 0 \\ 0 & 1 & 0 \\ 2 & 0 & 1 \end{bmatrix}

Expand this $3 \times 3$ along column 3:

= (+1)\left( 0 \cdot C_{13} + 0 \cdot C_{23} + 1 \cdot C_{33} \right) = \det\begin{bmatrix} 1 & 2 \\ 0 & 1 \end{bmatrix} = 1

So $\det(A) = 1$ .

Properties of the Determinant

(Multiplicative Property)

For square matrices $A$ and $B$ of the same size:

\det(AB) = \det(A) \det(B)

Interpretation: If $A$ scales volume by $\det(A)$ and $B$ scales by $\det(B)$ , then $AB$ scales by the product.

Consequence: $\det(A^k) = (\det(A))^k$

(Transpose)

\det(A^T) = \det(A)

Rows and columns play symmetric roles in the determinant.

(Inverse)

If $A$ is invertible:

\det(A^{-1}) = \frac{1}{\det(A)}

Proof: $\det(A)\det(A^{-1}) = \det(AA^{-1}) = \det(I) = 1$

(Scalar Multiplication)

For an $n \times n$ matrix:

\det(cA) = c^n \det(A)

Each of the $n$ rows gets multiplied by $c$ , contributing a factor of $c$ each.

Row Operations and the Determinant

The determinant responds predictably to row operations:

(Row Swap)

Swapping two rows negates the determinant:

\det(\text{swap rows } i, j) = -\det(A)

Intuition: Swapping reverses orientation.

(Row Scaling)

Multiplying a row by $c$ scales the determinant by $c$ :

\det(\text{row } i \to c \cdot \text{row } i) = c \cdot \det(A)

(Row Replacement)

Adding a multiple of one row to another preserves the determinant:

\det(\text{row } i \to \text{row } i + c \cdot \text{row } j) = \det(A)

This is why row reduction is useful for computing determinants.

(Computing via Row Reduction)

To find $\det(A)$ :

Row reduce to echelon form, tracking operations
For each row swap, multiply by $-1$
For each row scaling by $c$ , divide by $c$
The determinant of an echelon matrix is the product of diagonal entries

Example:

\begin{bmatrix} 2 & 6 \\ 1 & 4 \end{bmatrix} \xrightarrow{R_1 \leftrightarrow R_2} \begin{bmatrix} 1 & 4 \\ 2 & 6 \end{bmatrix} \xrightarrow{R_2 - 2R_1} \begin{bmatrix} 1 & 4 \\ 0 & -2 \end{bmatrix}

Echelon form has diagonal product $1 \times (-2) = -2$ . One row swap means $\det(A) = -(-2) = 2$ .

Check: $2(4) - 6(1) = 2$ ✓

Determinant and Invertibility

(The Fundamental Characterization)

For a square matrix $A$ :

A \text{ is invertible} \iff \det(A) \neq 0

Why?

$\det(A) = 0$ means the columns are linearly dependent, which means:

The transformation collapses some dimension
$A\mathbf{x} = \mathbf{0}$ has nontrivial solutions
$A$ cannot be inverted (no way to “uncollapse”)

$\det(A) \neq 0$ means the columns are linearly independent, which means:

The transformation preserves all dimensions
The kernel is trivial
$A$ is invertible

(Equivalent Conditions)

For an $n \times n$ matrix $A$ , the following are equivalent:

$\det(A) \neq 0$
$A$ is invertible
$\text{rank}(A) = n$
Columns of $A$ are linearly independent
Columns of $A$ span $\mathbb{R}^n$
$A\mathbf{x} = \mathbf{b}$ has a unique solution for every $\mathbf{b}$
$\ker(A) = \{\mathbf{0}\}$
$\text{rref}(A) = I_n$

Special Matrices

(Triangular Matrices)

For upper or lower triangular matrices, the determinant is the product of diagonal entries:

\det\begin{bmatrix} a_{11} & * & * \\ 0 & a_{22} & * \\ 0 & 0 & a_{33} \end{bmatrix} = a_{11} a_{22} a_{33}

This follows from cofactor expansion,each step picks up one diagonal entry.

(Diagonal Matrices)

\det\begin{bmatrix} d_1 & & \\ & d_2 & \\ & & d_3 \end{bmatrix} = d_1 d_2 d_3

The determinant is the product of eigenvalues (for diagonal matrices, the diagonal entries are the eigenvalues).

(Block Triangular Matrices)

If $A = \begin{bmatrix} B & C \\ 0 & D \end{bmatrix}$ where $B$ and $D$ are square:

\det(A) = \det(B) \det(D)

The Determinant Formula (Advanced)

(Leibniz Formula)

The determinant can be written as a sum over all permutations:

\det(A) = \sum_{\sigma \in S_n} \text{sgn}(\sigma) \prod_{i=1}^{n} a_{i, \sigma(i)}

where $S_n$ is the set of all permutations of $\{1, 2, \ldots, n\}$ and $\text{sgn}(\sigma) = \pm 1$ is the sign of the permutation.

Interpretation: Each term picks one entry from each row and each column. The sign depends on whether the permutation is even or odd.

For $n = 2$ : two permutations give $a_{11}a_{22} - a_{12}a_{21}$ .

For $n = 3$ : six permutations give the Sarrus formula.

For larger $n$ : there are $n!$ terms, which is why direct computation is impractical.

Why the Determinant Matters

The determinant answers fundamental questions:

Is this matrix invertible? Check if $\det \neq 0$
How does this transformation scale volume? That’s $|\det|$
Does it preserve orientation? Check the sign
Are these vectors linearly independent? Put them as columns and check $\det \neq 0$

The determinant compresses a matrix into a single number,but that number encodes deep geometric and algebraic information about what the matrix does.