Chapter 05

Medium

Determinants

00 — Symbol Glossary

01 — What is the Determinant?

Definition

The determinant of a square matrix $A\in\mathbb{R}^{n\times n}$ is a scalar $\det(A)\in\mathbb{R}$ that encodes the signed scaling factor of the linear transformation represented by $A$ .

\det(A) = \text{signed volume of the parallelepiped spanned by the column vectors of } A

Geometrically:

$\lvert\det(A)\rvert$ is the factor by which $A$ scales areas (2-D) or volumes (3-D).
$\det(A)>0$ : orientation is preserved.
$\det(A)<0$ : orientation is reversed (a reflection is involved).
$\det(A)=0$ : the transformation collapses space into a lower dimension — the matrix is singular (non-invertible).

Note

The determinant is defined only for square matrices. For non-square matrices, singular-value decomposition plays an analogous role.

2×2 Determinant

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

\det(A)=ad-bc

The formula $ad-bc$ is the area of the parallelogram formed by the two column vectors $\begin{pmatrix}a\\c\end{pmatrix}$ and $\begin{pmatrix}b\\d\end{pmatrix}$ .

Example

Let $A=\begin{pmatrix}3&2\\1&4\end{pmatrix}$ .

$\det(A)=3\cdot4-2\cdot1=12-2=10$

The columns $\begin{pmatrix}3\\1\end{pmatrix}$ and $\begin{pmatrix}2\\4\end{pmatrix}$ span a parallelogram of area $10$ .

Common mistake

Wrong: $\det\begin{pmatrix}a&b\\c&d\end{pmatrix}=ab-cd$ (multiplying rows instead of diagonals).
Why it happens: Confusing row products with diagonal products.
Correct: Multiply the main diagonal ( $ad$ ) and subtract the anti-diagonal ( $bc$ ).
Check: $\det(I)=1\cdot1-0\cdot0=1$ . Any other formula gives the wrong answer on the identity.

02 — Cofactor Expansion (3×3 and beyond)

For an $n\times n$ matrix, expand along any row or column. Expanding along row 1:

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

The sign pattern for cofactors is:

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

Cofactor expansion of a 3×3 matrix — inputs: $A=\begin{pmatrix}1&2&3\\0&4&5\\1&0&6\end{pmatrix}$

Row 1 signs: $+,\,-,\,+$ (since $(-1)^{1+1}=+1$ , $(-1)^{1+2}=-1$ , $(-1)^{1+3}=+1$ ).

$M_{11}=\det\begin{pmatrix}4&5\\0&6\end{pmatrix}=4\cdot6-5\cdot0=24-0=24$ — The $4$ is $a_{22}$ ; the $6$ is $a_{33}$ ; the $5$ and $0$ are off-diagonal.

$M_{12}=\det\begin{pmatrix}0&5\\1&6\end{pmatrix}=0\cdot6-5\cdot1=0-5=-5$ — The $0$ is $a_{21}$ ; the $6$ is $a_{33}$ ; the $5$ and $1$ are off-diagonal.

$M_{13}=\det\begin{pmatrix}0&4\\1&0\end{pmatrix}=0\cdot0-4\cdot1=0-4=-4$ — The $0$ s are $a_{21}$ and $a_{32}$ ; the $4$ is $a_{22}$ ; the $1$ is $a_{31}$ .

$\det(A)=a_{11}\cdot(+1)\cdot M_{11}+a_{12}\cdot(-1)\cdot M_{12}+a_{13}\cdot(+1)\cdot M_{13}$ $=1\cdot(+24)+2\cdot(-1)(-5)+3\cdot(-4)$ — The $1,2,3$ are the row-1 entries; the signs $+1,-1,+1$ are cofactor signs; the $24,-5,-4$ are the minors.

$=24+10-12=\boxed{22}$ — $24$ from step 2, $+10$ from $2\cdot(-1)\cdot(-5)=10$ , $-12$ from $3\cdot(-4)$ .

Note

Expand along a row or column with the most zeros to minimise computation. If a row has two zeros, only one minor needs to be computed.

Common mistake

Wrong: $\det(A)=a_{11}M_{11}+a_{12}M_{12}+a_{13}M_{13}$ (all signs positive).
Why it happens: The sign pattern $(-1)^{i+j}$ is forgotten.
Correct: $\det(A)=a_{11}M_{11}-a_{12}M_{12}+a_{13}M_{13}$ for row-1 expansion.
Check: $\det\begin{pmatrix}0&1&0\\1&0&0\\0&0&1\end{pmatrix}$ should equal $-1$ (row swap). If your formula gives $+1$ , you dropped the signs.

03 — Properties of Determinants

Definition

Let $A,B\in\mathbb{R}^{n\times n}$ . The following hold:

Row swap: swapping two rows multiplies $\det$ by $-1$ .
Scalar multiple: multiplying one row by $k$ multiplies $\det$ by $k$ .
Row addition: adding a multiple of one row to another leaves $\det$ unchanged.
Multiplicativity: $\det(AB)=\det(A)\det(B)$ .
Transpose: $\det(A^\top)=\det(A)$ .
Identity: $\det(I)=1$ .
Inverse: If $A$ is invertible, $\det(A^{-1})=\dfrac{1}{\det(A)}$ .
Triangular matrix: $\det(A)=\prod_{i=1}^{n}a_{ii}$ (product of diagonal entries).

Example

$\det\begin{pmatrix}3&7&2\\0&-1&5\\0&0&4\end{pmatrix}=3\cdot(-1)\cdot4=-12$ No expansion needed — just multiply the diagonal.

Example

If $\det(A)=3$ and $\det(B)=2$ , then $\det(AB)=6$ . This means composing two transformations multiplies their area-scaling factors.

Note

Property 8 is why Gaussian elimination is efficient for computing determinants: reduce to upper triangular form while tracking row-swap signs, then multiply the diagonal.

04 — Determinants via Row Reduction

Row reduction is computationally faster than cofactor expansion for large matrices. Track how each row operation changes the determinant.

Determinant by row reduction — inputs: $A=\begin{pmatrix}2&1&3\\4&5&1\\2&3&7\end{pmatrix}$

$\det(A)=?$ ; start with multiplier $k=1$ (no operations yet).

$\begin{pmatrix}2&1&3\\0&3&-5\\2&3&7\end{pmatrix}$ — $4-2\cdot2=0$ ; $5-2\cdot1=3$ ; $1-2\cdot3=-5$ . Row addition: $\det$ unchanged, so $k$ stays $1$ .

$\begin{pmatrix}2&1&3\\0&3&-5\\0&2&4\end{pmatrix}$ — $2-2=0$ ; $3-1=2$ ; $7-3=4$ . Row addition: $k$ stays $1$ .

$\begin{pmatrix}2&1&3\\0&3&-5\\0&0&\tfrac{22}{3}\end{pmatrix}$ — $2-\tfrac{2}{3}\cdot3=0$ ; $4-\tfrac{2}{3}\cdot(-5)=4+\tfrac{10}{3}=\tfrac{22}{3}$ . Row addition: $k$ stays $1$ .

$\det(A)=k\cdot2\cdot3\cdot\tfrac{22}{3}=1\cdot2\cdot3\cdot\tfrac{22}{3}=44$ — The $2$ is $a_{11}$ after reduction; $3$ is the pivot in row 2; $\tfrac{22}{3}$ is the pivot in row 3.

05 — Invertibility and Cramer's Rule

Definition

A square matrix $A\in\mathbb{R}^{n\times n}$ is invertible (non-singular) if and only if

$\det(A)\neq 0$

If $\det(A)=0$ , the system $A\mathbf{x}=\mathbf{b}$ either has no solution or infinitely many.

Definition

For a system $A\mathbf{x}=\mathbf{b}$ with $\det(A)\neq0$ and $A\in\mathbb{R}^{n\times n}$ :

$x_i=\frac{\det(A_i)}{\det(A)}$

where $A_i$ is $A$ with its $i$ -th column replaced by $\mathbf{b}$ .

Example

Solve $2x+y=5$ , $x+3y=7$ .

$A=\begin{pmatrix}2&1\\1&3\end{pmatrix}$ , $\mathbf{b}=\begin{pmatrix}5\\7\end{pmatrix}$ .

$\det(A)=2\cdot3-1\cdot1=5$ $x=\frac{\det\begin{pmatrix}5&1\\7&3\end{pmatrix}}{5}=\frac{15-7}{5}=\frac{8}{5}$ $y=\frac{\det\begin{pmatrix}2&5\\1&7\end{pmatrix}}{5}=\frac{14-5}{5}=\frac{9}{5}$

Note

Cramer's Rule is elegant but $O(n!)$ in theory — never use it for large systems. Gaussian elimination ( $O(n^3)$ ) is always preferred computationally.

Common mistake

Wrong: proceeding to compute $x_i=\det(A_i)/\det(A)$ when $\det(A)=0$ .
Why it happens: forgetting to check invertibility first.
Correct: check $\det(A)\neq0$ before applying Cramer's Rule. If $\det(A)=0$ , use row reduction to classify the system.
Check: a zero determinant means the columns are linearly dependent — the system cannot have a unique solution.

06 — Geometric Meaning and Quant Application

Area and Volume

\text{Area of parallelogram spanned by } \mathbf{u},\mathbf{v} = \left|\det\begin{pmatrix}u_1&v_1\\u_2&v_2\end{pmatrix}\right|

\text{Volume of parallelepiped spanned by } \mathbf{u},\mathbf{v},\mathbf{w} = \left|\det\begin{pmatrix}u_1&v_1&w_1\\u_2&v_2&w_2\\u_3&v_3&w_3\end{pmatrix}\right|

Quant Application — Change of Variables in Probability

When transforming a joint density $f_{X,Y}(x,y)$ to new variables $(u,v)=g(x,y)$ , the Jacobian determinant scales the probability mass:

$f_{U,V}(u,v)=f_{X,Y}(x(u,v),\,y(u,v))\cdot\left|\det\mathbf{J}\right|$

where $\mathbf{J}=\dfrac{\partial(x,y)}{\partial(u,v)}$ is the Jacobian matrix.

In options pricing, converting from the risk-neutral measure $\mathbb{Q}$ to the physical measure $\mathbb{P}$ via a Radon-Nikodym derivative is the infinite-dimensional analogue of this determinant scaling.

In portfolio optimisation, the determinant of the covariance matrix $\Sigma$ measures the "generalised variance" of a portfolio. A near-zero $\det(\Sigma)$ signals near-multicollinearity among assets — the portfolio has essentially fewer degrees of freedom than the number of positions.

Exercises

EXERCISE 5.1

Use $\det\begin{pmatrix}a&b\\c&d\end{pmatrix}=ad-bc$ . Identify $a,b,c,d$ carefully — do not multiply row entries.

$A=\begin{pmatrix}-3&5\\2&-1\end{pmatrix}$ .

Step 1 — Identify: $a=-3$ , $b=5$ , $c=2$ , $d=-1$ .

Step 2 — Apply formula: $\det(A)=(-3)(-1)-(5)(2)=3-10=-7$ .

The determinant is $-7$ . The negative sign means the transformation reverses orientation.

Compute $\det\begin{pmatrix}-3&5\\2&-1\end{pmatrix}$ .

EXERCISE 5.2

Expand along row 1 with signs $+,-,+$ . Look for the column with a zero to skip one minor.

$B=\begin{pmatrix}1&0&2\\3&1&-1\\0&2&4\end{pmatrix}$ .

Expand along row 1 (signs $+,-,+$ ):

$M_{11}=\det\begin{pmatrix}1&-1\\2&4\end{pmatrix}=4-(-2)=6$ .

$M_{12}=\det\begin{pmatrix}3&-1\\0&4\end{pmatrix}=12-0=12$ . (Multiplied by $a_{12}=0$ , so this term vanishes.)

$M_{13}=\det\begin{pmatrix}3&1\\0&2\end{pmatrix}=6-0=6$ .

$\det(B)=1\cdot(+1)\cdot6+0\cdot(-1)\cdot12+2\cdot(+1)\cdot6=6+0+12=\boxed{18}$ .

Compute $\det\begin{pmatrix}1&0&2\\3&1&-1\\0&2&4\end{pmatrix}$ using cofactor expansion.

EXERCISE 5.3

The matrix is upper triangular. The determinant of a triangular matrix is the product of its diagonal entries — no expansion needed.

$C=\begin{pmatrix}5&3&-2&1\\0&-2&7&4\\0&0&3&9\\0&0&0&-1\end{pmatrix}$ .

Upper triangular: $\det(C)=5\cdot(-2)\cdot3\cdot(-1)$ .

$=5\cdot(-2)=-10$ ; then $-10\cdot3=-30$ ; then $-30\cdot(-1)=30$ .

$\det(C)=\boxed{30}$ .

Compute $\det\begin{pmatrix}5&3&-2&1\\0&-2&7&4\\0&0&3&9\\0&0&0&-1\end{pmatrix}$ .

EXERCISE 5.4

Use properties: $\det(A^{-1})=1/\det(A)$ ; $\det(kA)=k^n\det(A)$ for $n\times n$ ; $\det(AB)=\det(A)\det(B)$ ; $\det(A^\top)=\det(A)$ . Apply each one in turn.

Given $A$ is $3\times3$ with $\det(A)=4$ .

(a) $\det(A^{-1})=1/\det(A)=1/4$ .

(b) $\det(2A)=2^3\det(A)=8\cdot4=32$ . (Each of 3 rows is scaled by 2, so $\det$ is multiplied by $2^3$ .)

(d) $\det(A^\top)=\det(A)=4$ . The transpose does not change the determinant.

Let $A$ be a $3\times3$ matrix with $\det(A)=4$ . Compute: (a) $\det(A^{-1})$ , (b) $\det(2A)$ , (c) $\det(A^2)$ , (d) $\det(A^\top)$ .

EXERCISE 5.5

The columns are linearly dependent if and only if $\det=0$ . Check whether one column is a linear combination of the others, or compute the determinant directly.

$D=\begin{pmatrix}1&2&3\\2&4&6\\0&1&1\end{pmatrix}$ .

Notice: row 2 $= 2\times$ row 1. A row operation $R_2\leftarrow R_2-2R_1$ gives a zero row.

A matrix with a zero row has $\det=0$ by linearity of the determinant in each row.

Therefore $\det(D)=0$ and $D$ is singular — not invertible.

Geometrically, the three column vectors are coplanar (they lie in a 2-D subspace), so the parallelepiped they span has zero volume.

Determine whether $\begin{pmatrix}1&2&3\\2&4&6\\0&1&1\end{pmatrix}$ is invertible. Explain the geometric meaning.

EXERCISE 5.6

A $2\times2$ covariance matrix $\Sigma=\begin{pmatrix}\sigma_1^2&\rho\sigma_1\sigma_2\\\rho\sigma_1\sigma_2&\sigma_2^2\end{pmatrix}$ . Compute $\det(\Sigma)=\sigma_1^2\sigma_2^2-\rho^2\sigma_1^2\sigma_2^2$ and factor.

$\Sigma=\begin{pmatrix}\sigma_1^2&\rho\sigma_1\sigma_2\\\rho\sigma_1\sigma_2&\sigma_2^2\end{pmatrix}$ .

$\det(\Sigma)=\sigma_1^2\sigma_2^2-(\rho\sigma_1\sigma_2)^2=\sigma_1^2\sigma_2^2-\rho^2\sigma_1^2\sigma_2^2=\sigma_1^2\sigma_2^2(1-\rho^2)$ .

When $\rho=0$ : $\det(\Sigma)=\sigma_1^2\sigma_2^2 > 0$ — full 2-D spread, assets independent.

When $|\rho|\to1$ : $\det(\Sigma)\to0$ — perfect correlation means both assets move on a 1-D line; the covariance matrix is singular.

In portfolio construction, $\det(\Sigma)\approx0$ means the portfolio's effective dimension is less than 2 — there is no diversification benefit from holding both assets.

A two-asset portfolio has return covariance matrix $\Sigma=\begin{pmatrix}\sigma_1^2&\rho\sigma_1\sigma_2\\\rho\sigma_1\sigma_2&\sigma_2^2\end{pmatrix}$ . Compute $\det(\Sigma)$ and explain what happens as $|\rho|\to1$ .

Chapter Summary

Concept	Formula / Rule
$2\times2$ determinant	$\det\begin{pmatrix}a&b\\c&d\end{pmatrix}=ad-bc$
Cofactor	$C_{ij}=(-1)^{i+j}M_{ij}$
Cofactor expansion (row $i$ )	$\det(A)=\sum_j a_{ij}C_{ij}$
Triangular shortcut	$\det(A)=\prod_i a_{ii}$
Row swap	Multiplies $\det$ by $-1$
Row scaling	Multiplies $\det$ by the scalar
Row addition	No change to $\det$
Multiplicativity	$\det(AB)=\det(A)\det(B)$
Invertibility	$A$ invertible $\iff$ $\det(A)\neq0$
Geometric meaning	$\lvert\det(A)\rvert$ = area/volume scaling factor

Next chapter: Chapter 06 — Eigenvalues & Eigenvectors, where the determinant plays a central role in computing the characteristic polynomial $\det(A-\lambda I)=0$ .