Determinants
Anton, Rorres, & Kaul Ch. 2
Assignment 2 due 9/25
October 9, 2026
\[ \operatorname{vol}_n\bigl(T_{\mat{A}}(\mathcal S)\bigr) =\abs{\det(\mat{A})}\operatorname{vol}_n(\mathcal S) \]
Determinant magnitude scales volume; determinant sign records orientation
One scalar answers three questions about a square matrix transformation
Let \(\mat{A}\in\reals^{n\times n}\) and \(T_{\mat{A}}(\vct{x})=\mat{A}\vct{x}\)
Let \(\mathcal S\subseteq\reals^n\) be a measurable region: a set of column-vector points, not a matrix
Its image under the transformation is
\[ T_{\mat{A}}(\mathcal S) =\{\mat{A}\vct{x}:\vct{x}\in\mathcal S\} \]
Then
\[ \operatorname{vol}_n\bigl(T_{\mat{A}}(\mathcal S)\bigr) =\abs{\det(\mat{A})}\operatorname{vol}_n(\mathcal S) \]
The determinant records scale, orientation, and collapse in one number
| Transformation in \(\reals^2\) | Determinant | Geometric effect |
|---|---|---|
| \(\operatorname{diag}(s,t)\) | \(st\) | area scales by \(\abs{st}\) |
| \(\begin{bmatrix}1&c\\0&1\end{bmatrix}\) | \(1\) | a shear changes shape, not area |
| rotation \(\mat{R}_\theta\) | \(1\) | area and orientation are preserved |
| reflection \(\mat{H}\) | \(-1\) | area is preserved; orientation reverses |
| projection onto a line \(\mat{Q}\) | \(0\) | the plane collapses to a line |
The sign distinguishes transformations that have the same area scale but opposite orientation
The columns are the images of the standard basis vectors:
\[ \mat{A} =\begin{bmatrix}a&b\\c&d\end{bmatrix} =\begin{bmatrix}\vct{A}_1&\vct{A}_2\end{bmatrix} \]
The image of the unit square is the parallelogram with edge vectors \(\vct{A}_1\) and \(\vct{A}_2\), and
\[ \det(\mat{A})=ad-bc, \qquad \text{parallelogram area}=\abs{ad-bc} \]
Use \(\vct{A}_1=(a,c)^{\mathsf T}\) as the base edge and \(\vct{A}_2=(b,d)^{\mathsf T}\) as the other edge
For base length \(L>0\), choose a perpendicular unit vector
\[\begin{align*} L&=\sqrt{a^2+c^2}, &\vct{n}&=\frac1L\begin{bmatrix}-c\\a\end{bmatrix},\\ \vct{A}_1^{\mathsf T}\vct{n}&=0, &\lVert\vct{n}\rVert&=1,\\[2mm] h\;\text{(height)}&=\abs{\vct{A}_2^{\mathsf T}\vct{n}} =\frac{\abs{-bc+ad}}{L} \end{align*}\]
\[\begin{align*} \text{Area}&=(\text{base length})(\text{height})=Lh\\ &=L\frac{\abs{ad-bc}}{L}=\abs{\det(\mat{A})} \end{align*}\]
The determinant follows the same row operations already used for systems, inverses, and PLU
| Row operation | Effect on the determinant |
|---|---|
| \(R_i\leftrightarrow R_j\) | multiply by \(-1\) |
| \(R_i\leftarrow cR_i\), \(c\ne0\) | multiply by \(c\) |
| \(R_i\leftarrow R_i+cR_j\), \(i\ne j\) | no change |
The analogous rules hold for column operations
Use row replacement freely; track every row swap and scaling
For an upper-triangular matrix,
\[ \mat{U}= \begin{bmatrix} u_{11}&u_{12}&\cdots&u_{1n}\\ 0&u_{22}&\cdots&u_{2n}\\ \vdots&\ddots&\ddots&\vdots\\ 0&\cdots&0&u_{nn} \end{bmatrix}, \qquad \det(\mat{U})=\prod_{j=1}^n u_{jj} \]
The same diagonal-product rule holds for lower-triangular matrices
To compute \(\det(\mat{A})\):
Return to the example from Matrix Transformations:
\[ \mat{A}= \begin{bmatrix} 0&0&1\\ 2&2&0\\ 1&3&1 \end{bmatrix} \quad\xrightarrow[\text{row replacement}]{\text{two row swaps}}\quad \mat{U}= \begin{bmatrix} 2&2&0\\ 0&2&1\\ 0&0&1 \end{bmatrix} \]
Therefore
\[ \det(\mat{A})=(-1)^2(2)(2)(1)=4 \]
For square matrices of the same size, \(\det(\mat{B}\mat{C})=\det(\mat{B})\det(\mat{C})\)
Apply this product rule to \(\mat{A}=\mat{P}\mat{L}\mat{U}\):
\[ \det(\mat{A}) =\det(\mat{P})\det(\mat{L})\det(\mat{U}) =(-1)^s\prod_{j=1}^n u_{jj} \]
Read column \(j\): its \(1\) in row \(i\) means \(\mat{P}\vct{e}_j=\vct{e}_i\)
\[ \mat{P}=\begin{bmatrix}0&0&1\\1&0&0\\0&1&0\end{bmatrix} \qquad 1\longmapsto2\longmapsto3\longmapsto1 \]
For \(c\) disjoint cycles on \(n\) indices: \(\det(\mat{P})=(-1)^{n-c}\)
No need to carry out row swaps back to \(\mat{I}\)
\[ \mat{B}= \begin{bmatrix} 2&1&0\\ 4&3&1\\ 2&2&3 \end{bmatrix}, \]
\(\exstar\) Compute \(\det(\mat{B})\) using row replacements and swaps, but no pivot-row scaling Which operations change the determinant? Does your result show that \(\mat{B}\) is invertible?
Compute each determinant by elimination without pivot-row scaling
\[ \mat{C}= \begin{bmatrix} 0&1&2\\ 1&1&0\\ 2&3&1 \end{bmatrix} \]
\(\exstar\) Compute \(\det(\mat{C})\) and decide whether \(\mat{C}\) is invertible
\[ \mat{D}= \begin{bmatrix} 1&2&1\\ 2&5&3\\ 3&7&4 \end{bmatrix} \]
\(\exstar\) Compute \(\det(\mat{D})\) and decide whether \(\mat{D}\) is invertible
The value connects computation, geometry, and matrix algebra
For a square matrix \(\mat{A}\in\reals^{n\times n}\),
\[\begin{align*} \det(\mat{A})\ne0 &\iff \mat{A}\text{ has }\,n\text{ pivots}\\ &\iff \mat{A}^{-1}\text{ exists}\\ &\iff \text{for every }\vct{b}\in\reals^n, \ \mat{A}\vct{x}=\vct{b}\text{ has exactly one solution} \end{align*}\]
Likewise,
\[ \det(\mat{A})=0 \iff \mat{A}\vct{z}=\vct{0} \text{ for some }\vct{z}\ne\vct{0} \]
That nonzero direction collapses to \(\vct{0}\)
The determinant summarizes a conclusion that elimination already reveals
First \(\mat{A}\) acts, then \(\mat{B}\) acts:
\[ \vct{x}\longmapsto\mat{A}\vct{x} \longmapsto\mat{B}\mat{A}\vct{x} \]
Their signed scale factors multiply:
\[ \det(\mat{B}\mat{A}) =\det(\mat{B})\det(\mat{A}) \]
Consequently, for \(k=1,2,\ldots\),
\[ \det(\mat{I})=1, \qquad \det(\mat{A}^k)=\det(\mat{A})^k \]
If \(\mat{A}\) is invertible, then \(\det(\mat{A}^{-1})=1/\det(\mat{A})\)
An eigenvector is a nonzero vector that \(\mat{A}\) maps to a scalar multiple of itself:
\[ \mat{A}\vct{v}=\lambda\vct{v}, \qquad \vct{v}\ne\vct{0} \]
The scale factor \(\lambda\) is the corresponding eigenvalue
Eigenvalues will develop this idea
Counting repeated eigenvalues and allowing complex values,
\[ \det(\mat{A})=\lambda_1\lambda_2\cdots\lambda_n \]
Use a determinant when the number itself answers the question:
Use elimination or PLU—not determinant formulas—to solve \(\mat{A}\vct{x}=\vct{b}\)
In floating-point computation, a small determinant depends strongly on scale
By itself, it is not a reliable test for a nearly singular matrix
The Determinants companion notebook explores these computations and geometric effects
A determinant summarizes structure
It does not replace the factorization that computed it
The entries may vary while the determinant keeps its familiar meaning
At each fixed \(t\), this is an ordinary numerical matrix:
\[ \mat{A}(t)= \begin{bmatrix} 1&t\\ t&1 \end{bmatrix}, \qquad \det\bigl(\mat{A}(t)\bigr)=1-t^2 \]
Therefore
\[\begin{align*} t\ne\pm1 &\implies \mat{A}(t)\text{ is invertible},\\ t=\pm1 &\implies \mat{A}(t)\text{ collapses a direction} \end{align*}\]
The determinant identifies the exceptional parameter values
For two differentiable functions,
\[ W(f_1,f_2)(t) =\det\begin{bmatrix}f_1(t)&f_2(t)\\f_1'(t)&f_2'(t)\end{bmatrix} \]
For \(f_1(t)=\cos t\) and \(f_2(t)=\sin t\),
\[ \det\begin{bmatrix}\cos t&\sin t\\-\sin t&\cos t\end{bmatrix}=1 \]
If \(c_1\cos t+c_2\sin t=0\) for every \(t\), differentiating gives a homogeneous system with this invertible coefficient matrix
Hence \(c_1=c_2=0\) is the only constant-coefficient relation
Bases and Coordinates will define this property as linear independence
Let \(\vct{x}=\vct{F}(\vct{u})\in\reals^n\), where \(\vct{u}\in\reals^n\) and \(\vct{F}\) may be nonlinear
Where its partial derivatives exist, the Jacobian has those vectors as columns:
\[ \mat{J}_{\vct{F}}(\vct{u}) =\begin{bmatrix} \dfrac{\partial F_1}{\partial u_1}&\cdots&\dfrac{\partial F_1}{\partial u_n}\\ \vdots&\ddots&\vdots\\ \dfrac{\partial F_n}{\partial u_1}&\cdots&\dfrac{\partial F_n}{\partial u_n} \end{bmatrix} \]
If \(\vct{F}\) is differentiable at \(\vct{u}_0\), then for a small displacement \(\vct{h}\) its Jacobian gives the first-order linear approximation
\[ \vct{F}(\vct{u}_0+\vct{h}) \approx \vct{F}(\vct{u}_0)+\mat{J}_{\vct{F}}(\vct{u}_0)\vct{h} \]
\(\abs{\det\bigl(\mat{J}_{\vct{F}}(\vct{u}_0)\bigr)}\) gives local volume scale; the sign records orientation, and zero signals first-order collapse
\[ \vct{F}(r,\theta)= \begin{bmatrix}r\cos\theta\\r\sin\theta\end{bmatrix} \]
\[ \mat{J}_{\vct{F}}(r,\theta)= \begin{bmatrix} \cos\theta&-r\sin\theta\\ \sin\theta&r\cos\theta \end{bmatrix}, \qquad \det\bigl(\mat{J}_{\vct{F}}\bigr)=r \]
For \(r>0\), a small coordinate rectangle has image area
\[ \text{area}\approx r\,\Delta r\,\Delta\theta \]
The Jacobian columns form the local edge vectors
The Jacobian determinant converts volume elements under a general coordinate change \(\vct{x}=\vct{F}(\vct{u})\):
\[\begin{align*} \dif x_1\cdots\dif x_n &=\abs{\det\bigl(\mat{J}_{\vct{F}}(\vct{u})\bigr)}\, \dif u_1\cdots\dif u_n, &\qquad \text{polar:}\quad \dif x\,\dif y &=r\,\dif r\,\dif\theta \end{align*}\]
Let \(\rho>0\), \(0<\phi<\pi\) be the angle from the positive \(z\)-axis, and \(\theta\) the azimuthal angle in the \(xy\)-plane
\[ \vct{F}(\rho,\phi,\theta)= \begin{bmatrix} \rho\sin\phi\cos\theta\\ \rho\sin\phi\sin\theta\\ \rho\cos\phi \end{bmatrix} \]
The Jacobian columns give the local directions of increasing \(\rho\), \(\phi\), and \(\theta\):
\[ \mat{J}_{\vct{F}}= \begin{bmatrix} \sin\phi\cos\theta&\rho\cos\phi\cos\theta&-\rho\sin\phi\sin\theta\\ \sin\phi\sin\theta&\rho\cos\phi\sin\theta&\rho\sin\phi\cos\theta\\ \cos\phi&-\rho\sin\phi&0 \end{bmatrix} \]
These columns are perpendicular, with lengths \(1\), \(\rho\), and \(\rho\sin\phi\)
\(\det\bigl(\mat{J}_{\vct{F}}\bigr)=\rho^2\sin\phi\), so \(\dif x\,\dif y\,\dif z=\rho^2\sin\phi\,\dif\rho\,\dif\phi\,\dif\theta\)
If we are clever about the row operations, the Jacobian becomes diagonal without dividing by any trigonometric functions
Write \(a=\sin\phi\), \(b=\cos\phi\), \(c=\cos\theta\), and \(s=\sin\theta\)
First rotate rows 1 and 2 by left multiplication:
\[ \mat{Q}_\theta\mat{J}_{\vct{F}} =\begin{bmatrix} c&s&0\\-s&c&0\\0&0&1 \end{bmatrix} \begin{bmatrix} ac&\rho bc&-\rho as\\ as&\rho bs&\rho ac\\ b&-\rho a&0 \end{bmatrix} = \begin{bmatrix} a&\rho b&0\\0&0&\rho a\\b&-\rho a&0 \end{bmatrix} \]
Then mix rows 1 and 3 and swap the last two rows:
\[ \mat{M}_\phi\mat{Q}_\theta\mat{J}_{\vct{F}} =\begin{bmatrix} a&0&b\\b&0&-a\\0&1&0 \end{bmatrix} \begin{bmatrix} a&\rho b&0\\0&0&\rho a\\b&-\rho a&0 \end{bmatrix} =\begin{bmatrix} 1&0&0\\0&\rho&0\\0&0&\rho a \end{bmatrix} \]
\(\det(\mat{Q}_\theta)=c^2+s^2=1\) and \(\det(\mat{M}_\phi)=a^2+b^2=1\)
\(\det\bigl(\mat{J}_{\vct{F}}\bigr)=1\cdot\rho\cdot\rho a=\rho^2\sin\phi\)
| Matrix of derivatives | What its columns represent | What the determinant tells us |
|---|---|---|
| Wronskian | functions and successive derivatives | rules out a constant-coefficient relation |
| Jacobian | responses to coordinate displacements | local scale, orientation, or collapse |
Both use derivatives, but their columns have different meanings
A map \(\vct{G}:D\subseteq\reals^2\to\reals^3\) assigns each pair \((u,v)\in D\) the point \[\vct{G}(u,v)=(x(u,v),y(u,v),z(u,v))\]
If its two tangent directions are independent, these points form a surface patch locally
For example, \(\vct{G}(u,v)=(u,v,u^2+v^2)\) traces the graph \(z=x^2+y^2\) as \((u,v)\) varies over \(\reals^2\)
The Jacobian columns \(\vct{G}_u\) and \(\vct{G}_v\) give tangent directions They approximate the edges of the image of a small parameter rectangle, whose area is approximately
\[ \sqrt{\det\bigl(\mat{J}_{\vct{G}}^{\mathsf T}\mat{J}_{\vct{G}}\bigr)} \,\Delta u\,\Delta v \]
Here \(\mat{J}_{\vct{G}}\) is \(3\times2\); the \(2\times2\) matrix \(\mat{J}_{\vct{G}}^{\mathsf T}\mat{J}_{\vct{G}}\) has determinant equal to the squared area scale
Solvability, computation, and geometry meet in one scalar
For \(n\times n\) \(\mat{A}\): \(\det(\mat{A})\ne0\) \(\iff\) \(n\) pivots \(\iff\) \(\mat{A}^{-1}\) exists \(\iff\) one solution for every \(\vct{b}\)
Determinants measure how a matrix transformation changes volume and orientation
Euclidean Spaces will make other geometric ideas systematic:
© 2026 Fred J. Hickernell · Illinois Tech · assisted by ChatGPT and Codex · Determinants · MATH 332 — Fall 2026 Website · \(\exstar\) = exercise