(Sec:DetExtras)=
# Miscellaneous applications of determinants
## Introduction
In this section we will address the following matters:
-
The determinant as a uniform scale factor for an arbitrary linear transformation from $\R^n$ to $\R^n$.
-
Cramer's rule. Seemingly the ultimate solution to almost all systems of $n$ linear equations in $n$ unknowns.
-
The generalisation of the formula
$$
\left(\begin{array}{cc} a & b \\ c & d\end{array} \right)^{-1}
= \dfrac{1}{ad-bc} \left(\begin{array}{cc} d & -b \\ -c & a\end{array} \right)
$$
to $n\times n$-matrices.
-
A certain generalisation of the cross product to $n$ dimensions.
(SubSec:DetExtras:Volume+Orientation)=
## Volume and orientation revisited
We have seen in {numref}`Sec:DetGeometric` how determinants arise in the context of areas of parallelograms and volumes of parallelepipeds.
In {numref}`Sec:DotProduct` we used the dot product to define length, distance and orthogonality in $\R^n$. Determinants permit to define the concepts of orientation and volume (at least, a beginning) in $n$ dimensions.
::::::{prf:definition} Volume in $\R^n$
:label: Dfn:DetExtras:VolumeRn
Let $\{\vect{v}_1, \ldots, \vect{v}_n\}$ be a set of $n$ vectors in $\R^n$.
The $n$-dimensional parallelepiped $\mathcal{P}$ spanned by $\vect{v}_1, \ldots, \vect{v}_n$ is the set
$$
\mathcal{P} = \mathcal{P}(\vect{v}_1, \ldots, \vect{v}_n) = \{c_1\vect{v}_1+c_2\vect{v}_2 + \cdots + c_1\vect{v}_n \,|\, c_i \in \R, 0 \leq c_i \leq 1\}.
$$
See {numref}`Figure %s ` for an illustration of such a set in $\R^2$.
The **volume** of such a parallelepiped is defined by
$$
\operatorname{Vol}(\mathcal{P}) = |\det{(\,\vect{v}_1\, \,\cdots \,\, \vect{v}_n\,)}|.
$$
So, it is the absolute value of a determinant.
::::::
Note that if the vectors $\vect{v}_1, \ldots, \vect{v}_n$ in {prf:ref}`Dfn:DetExtras:VolumeRn` are linearly dependent the volume automatically becomes $0$.
::::::{figure} Images/Fig-DetExtras-ParPed.svg
:name: Fig:DetExtras:Parped
:class: dark-light
The parallelepiped generated by two vectors is a parallelogram.
::::::
::::::{prf:remark}
:label: Rem:DetExtras:IdeaOfVolume
The intuition behind {prf:ref}`Dfn:DetExtras:VolumeRn` is the following.
To start with we can postulate that the $n$-dimensional unit cube, generated by the unit vectors $\{\vect{e}_1, \ldots, \vect{e}_n\}$, has volume $1$.
$$
\operatorname{Vol}\left({\mathcal{P}}(\vect{e}_1, \ldots, \vect{e}_n)\right) =
\begin{vmatrix}
1 & 0 & 0 & \cdots & 0 \\
0 & 1 & 0 & \cdots & 0 \\
0 & 0 & 1 & \cdots & 0 \\
\vdots & & \vdots & \ddots & \vdots \\
0 & 0 & 0 & \cdots & 1
\end{vmatrix} = 1.
$$
Next, if we scale one of the vectors with a factor $c$, we would want the volume to get a factor $|c|$,
$$
\operatorname{Vol}\left({\mathcal{P}}(\vect{v}_1, \ldots, {c} \vect{v}_k, \ldots , \vect{v}_n)\right) =
{|c|}\,\operatorname{Vol}\left({\mathcal{P}}(\vect{v}_1, \ldots, \vect{v}_k, \ldots , \vect{v}_n)\right).
$$
Third, if to one vector $\vect{v}_i$ a linear combination of the other vectors is added, the volume should not change. The underlying idea is borrowed from considerations in $\R^3$.
$$
\operatorname{Vol}\left({\mathcal{P}}(\vect{v}_1, \vect{v}_2, \vect{v}_3)\right) =
\operatorname{Vol}\left({\mathcal{P}}(\vect{v}_1, \vect{v}_2, \vect{v}_3 + a\vect{v}_1+b\vect{v}_2)\right),
$$
since the operation leaves the 'ground region' $ \mathcal{P}(\vect{v}_1, \vect{v}_2)$ and the 'height' invariant.
In fact, these two operations exactly match the first two rules of {prf:ref}`Prop:DetRowReduction:RowOps`, when applied to the columns of a determinant. The third rule of that proposition, swapping two columns in a determinant, leads to a sign change of the determinant, but should have no influence on the value of the volume, which we prefer to have non-negative.
::::::
The content of the next proposition is that (the absolute value of) the determinant of a matrix $A$ acts as a uniform *scaling factor* of the linear transformation that corresponds to $A$. We first consider the case of a $2\times 2$-matrix.
::::::{prf:proposition}
:label: Prop:DetExtras:DetAsScaleFactorR2
Suppose $T$ is a linear transformation from $\R^2$ to $\R^2$, with standard matrix $A = (\,\vect{a}_1 \,\, \vect{a}_2\,)$. So we have
$$
T(\vect{x}) = A\vect{x}, \quad \text{for} \,\, \vect{x} \,\text{ in }\, \R^2.
$$
Let $R$ be any region in $\R^2$ for which the area is well-defined, and let $S$ be the image of $R$ under $T$.
Then for the area of $S$ it holds that
$$
\operatorname{area}(S) = |\det{A}|\cdot \operatorname{area}(R).
$$
::::::
::::::{admonition} Proof of {prf:ref}`Prop:DetExtras:DetAsScaleFactorR2`
:class: tudproof
If the matrix $A$ is not invertible, the range of $T$, which is given by $\operatorname{Span}\{\vect{a}_1, \vect{a}_2\}$, is contained in a line.
Each region $R$ is then mapped onto a subset $S$ that is contained in this line, so
$$
\operatorname{area}(S) = 0 = 0\cdot \operatorname{area}(R) =|\det{A}|\cdot \operatorname{area}(R).
$$
Next suppose that $A$ is invertible. Then the unit grid is mapped onto a grid with as unit region the parallelogram with sides $\vect{a}_1 = T(\vect{e}_1) = A\vect{e}_1$ and $\vect{a}_2 = T(\vect{e}_2) = A\vect{e}_2$. See {numref}`Figure %s `.
::::{figure} Images/Fig-DetExtras-StandardGrid.svg
:name: Fig:DetExtras:Grid
:class: dark-light
The image of the standard grid.
::::
First we show that the formula holds if $R$ is the unit square, i.e., the parallelogram generated by $\vect{e}_1$ and $\vect{e}_2$. The unit square is mapped onto the parallelogram $S$ generated by $T(\vect{e}_1)=\vect{a}_1$ and $T(\vect{e}_2)=\vect{a}_2$. It follows
that
$$
\operatorname{area}(S) = |\det{(\,\vect{a}_1\,\, \vect{a}_2\,)}| = |\det{A}|,
$$
and since the area of $R$ is equal to $1$, we have
$$
|\det{A}| = |\det{A}| \cdot 1 = |\det{A}| \cdot \operatorname{area}(R).
$$
This then also holds for any square $R$ with sides of length $r$ that are parallel to the axes. Namely, such a square has area $r^2$ and can be described as the square with vertices
$$
\vect{p},\quad \vect{p}+ r\vect{e}_1, \quad \vect{p}+ r\vect{e}_1+r\vect{e}_2 \quad\text{and}\quad \vect{p}+ r\vect{e}_2.
$$
These are mapped to
$$
A\vect{p},\quad A\vect{p}+ rA\vect{e}_1, \quad A\vect{p}+ rA\vect{e}_1+rA\vect{e}_2 \quad\text{and}\quad A\vect{p}+ rA\vect{e}_2.
$$
This is a parallelogram with sides $rA\vect{e}_1 = r\vect{a}_1$ and $rA\vect{e}_2 =r \vect{a}_2$, which has area
$$
\operatorname{area}(S) = |\det{(\,r\vect{a}_1\,\, r\vect{a}_2\,)}| = r^2 |\det{A}| = |\det{A}|\cdot \operatorname{area}(R).
$$
See {numref}`Figure %s `
::::{figure} Images/Fig-DetExtras-ImageOfSquare.svg
:name: Fig:DetExtras:ImageOfSquare
:class: dark-light
The image of a square with 'corner' $\vect{p}$ and sides of length $r$.
::::
For a general (reasonable) region $R$ we sketch the idea and omit the technical details.
The region $R$ can be approximated arbitrarily close by a collection of smaller and smaller squares $R_i$ of which the interiors do not overlap.
See {numref}`Figure %s `.
The limit of the areas of these approximations when the grids get finer and finer gives the area of $R$.
::::{figure} Images/Fig-DetExtras-Subdivision.svg
:name: Fig:DetExtras:Subdivision
:class: dark-light
Approximating a region by smaller and smaller squares.
::::
The formula holds for each of the $R_i$. Since $T$ is one-to-one, the images $S_i = T(R_i)$ will not overlap either, and the images taken together will approximate the image $S = T(R)$ as well. We deduce that
$$
\begin{array}{rl}
\operatorname{area}(S) \approx \sum \operatorname{area}(S_i) \!\!\!&=\sum |\det{A}|\cdot \operatorname{area}(R_i) \\
&= |\det{A}| \sum \operatorname{area}(R_i) \approx |\det{A}|\cdot \operatorname{area}(R).
\end{array}
$$
By taking an appropriate limit one can show that in fact
$$
\operatorname{area}(S) = |\det{A}|\cdot \operatorname{area}(R).
$$
::::::
{prf:ref}`Prop:DetExtras:ScaleFactorR2` can be generalised to higher dimensions. For $n = 3$ area becomes volume.
For higher dimensions the starting point for volume is {prf:ref}`Dfn:DetExtras:VolumeRn` for the volume of parallelepids. To extend the definition to volumes of more general regions is by no means a trivial matter, and we will not consider it here.
In $\R^n$ we can at least generalise {prf:ref}`Prop:DetExtras:DetAsScaleFactorR2` to parallelepipeds.
::::::{prf:proposition}
:label: Prop:DetExtras:ScaleFactorRn
Suppose $T$ is a linear transformation from $\R^n$ to $\R^n$, with standard matrix $A$.
Then for any parallelepiped $\mathcal{P}$ generated by $\{\vect{v}_1, \ldots, \vect{v}_n\}$ , it holds that
$$
\operatorname{Vol}(S) = |\det{A}|\cdot \operatorname{Vol}(\mathcal{P}),
$$
where $S$ is the image of $\mathcal{P}$ under $T$.
::::::
::::::{admonition} Proof of {prf:ref}`Prop:DetExtras:ScaleFactorRn`
:class: tudproof
If $R$ is the $n$-dimensional parallelepiped $\mathcal{P}$ generated by $\{\vect{v}_1, \ldots, \vect{v}_n\}$ we have that $T(\mathcal{P})$ is generated by $\{T(\vect{v}_1), \ldots, T(\vect{v}_n)\}$.
Then
$$
\begin{array}{rcl}
\operatorname{Vol}(T(\mathcal{P})) &=& |\det{(\,T(\vect{v}_1)\, \,\cdots \,\, T(\vect{v}_n)\,)}| \\
&=& |\det{(\,A(\vect{v}_1)\, \,\cdots \,\, A(\vect{v}_n)\,)}| \\
&=& |\det{\left(A (\,\vect{v}_1\, \,\cdots \,\, \vect{v}_n\,)\right)}|\\
&=& |\det{A} \det{(\,\vect{v}_1\, \,\cdots \,\, \vect{v}_n\,)}|\\
&=& |\det{A}| \,\operatorname{Vol}(\mathcal{P}).
\end{array}
$$
%%For a more general region $R$ we would again have to work with approximations/subdivisions like in the proof %of {prf:ref}`Prop:DetExtras:ScaleFactorR2`. Then we would first have to extend the definition of %$n$-dimensional volume. We will not pursue that track.
::::::
To conclude our interpretation of the determinant of $A$ regarding the linear transformation $T(\vect{x}) = A\vect{x}$ we look at the _orientation_.
::::::{prf:definition} Orientation in $\R^n$
:label: Dfn:DetExtras:OrientationRn
Suppose the vectors $(\vect{v}_1, \ldots, \vect{v}_n)$ in $\R^n$ are linearly independent.
Then we say that the ordered set $(\vect{v}_1, \ldots, \vect{v}_n)$ is **positively orientated** if $ \det{(\vect{v}_1 \cdots \vect{v}_n)}>0$.
If this determinant is negative the set is called **negatively orientated**.
For vectors that are linearly dependent we do not define the orientation.
::::::
::::{prf:proposition}
:label: Prop:DetExtras:ScaleFactorR2
Suppose $A = (\,\vect{a}_1\,\,\vect{a}_2\,\,\cdots\,\,\vect{a}_n\, )$ is the standard matrix of the linear transformation $T: \R^n \to \R^n$. So we have
$$
T(\vect{x}) = A\vect{x}.
$$
Suppose $(\vect{v}_1,\,\vect{v}_2,\,\ldots\,,\,\vect{v}_n)$ is an ordered set of vectors in $\R^n$.
Then the following holds.
If $\det{A} > 0$, the (ordered) set $\big(T(\vect{v}_1),\,T(\vect{v}_2),\,\ldots\,,\,T(\vect{v}_n)\big)$ has the same orientation as the set $(\vect{v}_1,\,\vect{v}_2,\,\ldots\,,\,\vect{v}_n)$.
If $\det{A} < 0$ the set $\big(T(\vect{v}_1),\,T(\vect{v}_2),\,\ldots\,,\,T(\vect{v}_n)\big)$ has the opposite orientation as the set $(\vect{v}_1,\,\vect{v}_2,\,\ldots\,,\,\vect{v}_n)$.
In short: the transformation $T(\vect{x}) = A\vect{x}$ **preserves** the orientation if $\det{A} > 0$ and **reverses** the orientation if $\det{A} < 0$.
::::
If the determinant is $0$, then the set $\{T(\vect{v}_1), \ldots,T(\vect{v}_n) \}$ will be linearly dependent, and for such a set the orientation is not defined.
::::{admonition} Proof of {prf:ref}`Prop:DetExtras:ScaleFactorR2`
:class: tudproof
This too follows immediately from the product rule of determinants.
$$
\begin{array}{rcl}
\det{\left(\,T(\vect{v}_1)\,\,T(\vect{v}_2)\,\,\cdots\,\,T(\vect{v}_n)\, \right)} &=&
\det{\left(\,A\vect{v}_1\,\,A\vect{v}_2\,\,\cdots\,\,A\vect{v}_n\, \right)} \\
&=&
\det{\big(A\left(\,\vect{v}_1\,\,\vect{v}_2\,\,\cdots\,\,\vect{v}_n\, \right)\big)} \\
&=&
\det{A}\cdot\det{\left(\,\vect{v}_1\,\,\vect{v}_2\,\,\cdots\,\,\vect{v}_n\, \right)}.
\end{array}
$$
::::
A nice illustration of what this means in $\R^2$ is given by the following example.
::::{prf:example}
:label: Ex:DetExtras:Orientation
Consider the two linear transformations from $\R^2$ to $\R^2$ with matrices
$$
A = \begin{pmatrix} 1 & 3 \\ 3 & 1 \end{pmatrix}, \quad
B = \begin{pmatrix} 3 & 1 \\ 1 & 3 \end{pmatrix}.
$$
Note that
$$
\det{A} = -8 < 0 \quad \text{and} \quad \det{B} = 8 > 0.
$$
{numref}`Figure %s ` visualises what is going on.
```{applet}
:url: detextras/orientation
:fig: Images/Fig-DetExtras-Orientation.svg
:name: Fig:DetExtras:Orientation
:class: dark-light
Images under transformations with negative and positive determinant.
```
The images of a unit vector that rotates anticlockwise under transformation $A$ move around clockwise, i.e., in the _opposite_ orientation/direction. Under transformation $B$ the images will go around the origin anticlockwise, i.e., in the _same_ direction as the original vectors.
::::
(Subsec:DetExtra:Cramer)=
## Cramer's rule
We first introduce a new notation that will help to simplify formulas later.
::::::{prf:definition}
:label: Dfn:DetExtras:ReplaceColumn
Let $A$ be an $n\times n$-matrix, and $\vect{v}$ a vector in $\R^n$. Then $A^{(i)}(\vect{v})$ denotes the matrix that results when the $i$-th column of $A$ is replaced by the vector $\vect{v}$.
::::::
::::::{prf:example}
:label: Ex:DetExtras:ReplaceColumn
For the matrix $A = \begin{pmatrix} 1 & 3 & 1 \\ 1 & 4 & 2 \\ 3 & 1 & 5 \end{pmatrix}$
and the vector $\vect{v} = \begin{pmatrix} \class{blue}6 \\ \class{blue}7 \\ \class{blue}8 \end{pmatrix}$ we have that
$$
A^{(2)}(\vect{v}) = \begin{pmatrix} 1 & \class{blue}6 & 1 \\ 1 & \class{blue}7 & 2 \\ 3 & \class{blue}8 & 5 \end{pmatrix}.
$$
::::::
Suppose that $A$ is an invertible $n \times n$-matrix. Then we know that the linear system $A\vect{x} = \vect{b}$ has a unique solution for each $\vect{b}$ in $\R^n$. And we also know that the determinant of $A$ is not equal to zero.
The next proposition gives a ready-made formula for the solution.
::::::{prf:theorem} Cramer's Rule
:label: Thm:DetExtras:Cramer
Suppose $A$ is an invertible $n \times n$-matrix, and $\vect{b}$ a vector in $\R^n$.
The entries of $x_i$ of the unique solution $\vect{x}$ of the linear system
$$
A\vect{x} = \vect{b}
$$
are given by
:::::{math}
:label: Eq:DetExtras:Cramer
x_i = \dfrac{\det{\left(A^{(i)}(\vect{b})\right)}}{\det{A}}.
:::::
::::::
::::::{prf:example}
:label: Ex:DetExtras:Cramer
We use Cramer's rule to solve the system
$$
\left\lbrace
\begin{array}{rcc}
x_1 + 2x_2 + x_3 & = & 3 \\
x_1 - x_2 + 2x_3 & = & 4 \\
3x_1 + x_2 -5x_3 & = & 1
\end{array}
\right.
\quad\quad \text{i.e.,} \quad
\begin{pmatrix}
1 & 2 & 1 \\ 1 & -1 & 2 \\ 3 & 1 & -5
\end{pmatrix}
\left(\begin{array}{c} x_1 \\ x_2 \\ x_3 \end{array} \right)
=
\begin{pmatrix}3 \\ 4 \\ 1 \end{pmatrix}.
$$
First of all, the determinant of $A$ can be computed as follows (in the first step we use column reduction, with the boxed $1$ as a pivot):
$$
\left|\begin{array}{ccc} \fbox{$1$} & 2 & 1 \\ 1 & -1 & 2 \\ 3 & 1 & -5 \end{array} \right|=
\left|\begin{array}{ccc} 1 & 0 & 0 \\ 1 & -3 & 1 \\ 3 & -5 & -8 \end{array} \right|=
\left|\begin{array}{cc} -3 & 1 \\ -5 & -8 \end{array} \right|= 29 \neq 0,
$$
so the coefficient matrix is invertible and consequently the system has a unique solution.
According to Cramer's rule we find the first entry of the solution as follows (again we use the boxed $1$ as a pivot):
$$
x_1 = \dfrac{\begin{vmatrix} 3 & 2 & 1 \\ 4 & -1 & 2 \\ 1 & \fbox{$1$} & -5 \end{vmatrix}}{29} =
\dfrac{\begin{vmatrix} 1 & 0 & 11 \\ 5 & 0 & -3 \\ 1 & 1 & -5 \end{vmatrix}}{29} =
\dfrac{-\begin{vmatrix} 1 & 11 \\ 5 & -3 \end{vmatrix}}{29} = \dfrac{58}{29} = 2.
$$
Likewise we can compute the other two entries of the solution.
$$
x_2 = \dfrac{\begin{vmatrix} 1 & 3 & 1 \\ 1 & 4 & 2 \\ 3 & 1 & -5 \end{vmatrix}}{29} = 0
\quad \text{and} \quad
x_3 = \dfrac{\begin{vmatrix} 1 & 2 & 3 \\ 1 & -1 & 4 \\ 3 & 1 & 1 \end{vmatrix}}{29} = 1.
$$
::::::
The following proof of Cramer's rule rests rather nicely on properties of the determinant function.
::::::{admonition} Proof of {prf:ref}`Thm:DetExtras:Cramer`
:class: tudproof, dropdown
Suppose $\vect{x} = \vect{c} = \left(\begin{array}{c} c_1 \\ \vdots\\ c_n\end{array} \right)
$ is the unique solution to the linear system $A\vect{x} = \vect{b}$, with the invertible matrix $A = ( \vect{a}_1 \, \, \vect{a}_2 \, \cdots \,\vect{a}_n )$.
We show that Equation {eq}`Eq:DetExtras:Cramer` holds for $c_1$. The argument can be copied for the other $c_i$.
We first note that
$$
\begin{array}{ccl}
A\vect{c} = \vect{b} &\iff \quad & c_1\vect{a}_1+c_2\vect{a}_2 + \cdots + c_n\vect{a}_n =\vect{b} \\
&\iff \quad & c_1\vect{a}_1+c_2\vect{a}_2 + \cdots + c_n\vect{a}_n - \vect{b} = \vect{0}.
\end{array}
$$
The smart next move is to replace the first column of $A$ by the zero column disguised as
$$
c_1\vect{a}_1+c_2\vect{a}_2 + \cdots + c_n\vect{a}_n - \vect{b}.
$$
So we have
$$
\det{((c_1\vect{a}_1+ \,\cdots\, + c_n\vect{a}_n - \vect{b}) \,\,\,\vect{a}_2\,\, \cdots \,\, \vect{a}_n)}
=\det{(\vect{0} \,\,\vect{a}_2 \,\, \cdots\,\,\vect{a}_n)} = 0.
$$
By the linearity property (in all of the columns) of the determinant ({prf:ref}`Prop:DetRowReduction:SumofCols`) we may deduce
:::{math}
:label: Eq:DetExtras:ProofCramer
c_1\det{(A)} + c_2\det{(A^{(1)}(\vect{a}_2))} + \cdots + c_n\det{(A^{(1)}(\vect{a}_n))} - \det{(A^{(1)}(\vect{b}))} = 0.
:::
Now we note that
$$
\det{A^{(1)}(\vect{a}_i)} = 0, \quad i = 2,3,\ldots, n,
$$
since in the matrix $A^{(1)}(\vect{a}_i)$ the first column and the $i$-th column are identical. Hence all but the first and last determinant in Equation {eq}`Eq:DetExtras:ProofCramer` drop out and we can conclude that indeed
$$
c_1\det{(A)} - \det{(A^{(1)}(\vect{b}))} = 0
\quad \iff \quad c_1 = \dfrac{\det{(A^{(1)}(\vect{b}))}}{\det{(A)}}.
$$
::::::
(Subsec:DetExtra:Inverse)=
::::::{caution}
Cramer's formula seems the solution to all your linear systems. However, it has its drawbacks:
- Cramer's formula can only be used for a linear system with a coefficient matrix that is both **square** and **invertible**.
- For a system with two equations in two unknowns Cramer's rule may come in handy, but for solving larger systems it is **highly inefficient**. For instance, for a system of four equations in four unknowns, to find the solution using Cramer's rule, one needs to compute five $4 \times 4$-determinants. The good-old method using the augmented matrix $(\,A\,|\,\vect{b}\,)$ only asks for one row reduction process.
::::::
## The inverse of a matrix in terms of determinants
As an interesting corollary of Cramer's Theorem we can give a ready-made formula for the inverse of an invertible matrix. The following proposition considers the notation of the previous section for a special case.
::::::{prf:proposition}
:label: Prop:DetExtras:ReplaceColGivesCofactor
Let $A$ be an $n\times n$-matrix, and $\vect{e}_j$ the $j$-th vector of the standard basis of $\R^n$. Then
$$
\det{(A^{(i)}(\vect{e}_j))} = (-1)^{j+i} \det{A_{ji}} = C_{ji},
$$
where $A_{ji}$ is the submatrix and $C_{ji} = (-1)^{j+i} \det{\left(A_{ji}\right)} $ the cofactor as introduced in the definition of the $n \times n$-determinant ({prf:ref}`Dfn:DetCofactors:Determinant`).
::::::
The following example serves as an illustration of what is going on here.
::::::{prf:example}
:label: Ex:DetExtras:ReplaceColGivesCofactor
Let $A = \left(\begin{array}{rrrr}
a_{11} &a_{12} &a_{13} &a_{14} \\
a_{21} &a_{22} &a_{23} &a_{24} \\
a_{31} &a_{32} &a_{33} &a_{34} \\
a_{41} &a_{42} &a_{43} &a_{44}
\end{array} \right)
$ be any $4 \times 4$-matrix.
Then $
A^{(4)}(\vect{e}_2) = \left(\begin{array}{rrrr}
a_{11} &a_{12} &a_{13} &0 \\
a_{21} &a_{22} &a_{23} &1 \\
a_{31} &a_{32} &a_{33} &0 \\
a_{41} &a_{42} &a_{43} &0
\end{array} \right).$
Expanding along the fourth column gives
$$
\det{(A^{(4)}(\vect{e}_2))} = \left|\begin{array}{rrrr}
a_{11} &a_{12} &a_{13} &0 \\
a_{21} &a_{22} &a_{23} &1 \\
a_{31} &a_{32} &a_{33} &0 \\
a_{41} &a_{42} &a_{43} &0
\end{array} \right|= (-1)^{(2+4)} \left|\begin{array}{rrr}
a_{11} &a_{12} &a_{13} \\
a_{31} &a_{32} &a_{33} \\
a_{41} &a_{42} &a_{43}
\end{array} \right|= C_{24}.
$$
::::::
::::::{prf:proposition}
:label: Prop:DetExtras:Inverse
If $A$ is an invertible $n \times n$-matrix then the inverse of $A$ is given by
:::::{math}
:label: Eq:DetExtras:Inverse
A^{-1} = \dfrac{1}{\det{A}} \left(\begin{array}{ccccc}
C_{11} &C_{21} &C_{31} & \cdots &C_{n1} \\
C_{12} &C_{22} &C_{32} & \cdots &C_{n2} \\
C_{13} &C_{23} &C_{33} & \cdots &C_{n3} \\
\vdots & \vdots &\vdots & \ddots & \vdots \\
C_{1n} &C_{2n} &C_{3n} & \cdots &C_{nn} \\
\end{array} \right),
:::::
where again the $C_{ij}$ are the cofactors as defined in {prf:ref}`Dfn:DetCofactors:Determinant`.
::::::
::::::{admonition} Proof of {prf:ref}`Prop:DetExtras:Inverse`
:class: tudproof
The $j$-th column $\vect{b}_j$ of $B = A^{-1}$ is the solution to the linear system $A\vect{x} = \vect{e}_j$.
Cramer's rule then gives that $b_{ij}$, the $i$-th entry of this column, is equal to
$$
b_{ij} = \dfrac{\det{\left(A^{(i)}(\vect{e}_j)\right)}}{\det{A}} = \dfrac{C_{ji}}{\det{A}}.
$$
For the last step we used {prf:ref}`Prop:DetExtras:ReplaceColGivesCofactor`.
::::::
::::::{prf:definition}
:label: Dfn:DetExtras:CofactorMatrix
For an $n \times n$-matrix $A$ the matrix
$$
\left(\begin{array}{ccccc}
C_{11} &C_{12} &C_{13} & \cdots &C_{1n} \\
C_{21} &C_{22} &C_{23} & \cdots &C_{2n} \\
C_{31} &C_{32} &C_{33} & \cdots &C_{3n} \\
\vdots & \vdots &\vdots & \ddots & \vdots \\
C_{n1} &C_{n2} &C_{n3} & \cdots &C_{nn} \\
\end{array} \right)
$$
is called its **cofactor matrix**.
The **adjugate matrix** of $A$ is defined as the transpose of the cofactor matrix. So
$$
\operatorname{Adj}(A) = \left(\begin{array}{ccccc}
C_{11} &C_{21} &C_{31} & \cdots &C_{n1} \\
C_{12} &C_{22} &C_{32} & \cdots &C_{n2} \\
C_{13} &C_{23} &C_{33} & \cdots &C_{n3} \\
\vdots & \vdots &\vdots & \ddots & \vdots \\
C_{1n} &C_{2n} &C_{3n} & \cdots &C_{nn} \\
\end{array} \right)
.
$$
::::::
Thus {prf:ref}`Prop:DetExtras:Inverse` states that
$$
A^{-1} = \dfrac{1}{\det{A}} \operatorname{Adj}(A),
$$
provided that $A$ is invertible.
In fact a slightly more general formula holds for **any** square matrix.
::::::{prf:proposition}
:label: Prop:DetExtras:AdjointProperty
For any square matrix $A$ the following identity holds:
$$
A\,\operatorname{Adj}(A) = \operatorname{Adj}(A)\, A = (\det{A})\,I.
$$
::::::
Note that the first two products are matrix products and the third product is a scalar times a matrix.
The proof we think, is short and instructive.
::::::{admonition} Proof of {prf:ref}`Prop:DetExtras:AdjointProperty`
:class: tudproof, dropdown
For an invertible matrix the statement follows immediately from {prf:ref}`Prop:DetExtras:Inverse`.
However, we can give an 'elementary' proof, that includes the non-invertible case where $\det{A}=0$. We will use two properties of determinants from earlier sections.
First {prf:ref}`Thm:DetCofactors:RowOrColumnExpansion`, that states that the determinant of a matrix can be found by expansion along an arbitrary column
$$
\det{A} = \sum_{i=1}^n (-1)^{i+j} a_{ij}\det{A_{ij}} = \sum_{i=1}^n a_{ij} C_{ij}.
$$
And second {prf:ref}`Cor:DetRowReduction:EqualRows`: the determinant of a matrix with two equal rows (or columns) is equal to 0.
Let us consider the product $\operatorname{Adj}(A) A$ very carefully:
$$
\left(\begin{array}{ccccc}
C_{11} &C_{21} &C_{31} & \cdots &C_{n1} \\
C_{12} &C_{22} &C_{32} & \cdots &C_{n2} \\
C_{13} &C_{23} &C_{33} & \cdots &C_{n3} \\
\vdots & \vdots &\vdots & \ddots & \vdots \\
C_{1n} &C_{2n} &C_{3n} & \cdots &C_{nn} \\
\end{array} \right)
\left(\begin{array}{ccccc}
a_{11} &a_{12} &a_{13} & \cdots &a_{1n} \\
a_{21} &a_{22} &a_{23} & \cdots &a_{2n} \\
a_{31} &a_{32} &a_{33} & \cdots &a_{3n} \\
\vdots & \vdots &\vdots & \ddots & \vdots \\
a_{n1} &a_{n2} &a_{n3} & \cdots &a_{nn} \\
\end{array} \right).
$$
On the diagonal we see that the $j$-th entry is equal to
$$
C_{1j}a_{1j} + C_{2j}a_{2j} + \cdots + C_{nj}a_{nj} = \sum_{i=1}^n a_{ij} C_{ij} = \det{A}.
$$
For the off-diagonal elements we find as product of the $j$-th row of $\operatorname{Adj}(A)$ with the $k$-th column of $A$ the sum
$$
C_{1j}a_{1k} + C_{2j}a_{2k} + \cdots + C_{nj}a_{nk} = \sum_{i=1}^n a_{ik} C_{ij}.
$$
This expression can be interpreted as the expansion along the $k$-th row of the determinant of the matrix $A^{(j)}(\vect{a}_k)$ that results if the $j$-th column of
$A$ is replaced by the $k$-th column of $A$. Since this matrix has two equal columns, its determinant must be zero!
::::::
For $n = 2$ {prf:ref}`Prop:DetExtras:Inverse` gives us back the formula for the inverse
of {prf:ref}`Prop:MatrixInv:Inverse2x2`. That is, if we define the determinant of
a $1 \times 1$-matrix $A = (a)$ as the number $a$.
For an arbitrary invertible $3 \times 3$-matrix
$A=\left(\begin{array}{ccc} a_{11} &a_{12} &a_{13} \\ a_{21} &a_{22} &a_{23} \\ a_{31} &a_{32} &a_{33} \end{array} \right) $ the formula yields
$$
A^{-1} = \dfrac{1}{\begin{vmatrix}
a_{11} &a_{12} &a_{13} \\
a_{21} &a_{22} &a_{23} \\
a_{31} &a_{32} &a_{33}
\end{vmatrix}}
\left(\begin{array}{ccc}
\begin{vmatrix} a_{22} & a_{23} \\ a_{32} & a_{33} \end{vmatrix} &
- \begin{vmatrix} a_{12} & a_{13} \\ a_{32} & a_{33} \end{vmatrix} &
\begin{vmatrix} a_{12} & a_{13} \\ a_{22} & a_{23} \end{vmatrix} \\
- \begin{vmatrix} a_{21} & a_{23} \\ a_{31} & a_{33} \end{vmatrix} &
\begin{vmatrix} a_{11} & a_{13} \\ a_{31} & a_{33} \end{vmatrix} &
- \begin{vmatrix} a_{11} & a_{13} \\ a_{21} & a_{23} \end{vmatrix} \\
\begin{vmatrix} a_{21} & a_{22} \\ a_{31} & a_{32} \end{vmatrix} &
- \begin{vmatrix} a_{11} & a_{12} \\ a_{31} & a_{32} \end{vmatrix} &
\begin{vmatrix} a_{11} & a_{12} \\ a_{21} & a_{22} \end{vmatrix}
\end{array}
\right).
$$
::::::{caution}
Like Cramer's rule, the formula for the inverse is **highly inefficient**.
The comparison between the efforts required to compute the inverse via the adjugate matrix versus row reduction of the augmented matrix $(\,A\,|\,I\,)$ works out rather favorably for the latter. A glimpse of this inefficiency is shown by the above formula for the inverse of a $3 \times 3$-matrix.
::::::
(Subsec-DetExtras-DetAndCrossProd)=
## Determinant and cross product
In {numref}`Sec:CrossProduct` the cross product of two vectors $\mathbf{u}$ and $\mathbf{v}$ in $\R^3$ is defined. It is the unique vector $\mathbf{w}$ that is (1) orthogonal to $\mathbf{u}$ and $\mathbf{v}$, with (2) length equal to the area of the parallelogram with sides $\mathbf{u}$ and $\mathbf{v}$, and (3) such that the triple $(\mathbf{u},\mathbf{v},\mathbf{w})$ is 'righthanded' (= positively orientated).
In {numref}`Sec:DetGeometric`
we defined the determinant of the ordered set $(\vect{a},\vect{b},\vect{c})$ in $\R^3$ via
$$
\begin{array}{rcl}
\det{(\vect{a},\vect{b},\vect{c})} &=& (\vect{a}\times\vect{b})\ip\vect{c} =
\left|\begin{array}{ccc} a_1 & b_1 &c_1 \\ a_2 & b_2 &c_2 \\ a_3 & b_3 & c_3 \end{array}\right|\\
&=&
\left|\begin{array}{cc} a_2 & b_2 \\a_3 & b_3 \end{array}\right| c_1
- \left|\begin{array}{cc} a_1 & b_1 \\ a_3& b_3 \end{array}\right| c_2
+ \left|\begin{array}{cc} a_1 & b_1 \\ a_2 & b_2 \end{array}\right| c_3.
\end{array}
$$
Conversely, we can write the cross product in terms containing determinants.
:::{math}
:label: Eq:DetExtras:DetCrossProd
\begin{array}{rcl}
\left(\begin{array}{c} a_1 \\ a_2 \\ a_3 \end{array}\right) \times
\left(\begin{array}{c}b_1 \\ b_2 \\ b_3 \end{array}\right)
&=& \left(\begin{array}{c}a_2b_3-a_3b_2 \\ a_3b_1 - a_1b_3 \\ a_1b_2-a_2b_1 \end{array}\right) \\
&=&
\left|\begin{array}{cc} a_2 & b_2 \\a_3 & b_3 \end{array}\right|\vect{e}_1
- \left|\begin{array}{cc} a_1 & b_1 \\ a_3 & b_3 \end{array}\right|\vect{e}_2
+ \left|\begin{array}{cc} a_1 & b_1 \\a_2 & b_2 \end{array}\right|\vect{e}_3.
\end{array}
:::
The last expression can formally be written as
$$
\left|\begin{array}{ccc} a_1 & b_1 &\vect{e}_1 \\ a_2 & b_2 &\vect{e}_2 \\ a_3 & b_3 & \vect{e}_3 \end{array}\right|.
$$
In exactly the same fashion, we can, for $n-1$ vectors $\vect{a}_1, \ldots, \vect{a}_{n-1}$ in $\R^n$, say
$$
\vect{a}_1 = \left(\begin{array}{c} a_{11} \\ a_{21} \\ \vdots \\ a_{n1} \end{array}\right), \quad
\vect{a}_2 = \left(\begin{array}{c} a_{12} \\ a_{22} \\ \vdots \\ a_{n2} \end{array}\right), \quad
\ldots \quad , \quad
\vect{a}_{n-1} = \left(\begin{array}{c} a_{1,(n-1)} \\ a_{2,(n-1)} \\ \vdots \\ a_{n,(n-1)} \end{array}\right)
$$
define
:::{math}
:label: Eq:DetExtras:DetCrossProd-ndim
\vect{a}^{\ast}_n = \vect{N}(\vect{a}_1, \ldots, \vect{a}_{n-1}) = \left|\begin{array}{ccccc}
a_{11} & a_{12} & \cdots & a_{1,(n-1)} & \vect{e}_1 \\
a_{21} & a_{22} & \cdots & a_{2,(n-1)} & \vect{e}_2 \\
\vdots & \vdots & & \vdots & \vdots \\
a_{n1} & a_{n2} & \cdots & a_{n,(n-1)} & \vect{e}_n
\end{array}\right|.
:::
Here $\vect{e}_1, \ldots , \vect{e}_n$ denote the vectors of the standard basis for $\R^n$.
With some effort it can be shown that the following properties hold.
::::::{prf:proposition}
:label: Prop:DetExtras:Properties-ndimCrossProd
Suppose that $\vect{a}_1, \ldots, \vect{a}_{n-1}$ are vectors in $\R^n$ and $\vect{a}^{\ast}_n$ is defined as in Equation {eq}`Eq:DetExtras:DetCrossProd-ndim`. Then the following properties hold.
-
$\vect{a}^{\ast}_n \perp \vect{a}_i$, for $i = 1,2,\ldots, n-1$.
-
$ \{\vect{a}_1, \, \ldots, \,\vect{a}_{n-1}\}$ is linearly dependent if and only if $ \vect{a}^{\ast}_n = \vect{0}$.
-
If $ \{\vect{a}_1, \ldots, \vect{a}_{n-1}\}$ is linearly independent, then $\det{\left(\,\vect{a}_1, \ldots, \vect{a}_{n-1}, \vect{a}^{\ast}_n\,\right)
} > 0$.
-
The norm of the vector $\vect{a}^{\ast}_n$ is equal to the $(n-1)$-dimensional volume of the $(n-1)$-dimensional parallelepiped generated by $\vect{a}_1, \ldots, \vect{a}_{n-1}$.
::::::
For an independent set of vectors $\{\vect{a}_1, \ldots, \vect{a}_{n-1}\}$ in $\R^n$, the properties of {prf:ref}`Prop:DetExtras:Properties-ndimCrossProd` uniquely determine $\vect{a}^{\ast}_n$
as the vector $\vect{v}$ that is orthogonal to $ \vect{a}_1, \ldots, \vect{a}_{n-1}$, has a prescribed length, and makes the ordered set
$(\vect{a}_1, \ldots, \vect{a}_{n-1}, \vect{v}) $ positively orientated.
For a linearly dependent set of vectors property iv. implies that $\vect{a}^{\ast}_n = \vect{0}$.
::::::{prf:example}
:label: Ex:DetExtras:CrossProd2D
For $n = 2$ we get, for an arbitrary vector $\vect{v} =
\left(\begin{array}{c} a \\ b \end{array}\right)
\neq \left(\begin{array}{c} 0\\0 \end{array}\right)
$:
$$
\vect{w} = \vect{N}\left(\vect{v}\right)
=
\left|\begin{array}{cc}
a & \vect{e}_1\\
b & \vect{e}_2
\end{array}\right|= a\vect{e}_2 - b\vect{e}_1 = \left(\begin{array}{c} -b \\ a \end{array}\right).
$$
This is indeed a vector orthogonal to $\vect{v}$ with the same 'one-dimensional volume', i.e., length, as the vector $\vect{v}$.
Moreover, $\left(\vect{v}, \vect{w}\right)
= \left(\left(\begin{array}{c} a \\ b \end{array}\right)
, \left(\begin{array}{c} -b \\ a \end{array}\right)
\right)
$
is positively orientated, as can be seen by making a sketch.
This shows that the construction also works in $\R^2$.
::::::
::::::{prf:example}
:label: Ex:DetExtras:CrossProd3D
We will find the vector $\mathbf{v} = \vect{a}^{\ast}_4 = N(\vect{a}_1, \vect{a}_2, \vect{a}_3)$ for the columns of the matrix
$$
A = \left(\begin{array}{ccc}
1 & 1 & 3 \\
1 & -1 & 1 \\
1 & 1 & -3 \\
-1 & 1 & 1
\end{array}\right).
$$
The first entry $v_1$ is computed as
$$
v_1 = (-1)^{1+4}\left|\begin{array}{ccc}
1 & -1 & 1 \\
1 & 1 & -3 \\
-1 & 1 & 1
\end{array}\right| =
-\left|\begin{array}{ccc}
1 & -1 & 1 \\
0 & 2 & -4 \\
0 & 0 & 2
\end{array}\right| = -4.
$$
All in all we find
$$
\vect{v} = \left(\begin{array}{c}
-4 \\ 12 \\ 4 \\ 12
\end{array}\right).
$$
By taking inner products, or by computing $A^T\vect{v}$, it is checked that indeed
$\vect{v} \perp \vect{a}_i$ for each column $\vect{a}_i$. So property i. of {prf:ref}`Prop:DetExtras:Properties-ndimCrossProd` is satisfied.
Since the three columns of $A$ are orthogonal, the 'rectangular box' in $\R^4$ they generate will have 3d-volume
$$
\norm{\vect{a}_1} \cdot \norm{\vect{a}_2} \cdot \norm{\vect{a}_3} =
\sqrt{4}\cdot \sqrt{4}\cdot \sqrt{20} = 8\sqrt{5}.
$$
This is indeed equal to
$$
\norm{\vect{v}} = \sqrt{4^2+12^2+4^2+12^2} = \sqrt{320},
$$
so property iv. is satisfied too.
::::::
We end the chapter with a proof of {prf:ref}`Prop:DetExtras:Properties-ndimCrossProd`.
So, if you are interested, push the button on the right.
::::::{admonition} Proof of {prf:ref}`Prop:DetExtras:Properties-ndimCrossProd`
:class: tudproof, dropdown
The properties follow from the observation that for each vector $\vect{v}$ in $\R^n$
:::::{math}
:label: Eq:DetExtras:Deta1anv
\begin{array}{rcl}
\vect{N}(\vect{a}_1, \ldots, \vect{a}_{n-1})\ip\vect{v}
&=& \left|\begin{array}{ccccc}
a_{11} & a_{12} & \cdots & a_{1,(n-1)} & v_1 \\
a_{21} & a_{22} & \cdots & a_{2,(n-1)} & v_2 \\
\vdots & \vdots & & \vdots & \vdots \\
a_{n1} & a_{n2} & \cdots & a_{n,(n-1)} & v_n
\end{array}\right|\\
&=& \det{(\,\vect{a}_1\,\, \cdots\,\, \vect{a}_{n-1}\,\,\vect{v}\,)}.
\end{array}
:::::
This immediate generalisation of the identity $(\vect{a}\times\vect{b})\ip\vect{c} = \det{(\,\vect{a}\,\,\vect{b}\,\,\vect{c}\,) }$ follows if
we write Equation {eq}`Eq:DetExtras:DetCrossProd-ndim` as in Equation {eq}`Eq:DetExtras:DetCrossProd`.
-
Take any of the vectors $\vect{a}_j$. Then (by Equation {eq}`Eq:DetExtras:Deta1anv`)
$$
\vect{N}(\vect{a}_1, \ldots, \vect{a}_{n-1}) \ip \vect{a}_j= \det{ \left(\,\vect{a}_1\,\, \cdots\,\, \vect{a}_{n-1}\,\,\vect{a}_j\, \right)
} = 0,
$$
since the determinant has two equal columns.
So indeed
$$
\vect{N}(\vect{a}_1, \ldots, \vect{a}_{n-1}) \perp \vect{a}_j,\quad j = 1, \ldots, n-1.
$$
-
First suppose that the columns of the matrix
$$
\begin{pmatrix}
a_{11} & a_{12} & \cdots & a_{1,(n-1)} \\
a_{21} & a_{22} & \cdots & a_{2,(n-1)} \\
\vdots & \vdots & & \vdots \\
a_{n1} & a_{n2} & \cdots & a_{n,(n-1)}
\end{pmatrix}
$$
are linearly dependent. Then for each vector $\vect{v}$ in $\R^n$
$$
\vect{N}(\vect{a}_1, \ldots, \vect{a}_{n-1}) \ip \vect{v} = \det{ \left(\,\vect{a}_1\,\, \cdots\,\, \vect{a}_{n-1}\,\,\vect{v}\, \right)
} = 0.
$$
Namely, the first $n-1$ columns in the determinant are already linearly dependent.
This implies that $\vect{N}(\vect{a}_1, \ldots, \vect{a}_{n-1}) $ must be the zero vector.
To prove the other implication, suppose the vectors $\{ \vect{a}_1, \,\ldots\, \, , \vect{a}_{n-1} \}$ are linearly independent. Then the $n \times (n-1)$-matrix $A = (\,\vect{a}_1 \,\, \cdots \,\, \vect{a}_{n-1} \,) $ has rank $n-1$. The matrix $A$ must have $n-1$ linearly independent rows. Say, if we delete the $k$-th row we have an $(n-1) \times (n-1)$ sub-matrix with independent rows.
Then the coefficient of $\vect{e}_k$ in the expansion of
$ \vect{N} ( \vect{a}_1, \ldots, \vect{a}_{n-1})$,
which by the defining Equation {eq}`Eq:DetExtras:DetCrossProd-ndim` is precisely (plus or minus) the determinant of this submatrix, is non-zero.
-
This is a consequence of the observation (again using Equation {eq}`Eq:DetExtras:Deta1anv`)
$$
\begin{array}{rcl}
\det{\left(\,\vect{a}_1\,\, \cdots\,\, \vect{a}_{n-1}\,\,\vect{N}(\vect{a}_1, \ldots, \vect{a}_{n-1})\, \right)}
&=& \vect{N}(\vect{a}_1, \ldots, \vect{a}_{n-1}) \ip \vect{N}(\vect{a}_1, \ldots, \vect{a}_{n-1})\\
&=& \norm{\vect{N}(\vect{a}_1, \ldots, \vect{a}_{n-1})}^2 \geq 0,
\end{array}
$$
and the already established fact that $\vect{N}(\vect{a}_1, \ldots, \vect{a}_{n-1}) \neq \vect{0}$ if $\{\vect{a}_1,\, \ldots\,,\, \vect{a}_{n-1}\}$ is linearly independent.
-
We sketch the idea, which we borrow from volume versus area considerations in $\R^2$ and $\R^3$.
We defined the volume of the $n$-dimensional parallelepiped $\mathcal{P} \left(\vect{a}_1, \ldots, \vect{a}_{n} \right) $ generated by the $n$ vectors $\vect{a}_1, \ldots, \vect{a}_{n}$ as the absolute value of a determinant:
$$
\operatorname{Vol}_n\!\left(\mathcal{P}(\vect{a}_1, \ldots, \vect{a}_{n}) \right)
= |\det{\left(\,\vect{a}_1\,\, \cdots\,\, \,\vect{a}_{n}\,\right)
}|.
$$
The height times base principle in $\R^n$ must be:
if
$$
a_{n} \perp \mathcal{P}(\vect{a}_1, \ldots, \vect{a}_{n-1})
$$
then
$$
\operatorname{Vol}_n\!\left(\mathcal{P}(\vect{a}_1, \ldots, \vect{a}_{n}) \right)
=
\operatorname{Vol}_{n-1}\! \left(\mathcal{P}(\vect{a}_1, \ldots, \vect{a}_{n-1}) \right)
\cdot \norm{\vect{a}_{n}}.
$$
where $\operatorname{Vol}_{n-1}$ denotes the $(n-1)$-dimensional volume of an $(n-1)$-dimensional subset of $\R^n$.
We apply this principle to the vector $\vect{a}_{n} = \vect{a}^{\ast}_n = \vect{N}(\vect{a}_1, \ldots, \vect{a}_{n-1})$.
We know that $\vect{a}^{\ast}_n$ is orthogonal to all vectors $\vect{a}_1, \ldots, \vect{a}_{n-1}$. So the 'height' of $\mathcal{P}(\vect{a}_1, \ldots, \vect{a}_{n-1}, \vect{a}^{\ast}_n)$ is equal to $\norm{\vect{a}^{\ast}_n}$.
On the one hand we then have that
$$
\operatorname{Vol}_n\!\left(\mathcal{P}(\vect{a}_1, \ldots, \vect{a}_{n-1}, \vect{a}^{\ast}_n) \right)
=
\operatorname{Vol}_{n-1}\!\left(\mathcal{P}(\vect{a}_1, \ldots, \vect{a}_{n-1}) \right)
\cdot \norm{\vect{a}^{\ast}_n}
$$
and on the other hand
$$
\begin{array}{rcl}
\operatorname{Vol}_n\!\left(\mathcal{P}(\vect{a}_1, \ldots, \vect{a}_{n-1}, \vect{a}^{\ast}_n) \right)
&=&
|\det{ \left(\,\vect{a}_1\,\, \cdots\,\, \vect{a}_{n-1}\,\, \vect{a}^{\ast}_n\, \right)
}| \\
&=&
| \vect{a}^{\ast}_n\ip \vect{a}^{\ast}_n| = \norm{\vect{a}^{\ast}_n}^2.
\end{array}.
$$
Equating the two expressions for $\operatorname{Vol}_n\! \left(\mathcal{P} (\vect{a}_1, \ldots, \vect{a}_{n-1}, \vect{a}^{\ast}_n) \right) $
we conclude that indeed
$$
\norm{\vect{a}^{\ast}_n} = \operatorname{Vol}_{n-1}\! \left(\mathcal{P} (\vect{a}_1, \ldots, \vect{a}_{n-1}) \right).
$$
::::::
## Grasple exercises
::::::{grasple}
:iframeclass: dark-light
:url: https://embed.grasple.com/exercises/ddb8daf3-3773-44c9-8df0-fe3084a6e7c4?id=93170
:label: grasple_exercise_5_4_1
:dropdown:
:description: Area of a triangle under a linear transformation.
::::::
::::::{grasple}
:iframeclass: dark-light
:url: https://embed.grasple.com/exercises/c630cded-e9cd-482e-a7fe-68ff75430f38?id=121803
:label: grasple_exercise_5_4_2
:dropdown:
:description: Area of a rectangle under a linear transformation.
::::::
%::::::{grasple}
%:iframeclass: dark-light
%:url: https://embed.grasple.com/exercises/ddb8daf3-3773-44c9-8df0-fe3084a6e7c4?id=93170
%:label: grasple_exercise_5_4_1
%:dropdown:
%:description: To compute the area of a triangle with sides $\vect{u}$ and $\vect{v}$ in the plane.
%::::::
::::::{grasple}
:iframeclass: dark-light
:url: https://embed.grasple.com/exercises/8d7a0672-6283-4bb1-9b43-b41a03067e40?id=93171
:label: grasple_exercise_5_4_3
:dropdown:
:description: To find a point $C$ on a line, such that the area of a triangle $ABC$ has a given value.
::::::
::::::{grasple}
:iframeclass: dark-light
:url: https://embed.grasple.com/exercises/62770bc4-da31-4212-a713-bb2843b0e580?id=93172
:label: grasple_exercise_5_4_4
:dropdown:
:description: Which points lie on the same side of a plane?
::::::
::::::{grasple}
:iframeclass: dark-light
:url: https://embed.grasple.com/exercises/f787e084-9a77-40b4-b755-97890b98cfb6?id=93176
:label: grasple_exercise_5_4_5
:dropdown:
:description: To solve a $3\times3$-system using Cramer's rule.
::::::
::::::{grasple}
:iframeclass: dark-light
:url: https://embed.grasple.com/exercises/3add427a-88a3-4da0-8f0a-2bf8bb8781dd?id=93179
:label: grasple_exercise_5_4_6
:dropdown:
:description: Finding two entries in the inverse of a $4\times4$-matrix (using the adjoint matrix).
::::::
::::::{grasple}
:iframeclass: dark-light
:url: https://embed.grasple.com/exercises/8d4a98f7-50ac-4705-8b34-680b7b8395d9?id=93181
:label: grasple_exercise_5_4_7
:dropdown:
:description: To find a vector orthogonal to $\vect{v}_1,\vect{v}_2,\vect{v}_3$ in $\mathbb{R}^4$, with good orientation.
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::::::{grasple}
:iframeclass: dark-light
:url: https://embed.grasple.com/exercises/bc3df113-95b3-470a-a730-3ad8faab08f5?id=93183
:label: grasple_exercise_5_4_8a
:dropdown:
:description: To compute the normal vector $N(\vect{a}_1,\vect{a}_2,\vect{a}_3)$ as in {numref}`Subsection %s `.
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::::::{grasple}
:iframeclass: dark-light
:url: https://embed.grasple.com/exercises/45048350-326b-4f4a-bbbf-9f7edb947914?id=96120
:label: grasple_exercise_5_4_8
:dropdown:
:description: To compute the area of parallelogram, a triangle and the image of a parallelogram.
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