Determinants solve systems (Cramer's rule), invert matrices ($A^{-1}=\operatorname{adj}A/\det A$) and measure space: $|\det A|$ is the area or volume of the box spanned by the columns, and the factor by which $\mathbf{x}\mapsto A\mathbf{x}$ scales every area.
So far the determinant has been a test: zero or not zero, singular or invertible. In this section it becomes a formula and a measurement. First, the formula: when $\det A\ne0$, every entry of the solution of $A\mathbf{x}=\mathbf{b}$, and every entry of $A^{-1}$, is a ratio of two determinants. These formulas are rarely the fastest way to compute, but they show exactly how the answer depends on the data โ which is what you need when the data contain a parameter, or when you want one unknown and not the others.
Then the measurement. Surveyors find the area of a plot of land from the coordinates of its corners, crystallographers find the volume of a unit cell from three edge vectors, and computer-graphics programs need to know how much a transformation enlarges a shape. All three are determinants: the absolute value of a determinant is an area or a volume, and it is the factor by which a linear map stretches space.
Cramer's Rule
For an $n\times n$ matrix $A$ and a vector $\mathbf{b}$ in $\mathbb{R}^n$, write $A_i(\mathbf{b})$ for the matrix obtained from $A$ by replacing column $i$ with $\mathbf{b}$.
๐ Theorem 3.9 โ Cramer's rule
Let $A$ be an invertible $n\times n$ matrix (that is, $\det A\ne0$) and let $\mathbf{b}$ be any vector in $\mathbb{R}^n$. Then the unique solution $\mathbf{x}$ of $A\mathbf{x}=\mathbf{b}$ has entries
$$x_i=\frac{\det A_i(\mathbf{b})}{\det A},\qquad i=1,2,\dots,n.$$
Why. Let $I_i(\mathbf{x})$ be the identity matrix with column $i$ replaced by $\mathbf{x}$. Multiplying column by column, $A\,I_i(\mathbf{x})$ has columns $A\mathbf{e}_1,\dots,A\mathbf{x},\dots,A\mathbf{e}_n$, which is exactly $A_i(\mathbf{b})$ because $A\mathbf{x}=\mathbf{b}$. Expanding $\det I_i(\mathbf{x})$ along row $i$ gives $x_i$. So $\det A\cdot x_i=\det A_i(\mathbf{b})$ by Theorem 3.7, and we divide by $\det A\ne0$.
โ๏ธ Example 1 โ Cramer's rule for a 2ร2 system
Solve $\ 3x_1-2x_2=6,\ \ -5x_1+4x_2=8\ $ by Cramer's rule.
๐ก Show solutionโถ
det A $A=\begin{bmatrix}3&-2\\-5&4\end{bmatrix}$, $\mathbf{b}=\begin{bmatrix}6\\8\end{bmatrix}$: $\det A=12-10=2\ne0$, so Cramer's rule applies.
Replace $\det A_1(\mathbf{b})=\begin{vmatrix}6&-2\\8&4\end{vmatrix}=24+16=40$ and $\det A_2(\mathbf{b})=\begin{vmatrix}3&6\\-5&8\end{vmatrix}=24+30=54$.
Check The full solution (by elimination, or two more Cramer ratios) is $x_1=-\tfrac{10}{11}$, $x_2=\tfrac{18}{11}$, $x_3=\tfrac{38}{11}$; for example the first equation gives $-\tfrac{10}{11}+\tfrac{76}{11}=\tfrac{66}{11}=6$. โ
โ ๏ธ When to use Cramer's rule โ and when not to
Good uses: $2\times2$ and small $3\times3$ systems; finding one unknown without the others; systems whose coefficients contain a parameter, where the rule shows the answer as a formula (Exercise I.6); and theory, such as seeing that the solution depends continuously on the data.
Bad uses: large systems. Cramer's rule needs $n+1$ determinants of size $n$; even with row reduction for each, that is about $n$ times the work of solving the system once by elimination.
No use at all: when $\det A=0$. Then the system has no solution or infinitely many, and the formula divides by zero โ go back to row reduction to find out which.
There is a picture behind the rule. For a $2\times2$ system write $\mathbf{b}=x_1\mathbf{a}_1+x_2\mathbf{a}_2$. Replacing $\mathbf{a}_1$ by $\mathbf{b}$ changes the parallelogram on $\mathbf{a}_1,\mathbf{a}_2$ into the one on $\mathbf{b},\mathbf{a}_2$; since $\mathbf{b}-x_1\mathbf{a}_1$ is parallel to $\mathbf{a}_2$, this new parallelogram has the same base and height as the one on $x_1\mathbf{a}_1,\mathbf{a}_2$, whose area is $x_1$ times the original. So $x_1$ is a ratio of areas.
Interactive 1 โ Cramer's Rule as a Ratio of Areas
The blue parallelogram is spanned by the columns $\mathbf{a}_1,\mathbf{a}_2$ (signed area $\det A$). The orange one has the chosen column replaced by $\mathbf{b}$ (signed area $\det A_i(\mathbf{b})$). The dashed outline is the parallelogram on $x_i\mathbf{a}_i$ and the other column โ it has the same area as the orange one, because sliding a vertex parallel to a side keeps base and height. Drag $\mathbf{b}$, or press Play to move it, and watch $x_i=\det A_i(\mathbf{b})/\det A$.
$\mathbf{b}$โ
$\det A$ (blue)โ
$\det A_1(\mathbf{b})$, $\det A_2(\mathbf{b})$โ
$x_1$, $x_2$โ
The Adjugate and a Formula for the Inverse
Cramer's rule applied to $A\mathbf{x}=\mathbf{e}_j$ gives column $j$ of $A^{-1}$. Expanding each $\det A_i(\mathbf{e}_j)$ down column $i$ (it has a single $1$, in row $j$) gives the cofactor $C_{ji}$. That leads to a formula for the whole inverse.
๐ Definition 3.4 โ The adjugate (classical adjoint)
Let $A$ be an $n\times n$ matrix with cofactors $C_{ij}$. The cofactor matrix of $A$ is $[C_{ij}]$. Its transpose is the adjugate (also called the classical adjoint) of $A$:
$$\operatorname{adj}A=\begin{bmatrix}C_{11}&C_{21}&\cdots&C_{n1}\\C_{12}&C_{22}&\cdots&C_{n2}\\\vdots&\vdots&&\vdots\\C_{1n}&C_{2n}&\cdots&C_{nn}\end{bmatrix}.$$
Note the transpose: the $(i,j)$ entry of $\operatorname{adj}A$ is $C_{ji}$.
๐ Theorem 3.10 โ The adjugate formula for the inverse
For every $n\times n$ matrix $A$, $\ A\,(\operatorname{adj}A)=(\det A)\,I$. Consequently, if $\det A\ne0$,
$$A^{-1}=\frac{1}{\det A}\operatorname{adj}A .$$
Why. The $(i,i)$ entry of $A\,\operatorname{adj}A$ is $a_{i1}C_{i1}+\cdots+a_{in}C_{in}$, the expansion of $\det A$ along row $i$. For $i\ne j$, the $(i,j)$ entry $a_{i1}C_{j1}+\cdots+a_{in}C_{jn}$ is the expansion along row $j$ of the matrix obtained by copying row $i$ into row $j$ โ a matrix with two equal rows, whose determinant is $0$.
For $A=\begin{bmatrix}a&b\\c&d\end{bmatrix}$ the cofactors are $C_{11}=d$, $C_{12}=-c$, $C_{21}=-b$, $C_{22}=a$, so $\operatorname{adj}A=\begin{bmatrix}d&-b\\-c&a\end{bmatrix}$ and Theorem 3.10 is the familiar $2\times2$ inverse formula from Section 2.2.
โ๏ธ Example 3 โ Inverse by the adjugate
Find $\operatorname{adj}A$ and $A^{-1}$ for $A=\begin{bmatrix}2&1&3\\1&-1&1\\1&4&-2\end{bmatrix}$.
Determinant Along row 1, reusing the cofactors: $\det A=2(-2)+1(3)+3(5)=14$.
Adjugate Transpose the cofactor matrix, then divide by $14$: $$\operatorname{adj}A=\begin{bmatrix}-2&14&4\\3&-7&1\\5&-7&-3\end{bmatrix},\qquad A^{-1}=\frac1{14}\begin{bmatrix}-2&14&4\\3&-7&1\\5&-7&-3\end{bmatrix}.$$
Check Row 1 of $A$ times column 1 of $\operatorname{adj}A$: $2(-2)+1(3)+3(5)=14$; row 1 times column 2: $2(14)+1(-7)+3(-7)=0$. Continuing, $A\,\operatorname{adj}A=14I$, as Theorem 3.10 promises. โ
โ ๏ธ Adjugate pitfalls
1. Forgetting the transpose. $\operatorname{adj}A$ is the transpose of the cofactor matrix; without the transpose you get $(A^{-1})^{T}$ instead of $A^{-1}$ โ the error is invisible for symmetric matrices, which is why it survives.
2. Forgetting the signs. Cofactors, not minors: use the chessboard.
3. Using it for big matrices. The adjugate needs $n^2$ determinants of size $n-1$. For $n\ge4$, GaussโJordan elimination on $[A\,|\,I]$ (Section 2.2) is far quicker. The adjugate's value is the formula: for instance, an integer matrix with $\det A=\pm1$ has an inverse with integer entries.
Area and Volume
๐ Theorem 3.11 โ Area of a parallelogram and a triangle
If $\mathbf{u}$ and $\mathbf{v}$ are vectors in $\mathbb{R}^2$, the parallelogram with sides $\mathbf{u}$ and $\mathbf{v}$ (vertices $\mathbf{0},\mathbf{u},\mathbf{v},\mathbf{u}+\mathbf{v}$) has
$$\text{area}=\left|\det\begin{bmatrix}\mathbf{u}&\mathbf{v}\end{bmatrix}\right| .$$
The triangle with vertices $\mathbf{0},\mathbf{u},\mathbf{v}$ has half this area. For a triangle with vertices $(x_1,y_1)$, $(x_2,y_2)$, $(x_3,y_3)$,
$$\text{area}=\frac12\left|\det\begin{bmatrix}x_1&y_1&1\\x_2&y_2&1\\x_3&y_3&1\end{bmatrix}\right|.$$
Why. If $\mathbf{u},\mathbf{v}$ are parallel, both sides are $0$. Otherwise, a column replacement $\mathbf{v}\to\mathbf{v}+k\mathbf{u}$ shears the parallelogram: it keeps the base $\mathbf{u}$ and the height, so the area is unchanged โ and so is the determinant (Theorem 3.3 for columns). Column interchanges change only the sign. With these moves any such matrix can be brought to a diagonal matrix $\begin{bmatrix}p&0\\0&q\end{bmatrix}$, whose parallelogram is a rectangle of area $|pq|=|\det|$. For the $3\times3$ triangle formula, subtract row 1 from rows 2 and 3 and expand along column 3: you get $\tfrac12|\det|$ of the two edge vectors from $(x_1,y_1)$.
โ๏ธ Example 4 โ Areas from coordinates
(a) Find the area of the parallelogram spanned by $\mathbf{u}=(3,1)$ and $\mathbf{v}=(1,4)$. (b) Find the area of the triangle with vertices $P(1,1)$, $Q(4,2)$, $R(2,5)$.
(b) edges Move $P$ to the origin: $\overrightarrow{PQ}=(3,1)$ and $\overrightarrow{PR}=(1,4)$ โ the vectors from (a). The triangle is half the parallelogram: area $\tfrac{11}{2}=5.5$.
(b) formula $\dfrac12\left|\det\begin{bmatrix}1&1&1\\4&2&1\\2&5&1\end{bmatrix}\right|=\dfrac12\left|1(2-5)-1(4-2)+1(20-4)\right|=\dfrac12|{-3}-2+16|=\dfrac{11}{2}$. โ
Remark If the determinant had come out $0$, the three points would lie on one line (compare Exercise 3.1 IV.3).
๐ Theorem 3.12 โ Volume of a parallelepiped
If $\mathbf{u},\mathbf{v},\mathbf{w}$ are vectors in $\mathbb{R}^3$, the parallelepiped with edges $\mathbf{u},\mathbf{v},\mathbf{w}$ from the origin has
$$\text{volume}=\left|\det\begin{bmatrix}\mathbf{u}&\mathbf{v}&\mathbf{w}\end{bmatrix}\right| .$$
The tetrahedron with vertices $\mathbf{0},\mathbf{u},\mathbf{v},\mathbf{w}$ has volume $\tfrac16\left|\det\begin{bmatrix}\mathbf{u}&\mathbf{v}&\mathbf{w}\end{bmatrix}\right|$. In particular, $\mathbf{u},\mathbf{v},\mathbf{w}$ lie in one plane through the origin exactly when this determinant is $0$.
The reason is the same as in the plane: replacing $\mathbf{w}$ by $\mathbf{w}+k\mathbf{u}$ slides the top face parallel to the base, which keeps both the base area and the height, and leaves the determinant unchanged. Interactive 2 lets you try this.
โ๏ธ Example 5 โ A volume and a coplanarity test
(a) Find the volume of the parallelepiped with edges $\mathbf{u}=(1,2,0)$, $\mathbf{v}=(-1,1,3)$, $\mathbf{w}=(2,0,1)$. (b) Do the vectors $(1,2,3)$, $(4,5,6)$, $(7,8,9)$ lie in a plane through the origin?
๐ก Show solutionโถ
(a) Set up Put the vectors in as columns: $\det\begin{bmatrix}1&-1&2\\2&1&0\\0&3&1\end{bmatrix}$.
(a) Expand Along row 1: $1(1-0)-(-1)(2-0)+2(6-0)=1+2+12=15$. The volume is $15$ cubic units.
(b) $\det\begin{bmatrix}1&4&7\\2&5&8\\3&6&9\end{bmatrix}$: $R_2\to R_2-2R_1$, $R_3\to R_3-3R_1$ give rows $(0,-3,-6)$ and $(0,-6,-12)$, which are proportional. So the determinant is $0$: yes, the three vectors are coplanar (indeed $(7,8,9)=2(4,5,6)-(1,2,3)$).
The Determinant as a Scaling Factor
A $2\times2$ matrix $A$ defines the map $T(\mathbf{x})=A\mathbf{x}$ of the plane. It sends the unit square (area $1$) to the parallelogram spanned by $A\mathbf{e}_1$ and $A\mathbf{e}_2$ โ the columns of $A$ โ whose area is $|\det A|$ by Theorem 3.11. Every small square of side $h$ is sent to a parallelogram of area $|\det A|\,h^2$, and any region can be filled with small squares.
๐ Theorem 3.13 โ Linear maps scale area and volume by |det A|
Let $A$ be a $2\times2$ matrix and $T(\mathbf{x})=A\mathbf{x}$. For every region $S$ in the plane with a finite area,
$$\text{area of }T(S)=|\det A|\cdot\text{area of }S .$$
Likewise, if $A$ is $3\times3$, $\ \text{volume of }T(S)=|\det A|\cdot\text{volume of }S$ for every solid $S$ with a finite volume. The sign of $\det A$ tells whether $T$ preserves orientation ($\det A\gt 0$) or reverses it, like a mirror ($\det A\lt 0$).
This also explains Theorem 3.7 geometrically: doing $B$ and then $A$ scales areas first by $|\det B|$ and then by $|\det A|$ (Interactive 2 of Section 3.2).
โ๏ธ Example 6 โ The area of an ellipse
Use Theorem 3.13 to show that the ellipse $\dfrac{x^2}{a^2}+\dfrac{y^2}{b^2}\le1$ ($a,b\gt 0$) has area $\pi ab$. Then find the area of the image of this ellipse under $B=\begin{bmatrix}1&2\\-1&1\end{bmatrix}$ when $a=3$, $b=2$.
๐ก Show solutionโถ
Map Let $A=\begin{bmatrix}a&0\\0&b\end{bmatrix}$ and let $D$ be the unit disc $u^2+v^2\le1$. A point $(u,v)$ goes to $(x,y)=(au,bv)$, and $\dfrac{x^2}{a^2}+\dfrac{y^2}{b^2}=u^2+v^2\le1$. So $A$ maps the disc exactly onto the ellipse.
Scale The disc has area $\pi$ and $\det A=ab$, so the ellipse has area $ab\cdot\pi=\pi ab$.
Image With $a=3$, $b=2$ the ellipse has area $6\pi$. $\det B=1+2=3$, so its image has area $3\cdot6\pi=18\pi\approx56.55$ square units โ even though we never found the image's equation.
Interactive 2 โ A Parallelepiped and Its Determinant
The box spanned by $\mathbf{u}$ (purple), $\mathbf{v}$ (green) and $\mathbf{w}$ (orange) in an oblique view. Press Play to turn it. The shear slider replaces $\mathbf{w}$ by $\mathbf{w}+s\,\mathbf{u}$: the top face slides along, the base and the height stay the same โ and so does $\det\begin{bmatrix}\mathbf{u}&\mathbf{v}&\mathbf{w}+s\mathbf{u}\end{bmatrix}$. Choose the coplanar set to see the box collapse to volume $0$.
5 $\det A=2(3+2)-1(1-6)+(-1)(-1-9)=10+5+10=25$. $\det A_2=\begin{vmatrix}2&1&-1\\1&13&2\\3&4&1\end{vmatrix}=2(13-8)-1(1-6)+(-1)(4-39)=10+5+35=50$. $x_2=\tfrac{50}{25}=2$. (The full solution is $(1,2,3)$.)
6 $\det A=s^2-4$, so there is a unique solution exactly when $s\ne\pm2$. Then $x_1=\dfrac{\begin{vmatrix}1&1\\2&s\end{vmatrix}}{s^2-4}=\dfrac{s-2}{s^2-4}=\dfrac{1}{s+2}$ and $x_2=\dfrac{\begin{vmatrix}s&1\\4&2\end{vmatrix}}{s^2-4}=\dfrac{2s-4}{s^2-4}=\dfrac{2}{s+2}$. At $s=2$ the equations are $2x_1+x_2=1$ and $4x_1+2x_2=2$ โ the same line, infinitely many solutions. At $s=-2$ they are $-2x_1+x_2=1$ and $4x_1-2x_2=2$; multiplying the first by $-2$ gives $4x_1-2x_2=-2$, which contradicts the second โ no solution.
7 $\det A=4-4=0$, so the formula would divide by zero. Row reduction: $R_2\to R_2-2R_1$ gives $0=-1$, so the system is inconsistent (two parallel lines).
8 $\det A=30-4=26$; $\det A_1=\begin{vmatrix}12&-2\\4&6\end{vmatrix}=80$, $\det A_2=\begin{vmatrix}5&12\\-2&4\end{vmatrix}=44$. $I_1=\tfrac{80}{26}=\tfrac{40}{13}\approx3.08$ A and $I_2=\tfrac{44}{26}=\tfrac{22}{13}\approx1.69$ A.
II
The adjugate and the inverse.
1 Find $\operatorname{adj}A$ and $A^{-1}$ for $A=\begin{bmatrix}3&5\\1&2\end{bmatrix}$.
2 Find $\operatorname{adj}A$ and $A^{-1}$ for $A=\begin{bmatrix}1&2&0\\0&1&3\\0&0&1\end{bmatrix}$.
3 Find $\operatorname{adj}A$ and $A^{-1}$ for $A=\begin{bmatrix}1&0&1\\2&1&0\\0&1&1\end{bmatrix}$.
4 Without computing the whole inverse, find the $(2,3)$ entry of $A^{-1}$ for $A=\begin{bmatrix}2&1&0\\1&3&1\\0&1&2\end{bmatrix}$.
5 Prove that if $A$ is $n\times n$ and invertible, then $\det(\operatorname{adj}A)=(\det A)^{n-1}$.
6 $A$ is $3\times3$ with $\det A=4$. Find $\det(\operatorname{adj}A)$ and $\det\left(A^{-1}\right)$.
7 Explain why a square matrix with integer entries and $\det A=\pm1$ has an inverse with integer entries. Illustrate with $\begin{bmatrix}2&3\\1&2\end{bmatrix}$.
๐ก Show solutionsโถ
1 $\operatorname{adj}A=\begin{bmatrix}2&-5\\-1&3\end{bmatrix}$, $\det A=6-5=1$, so $A^{-1}=\begin{bmatrix}2&-5\\-1&3\end{bmatrix}$.
4 $(A^{-1})_{23}=\dfrac{C_{32}}{\det A}$ (note the swapped indices). $\det A=2(6-1)-1(2-0)+0=8$ and $C_{32}=-\begin{vmatrix}2&0\\1&1\end{vmatrix}=-2$, so the entry is $-\tfrac28=-\tfrac14$.
5 Take determinants of $A\,\operatorname{adj}A=(\det A)I_n$: $\det A\cdot\det(\operatorname{adj}A)=(\det A)^n$ (Theorem 3.8, part 3). Divide by $\det A\ne0$.
6 $\det(\operatorname{adj}A)=4^{2}=16$ and $\det(A^{-1})=\tfrac14$.
7 Every cofactor of an integer matrix is an integer (sums of products of integers), so $\operatorname{adj}A$ has integer entries, and $A^{-1}=\pm\operatorname{adj}A$. Example: $\det=4-3=1$ and $A^{-1}=\begin{bmatrix}2&-3\\-1&2\end{bmatrix}$.
III
Area and volume.
1 The parallelogram spanned by $(5,2)$ and $(1,3)$.
2 The parallelogram with vertices $(0,0)$, $(2,-1)$, $(5,3)$, $(3,4)$.
3 The triangle with vertices $(0,0)$, $(6,1)$, $(2,5)$.
4 The triangle with vertices $(-1,2)$, $(3,-1)$, $(4,5)$.
5Surveying. A plot of land has corners at $(0,0)$, $(4,0)$, $(5,3)$, $(1,4)$ (in units of $10$ m, in this order around the plot). Find its area in mยฒ.
6 The parallelepiped with edges $(1,0,2)$, $(0,3,1)$, $(2,1,0)$.
7 The tetrahedron with vertices $(0,0,0)$, $(1,2,0)$, $(0,1,3)$, $(2,0,1)$.
8 Do $(1,1,0)$, $(0,1,1)$, $(1,2,1)$ lie in one plane through the origin?
9 Find every $k$ for which the parallelepiped with edges $(1,0,k)$, $(2,1,0)$, $(0,1,3)$ has volume $10$.
4 Edges from $(-1,2)$: $(4,-3)$ and $(5,3)$. $\tfrac12\left|\begin{vmatrix}4&5\\-3&3\end{vmatrix}\right|=\tfrac12|12+15|=13.5$.
5 Split along the diagonal from $(0,0)$ to $(5,3)$: $\tfrac12\left|\begin{vmatrix}4&5\\0&3\end{vmatrix}\right|=6$ and $\tfrac12\left|\begin{vmatrix}5&1\\3&4\end{vmatrix}\right|=\tfrac{17}{2}$. Total $14.5$ square units; one unit is $10\ \text{m}\times10\ \text{m}=100$ mยฒ, so the plot is $1450$ mยฒ.
7 Edges from the origin $(1,2,0)$, $(0,1,3)$, $(2,0,1)$: $\det\begin{bmatrix}1&0&2\\2&1&0\\0&3&1\end{bmatrix}=1(1-0)-0+2(6-0)=13$. Volume $\tfrac{13}{6}$.
8 $\det\begin{bmatrix}1&0&1\\1&1&2\\0&1&1\end{bmatrix}=1(1-2)-0+1(1-0)=0$. Yes โ in fact $(1,2,1)=(1,1,0)+(0,1,1)$.
9 $\det\begin{bmatrix}1&2&0\\0&1&1\\k&0&3\end{bmatrix}=1(3-0)-2(0-k)+0=2k+3$. $|2k+3|=10$ gives $k=\tfrac72$ or $k=-\tfrac{13}{2}$.
IV
Linear maps scale area and volume.
1 $A=\begin{bmatrix}3&1\\0&2\end{bmatrix}$. Find the area of the image of the unit square, and of a disc of radius $2$.
2 Find the area of the image of the triangle with vertices $(0,0)$, $(1,0)$, $(0,1)$ under $\begin{bmatrix}2&-1\\1&3\end{bmatrix}$.
3 Show that every shear $\begin{bmatrix}1&k\\0&1\end{bmatrix}$ and every rotation preserves area. Does a reflection?
4 Find the area enclosed by the ellipse $\dfrac{x^2}{16}+\dfrac{y^2}{9}=1$.
5 A region of area $5$ is rotated by $30^\circ$ and then transformed by $\begin{bmatrix}2&0\\0&3\end{bmatrix}$. What is its final area?
6Image editing. A photo editor stretches an image to $150\%$ of its width and $80\%$ of its height, then applies a horizontal shear. By what factor does the area of every region change?
7 Find the volume of the ellipsoid $\dfrac{x^2}{4}+\dfrac{y^2}{9}+\dfrac{z^2}{16}\le1$, the image of the unit ball (volume $\tfrac43\pi$) under a diagonal matrix.
๐ก Show solutionsโถ
1 $\det A=6$. Unit square $\to$ area $6$; the disc has area $4\pi$, so its image (an ellipse) has area $24\pi\approx75.40$.
2 The triangle has area $\tfrac12$ and $\det=6+1=7$, so the image has area $\tfrac72$.
3 $\det\begin{bmatrix}1&k\\0&1\end{bmatrix}=1$ and $\det\begin{bmatrix}\cos\theta&-\sin\theta\\\sin\theta&\cos\theta\end{bmatrix}=1$, so areas are unchanged. A reflection such as $\begin{bmatrix}1&0\\0&-1\end{bmatrix}$ has determinant $-1$: it preserves area ($|-1|=1$) but reverses orientation.
4 Example 6 with $a=4$, $b=3$: area $12\pi\approx37.70$.
5 The rotation has determinant $1$ and the scaling $6$: final area $5\cdot1\cdot6=30$.
6 $\det\begin{bmatrix}1.5&0\\0&0.8\end{bmatrix}=1.2$ and a shear has determinant $1$, so every area is multiplied by $1.2$ (a $20\%$ increase).
7 The map is $\operatorname{diag}(2,3,4)$, with determinant $24$: volume $24\cdot\tfrac43\pi=32\pi\approx100.5$. In general an ellipsoid with semi-axes $a,b,c$ has volume $\tfrac43\pi abc$.
๐ฌ ResearchโTeaching NexusResearch-tutored ยท optional, about 30 minutes
Research question. How do land surveyors and GPS mapping apps compute the area of an irregular plot from the coordinates of its corners โ and how accurate is it?
Why it matters. Land registration, farm subsidies, property taxes and irrigation planning all depend on plot areas. In many countries plots are now mapped with handheld GPS, which records corner coordinates. The surveyor's (or shoelace) formula that converts them to an area is a sum of $2\times2$ determinants.
What to do.
For a polygon with corners $(x_1,y_1),\dots,(x_n,y_n)$ listed in order around the boundary, the area is $\tfrac12\left|\sum_{k=1}^{n}\begin{vmatrix}x_k&x_{k+1}\\y_k&y_{k+1}\end{vmatrix}\right|$ with $(x_{n+1},y_{n+1})=(x_1,y_1)$. Explain why, by splitting the polygon into triangles with one vertex at the origin (Theorem 3.11). Check it on Exercise III.5.
Sample case: a date-palm farm has corners (in metres, from a GPS app) $(0,0)$, $(120,10)$, $(150,90)$, $(80,140)$, $(10,100)$. Compute its area in mยฒ and in hectares.
GPS positions are accurate to about $\pm3$ m. Move one corner by $3$ m in a direction of your choice and recompute. By roughly what percentage does the area change?
What goes wrong if the corners are listed in a crossing order rather than around the boundary? Try it.
Tools: Spreadsheet (one row per corner, one column per determinant term); optionally a free online map tool that reports polygon areas.
Hand in: A half-page note: the formula with a short justification, the farm's area, your error estimate, and one practical recommendation for whoever records the corners.
Going further. Find the coordinates of a real plot, park or campus building near you on an online map, and compare your determinant-based area with the area the map tool reports.
Instructor note: the five terms are $0$, $120\cdot90-150\cdot10=9300$, $150\cdot140-80\cdot90=13800$, $80\cdot100-10\cdot140=6600$ and $0$; the area is $\tfrac12(29700)=14\,850$ mยฒ $=1.485$ ha. A $3$ m shift of one corner changes the area by at most a few hundred mยฒ (about $1$โ$2\%$). A crossing order gives a smaller, meaningless number because parts of the polygon are counted with opposite signs.
๐งช Self-check quiz
0/5 correct
Q1. In Cramer's rule for $A\mathbf{x}=\mathbf{b}$, $x_2=\dfrac{\det A_2(\mathbf{b})}{\det A}$, where $A_2(\mathbf{b})$ is
Q2. The parallelogram spanned by $(2,1)$ and $(1,3)$ has area
Q3. The $(1,2)$ entry of $\operatorname{adj}A$ is
Q4. The box with edges $(1,0,0)$, $(0,2,0)$, $(0,0,3)$ has volume
Q5. A linear map of the plane has matrix with $\det A=-3$. A region of area $4$ is mapped to a region of area