Matrices as transformations

The last chapter ended with the simplest map of the complex plane, multiplication by a number. Multiplying by $a = p + qi$ means rotating the plane about zero through the angle $\arg a$ and stretching it by a factor of $|a|$. One goes to the point $a$, that is, to $(p, q)$, and the imaginary unit $i$ goes to $ia = -q + pi$, the point $(-q, p)$: the same segment, turned through a right angle. A square grid goes to a square grid. The squares get bigger and turn, but they stay squares.

Such a map has two levers, and they’re linked: wherever you turn the first, the second follows, a quarter turn further on. Let’s break the link. Let the point $(1, 0)$ go anywhere at all, and the point $(0, 1)$ anywhere too, independently of the first. The squares stop being squares: they stretch, skew, turn inside out, or even squash into a segment. All these deformations, rotations, stretches, reflections, shears, projections, are written with four numbers in two columns. The chapter is laid out as a workshop: we’ll take apart how this notation works, learn to chain deformations together and undo them, and at the end go out into space.

One convention before we start. From this chapter on, vectors are written in bold, $\mathbf v$ instead of $\vec v$, as is usual in linear algebra. The coordinates of a vector are written as a column: $\mathbf v = \left(\begin{smallmatrix} v_1 \\ v_2 \end{smallmatrix}\right)$, and in running text, so as not to break the line, in a row with commas: $(1, -1)$ means the column $\left(\begin{smallmatrix} 1 \\ -1 \end{smallmatrix}\right)$. The unit vectors along the axes, called $\vec i$ and $\vec j$ in chapter 22, are now called $\mathbf e_1 = (1, 0)$ and $\mathbf e_2 = (0, 1)$.

Two levers

Which transformations of the plane are we going to study? Multiplication by a complex number suggests the answer. It respects addition and multiplication by real numbers: $a(z + w) = az + aw$ and $a(cz) = c(az)$. These are the two properties we’ll demand.

The definition immediately shows how such a transformation behaves. The origin stays put: $L\mathbf 0 = L(0 \cdot \mathbf v) = 0 \cdot L\mathbf v = \mathbf 0$. The line through the point $\mathbf p$ in the direction $\mathbf d$ consists of the points $\mathbf p + t\mathbf d$, and these go to the points $L\mathbf p + t\,L\mathbf d$. That’s again a line with direction $L\mathbf d$, if $L\mathbf d \ne \mathbf 0$, or a single point, if $L\mathbf d = \mathbf 0$. Parallel lines share a direction, so their images are parallel too. Points spaced at equal steps along a line, $t = 0, 1, 2, \dots$, go to points at equal steps. So a square grid turns into a grid of identical parallelograms, with the origin staying put. That’s the portrait of a linear transformation.

Which of these transformations of the plane isn’t linear?

Rotation through $30^\circ$ about the origin Translation by the vector $(1, 0)$ Reflection in the line $y = x$ Projection onto the $x$-axis: $(x, y) \mapsto (x, 0)$

A linear transformation has to keep the origin in place, while a translation moves it to the point $(1, 0)$. Checking against the definition: $L(\mathbf u + \mathbf v) = \mathbf u + \mathbf v + \mathbf e_1$, while $L\mathbf u + L\mathbf v = \mathbf u + \mathbf v + 2\mathbf e_1$. Translations belong to the neighbouring class of affine transformations: linear plus a shift by a constant vector.

The main property of a linear transformation: it’s determined entirely by where the two vectors $\mathbf e_1$ and $\mathbf e_2$ go. Two levers control the whole plane.

The trick is to read a vector as a route on a square grid. A linear transformation carries the route over to its own skewed grid, step by step.

The vector $\p3{\mathbf v} = v_1\mathbf e_1 + v_2\mathbf e_2$ is a route: $v_1$ steps along $\p1{\mathbf e_1}$, then $v_2$ steps along $\p2{\mathbf e_2}$. That’s how the coordinates of a vector were defined in chapter 22. Apply $L$. The vector $\mathbf e_1$ goes to $\p1{\mathbf a_1} = L\mathbf e_1$, the vector $\mathbf e_2$ to $\p2{\mathbf a_2} = L\mathbf e_2$. Grid lines go to lines, parallel ones to parallel ones, equal steps to equal steps: the square grid becomes a skewed one, spanned by $\p1{\mathbf a_1}$ and $\p2{\mathbf a_2}$. By the first property of linearity the image of a sum is the sum of the images: $L(v_1\mathbf e_1 + v_2\mathbf e_2) = L(v_1\mathbf e_1) + L(v_2\mathbf e_2)$. By the second, $L(v_1\mathbf e_1) = v_1\p1{\mathbf a_1}$ and $L(v_2\mathbf e_2) = v_2\p2{\mathbf a_2}$. The route is the same, $v_1$ steps and $v_2$ steps, only now it has to be walked on the skewed grid. The end of the route is the point $L\p3{\mathbf v} = v_1\p1{\mathbf a_1} + v_2\p2{\mathbf a_2}$. The converse is checked directly: if $L\mathbf v = v_1\mathbf a_1 + v_2\mathbf a_2$, then $L(\mathbf u + \mathbf v) = (u_1 + v_1)\mathbf a_1 + (u_2 + v_2)\mathbf a_2 = L\mathbf u + L\mathbf v$ and $L(c\mathbf v) = cv_1\mathbf a_1 + cv_2\mathbf a_2 = c\,L\mathbf v$ by the distributive laws for vectors, and $L\mathbf e_1 = 1\cdot\mathbf a_1 + 0\cdot\mathbf a_2 = \mathbf a_1$ and likewise $L\mathbf e_2 = \mathbf a_2$.

So there are exactly as many linear transformations of the plane as there are pairs of vectors $(\mathbf a_1, \mathbf a_2)$, and each is given by four numbers, the coordinates of these vectors. Take the levers in your hands.

A blueprint in two columns

Since a transformation is given by two vectors, it’s convenient to write their coordinates side by side: the first as a column on the left, the second as a column on the right.

The matrix and the transformation are denoted by the same letter: $A\mathbf v$ is the image of the vector $\mathbf v$. By the two-levers theorem it’s computed like this.

The first column of the matrix: where $\mathbf e_1$ goes. The second column: where $\mathbf e_2$ goes. The first coordinate of the vector: how many steps to take along the first column. The second coordinate: how many steps along the second. Example: $A = \left(\begin{smallmatrix} 2 & -1 \\ 1 & 1 \end{smallmatrix}\right)$, $\mathbf v = (3, 2)$. Then $A\mathbf v = 3\,(2, 1) + 2\,(-1, 1) = (6 - 2, 3 + 2) = (4, 5)$.

The right side of the formula gives the familiar “row times column” rule: to get the $i$th coordinate of $A\mathbf v$, go along the $i$th row of the matrix and down the column $\mathbf v$, multiplying the numbers in pairs, and add. That’s the dot product of the row with the vector. But it’s more useful to think in columns: the result is a combination of the columns of the matrix with coefficients from $\mathbf v$.

The humblest matrix moves nothing: it leaves $\mathbf e_1$ and $\mathbf e_2$ in place.

Find $A\mathbf v$ for $A = \left(\begin{smallmatrix} 1 & -2 \\ 3 & 0 \end{smallmatrix}\right)$ and $\mathbf v = (2, 1)$.

By columns: $A\mathbf v = 2\,(1, 3) + 1\,(-2, 0) = (2 - 2, 6 + 0) = (0, 6)$. By rows it’s the same: $1 \cdot 2 + (-2) \cdot 1 = 0$ and $3 \cdot 2 + 0 \cdot 1 = 6$.

A linear transformation takes $\mathbf e_1$ to $(2, 1)$ and $\mathbf e_2$ to $(-1, 3)$. Where does it take the vector $(3, -2)$?

No matrix is needed; the two-levers theorem is enough: the image is $3\,(2, 1) - 2\,(-1, 3) = (6 + 2, 3 - 6) = (8, -3)$.

A catalogue of deformations

Now any familiar transformation can be written as a matrix almost without calculation: it’s enough to ask where it sends $\mathbf e_1$ and $\mathbf e_2$. Rotation through the angle $\varphi$ takes $\mathbf e_1$ to $(\cos\varphi, \sin\varphi)$ and $\mathbf e_2$ to $(-\sin\varphi, \cos\varphi)$; that’s exactly what was proved in the chapter on symmetry. The other rows of the catalogue are obtained the same way.

Projection onto the line with unit vector $\mathbf u = (\cos\alpha, \sin\alpha)$ takes a vector $\mathbf v$ to its shadow $(\mathbf u \cdot \mathbf v)\,\mathbf u$ (chapter 22). For $\mathbf v = \mathbf e_1$ the shadow is $\cos\alpha \cdot \mathbf u = (\cos^2\alpha, \sin\alpha\cos\alpha)$: that’s the first column. Reflection takes $\mathbf v$ to $2(\mathbf u \cdot \mathbf v)\,\mathbf u - \mathbf v$: the shadow laid off twice, minus the vector itself; hence the columns $(2\cos^2\alpha - 1, 2\sin\alpha\cos\alpha) = (\cos 2\alpha, \sin 2\alpha)$ and $(\sin 2\alpha, -\cos 2\alpha)$, matching the formula from chapter 23. The last row is the previous chapter: $(p + qi)(x + yi) = (px - qy) + (qx + py)i$, so $\mathbf e_1$ goes to $(p, q)$ and $\mathbf e_2$ to $(-q, p)$.

The shear shows well how a linear transformation works. It doesn’t change area: a square turns into a parallelogram with the same base and the same height (chapter on area). But angles and lengths change, so a shear isn’t a rigid motion. In the workshop above it’s one of the presets.

The conveyor

Put two transformations one after the other: first $B$, then $A$. The vector $\mathbf v$ first goes to $B\mathbf v$, and that goes to $A(B\mathbf v)$. The chain is linear again: $A(B(\mathbf u + \mathbf v)) = A(B\mathbf u + B\mathbf v) = A(B\mathbf u) + A(B\mathbf v)$, and likewise with a factor. So it has a matrix of its own.

The letters are read from right to left, as in the notation $f(g(x))$ from the chapter on functions: whatever acts first stands closer to the vector.

The first column of $B$: where $B$ sends $\mathbf e_1$. Next, $A$ takes it to the first column of the product. The second column of $B$ and its image under $A$: the second column of the product. Example: $A = \left(\begin{smallmatrix} 0 & -1 \\ 1 & 0 \end{smallmatrix}\right)$ is rotation through $90^\circ$, $B = \left(\begin{smallmatrix} 1 & 1 \\ 0 & 1 \end{smallmatrix}\right)$ a shear. Then $AB = \left(\begin{smallmatrix} 0 & -1 \\ 1 & 1 \end{smallmatrix}\right)$, while $BA = \left(\begin{smallmatrix} 1 & -1 \\ 1 & 0 \end{smallmatrix}\right)$: shearing then rotating isn’t the same as rotating then shearing.

The example shows the main oddity of matrix arithmetic: $AB$ and $BA$ can differ. Putting on socks and then shoes isn’t the same as shoes and then socks. On the other hand, brackets can be placed any way you like.

$A$ is rotation through $90^\circ$ anticlockwise, $B$ reflection in the $x$-axis. Are $AB$ and $BA$ the same?

The same: in both cases it’s a rotation and a reflection No: one is reflection in the line $y = x$, the other in the line $y = -x$ No: one of them is a rotation The same, because multiplication is commutative

$AB$: the reflection leaves $\mathbf e_1$ in place, and the rotation sends it to $\mathbf e_2$; the reflection takes $\mathbf e_2$ to $-\mathbf e_2$, and the rotation takes that to $\mathbf e_1$. We get $AB = \left(\begin{smallmatrix} 0 & 1 \\ 1 & 0 \end{smallmatrix}\right)$, reflection in the line $y = x$. While $BA = \left(\begin{smallmatrix} 0 & -1 \\ -1 & 0 \end{smallmatrix}\right)$ is reflection in the line $y = -x$.

Sometimes the order doesn’t matter. Two rotations about the origin commute: the angles simply add. All matrices of the form $\left(\begin{smallmatrix} p & -q \\ q & p \end{smallmatrix}\right)$ commute too: they multiply like the complex numbers $p + qi$, and multiplication of complex numbers is commutative. So the complex numbers live inside the $2 \times 2$ matrices, as a small corner where the arithmetic is the familiar one.

The area meter

The matrix $A$ turns the unit square with sides $\mathbf e_1$ and $\mathbf e_2$ into the parallelogram on its columns $(a, c)$ and $(b, d)$. We already know how to find its area: chapter 22 proved that it equals $|ad - bc|$, and the number $ad - bc$ itself has a sign: plus if the second column points to the left of the first, minus if to the right. In the chapter on systems the same expression appeared as the determinant of a system, $\Delta$. It’s time to give it its name for good.

The first column: where $\mathbf e_1$ goes. The second column: where $\mathbf e_2$ goes. The area of the parallelogram on the columns, with the sign of the orientation: the unit square turns into a figure of area $|ad - bc|$. Example: $\left(\begin{smallmatrix} 3 & 1 \\ 1 & 2 \end{smallmatrix}\right)$ has determinant $3 \cdot 2 - 1 \cdot 1 = 5$. The unit square becomes a parallelogram of area $5$, and the circle of radius $1$ an ellipse of area $5\pi$. The reflection $\left(\begin{smallmatrix} 0 & 1 \\ 1 & 0 \end{smallmatrix}\right)$ has determinant $-1$: the areas are the same, the orientation is reversed. For the shear $\left(\begin{smallmatrix} 1 & 1 \\ 0 & 1 \end{smallmatrix}\right)$ it equals $1$.

The square is only the beginning. Why does the area of any figure change by the same factor? Because any figure can be assembled from small squares, and the transformation changes all the squares in the same way.

The trick is that the transformation changes all the little squares of a grid in the same way: each turns into a copy of the same parallelogram.

The unit square goes to the parallelogram on the columns $A\mathbf e_1$ and $A\mathbf e_2$. Its area is $|ad - bc| = |\det A|$, the formula from chapter 22. Take a little square with side $h$ and a corner at the point $\mathbf p$: its points are $\mathbf p + s\,h\mathbf e_1 + t\,h\mathbf e_2$ with $0 \le s, t \le 1$. By linearity they go to $A\mathbf p + s\,hA\mathbf e_1 + t\,hA\mathbf e_2$: the parallelogram on the vectors $hA\mathbf e_1$ and $hA\mathbf e_2$ with a corner at $A\mathbf p$. Its base and height are $h$ times those of the image of the unit square, so its area is $|\det A|\,h^2$. Cut the plane with a grid of little squares of side $h$. The squares lying entirely inside the figure (shaded) have total area $S_h^{\,\text{in}}$; together with the squares touching the edge (outlined) they give $S_h^{\,\text{all}}$. The figure contains the first and is contained in the second, and as $h \to 0$ both sums tend to its area $S$: that’s how area is defined (chapter 20). For $\det A \ne 0$ the transformation is one-to-one (we’ll prove this in the section on reversing), so the images of the inner squares don’t overlap and lie inside the image of the figure, while the images of all the squares cover it. The area of the image is squeezed between $|\det A|\,S_h^{\,\text{in}}$ and $|\det A|\,S_h^{\,\text{all}}$. As $h \to 0$ both bounds tend to $|\det A| \cdot S$, so the area of the image is $|\det A| \cdot S$. And if $\det A = 0$, the columns lie on one line, and the whole plane goes to that line (or to a point): the area of the image is zero, and the formula holds again.

A triangle of area $6$ undergoes the transformation with matrix $\left(\begin{smallmatrix} 4 & 1 \\ 2 & 3 \end{smallmatrix}\right)$. What’s the area of the image?

$\det A = 4 \cdot 3 - 1 \cdot 2 = 10$, so the area is multiplied by $10$ and becomes $60$. The image of the triangle is a triangle again: segments go to segments.

The last experiment in the widget is more than an observation. The determinant depends continuously on the columns, and to go from positive to negative it has to pass through zero. So no smooth movement of the levers will turn the boat into its mirror image without squashing it on the way. In the same way, a glove can’t be turned into one for the other hand by rotating it on a table top.

Now chains. If $B$ multiplies areas by $|\det B|$ and then $A$ by $|\det A|$, the whole chain multiplies them by the product. The same with orientation: two mirrors in a row restore it. All this fits into one equation.

The trick is to measure the area of a parallelogram in cells of the skewed grid: in it the columns of $AB$ stand where the columns of $B$ stood in the square grid.

The columns of $B$ are the vectors $\mathbf u = B\mathbf e_1$ and $\mathbf v = B\mathbf e_2$. The parallelogram $P$ on them has signed area $\det B$. The columns of $AB$ are the vectors $A\mathbf u$ and $A\mathbf v$ (the theorem on the columns of a product). So $\det(AB)$ is the signed area of the parallelogram $A(P)$. By the area factor theorem $|\det(AB)| = |\det A| \cdot |\det B|$. The sign remains. By the two-levers theorem $A\mathbf u = u_1\mathbf a_1 + u_2\mathbf a_2$ and $A\mathbf v = v_1\mathbf a_1 + v_2\mathbf a_2$, where $\mathbf a_1$, $\mathbf a_2$ are the columns of $A$: in the skewed grid on $\mathbf a_1$, $\mathbf a_2$ the vectors $A\mathbf u$, $A\mathbf v$ have the same coordinates as $\mathbf u$, $\mathbf v$ in the square one. The signed area $S(\mathbf x, \mathbf y) = x_1y_2 - x_2y_1$ is linear in each vector and changes sign if the vectors are swapped (you can see this straight from the formula), and $S(\mathbf a_1, \mathbf a_1) = 0$. Multiply out: $S(A\mathbf u, A\mathbf v) = u_1v_2\,S(\mathbf a_1, \mathbf a_2) + u_2v_1\,S(\mathbf a_2, \mathbf a_1) = (u_1v_2 - u_2v_1)\,S(\mathbf a_1, \mathbf a_2)$. Here $u_1v_2 - u_2v_1 = \det B$ is the number of cells in $P$, and $S(\mathbf a_1, \mathbf a_2) = \det A$ is the signed area of one skewed cell. So $\det(AB) = \det A \cdot \det B$, sign included. A check on the example from the conveyor: $\det A = 1$, $\det B = 1$ and $\det(AB) = 0 \cdot 1 - (-1) \cdot 1 = 1$.

How are $\det(2A)$ and $\det A$ related for a $2 \times 2$ matrix $A$?

$\det(2A) = 2\det A$ $\det(2A) = 4\det A$ $\det(2A) = \det A$ $\det(2A) = 8\det A$

The matrix $2A$ is $A$ followed by scaling by a factor of $2$, and the scaling quadruples areas: $\det(2A) = \det(2I)\det A = 4\det A$. Straight from the formula: $(2a)(2d) - (2b)(2c) = 4(ad - bc)$. In space, scaling multiplies volumes by $8$, and for $3 \times 3$ matrices $\det(2A) = 8\det A$.

A special case is a zero determinant. Then the columns lie on one line, the image of the square is squashed into a segment, and the whole plane into a line (or, for the zero matrix, into a point).

One more operation will be needed in the coming chapters. If the rows and columns of a matrix swap roles, we get the transpose $A^{\mathsf T}$: $\left(\begin{smallmatrix} a & b \\ c & d \end{smallmatrix}\right)^{\mathsf T} = \left(\begin{smallmatrix} a & c \\ b & d \end{smallmatrix}\right)$. A column vector becomes a row after transposing, and the dot product can be written as a product of matrices: $\mathbf x^{\mathsf T}\mathbf y = x_1y_1 + x_2y_2$. In the university chapters it’s written $\langle \mathbf x, \mathbf y\rangle$.

Reverse gear

Can every transformation be undone? A rotation through $30^\circ$ is undone by a rotation through $-30^\circ$, a threefold stretch by a threefold squeeze, a shear to the right by a shear to the left. But projection onto the $x$-axis can’t be undone: the points $(2, 5)$ and $(2, -1)$ both land on $(2, 0)$, and from the image there’s no telling where the point came from.

The determinant. The inverse transformation multiplies areas by $\frac{1}{\det A}$, and if $\det A = 0$, there’s no inverse. The diagonal entries swap places. The off-diagonal entries stay where they are but change sign. Example: $A = \left(\begin{smallmatrix} 3 & 1 \\ 1 & 2 \end{smallmatrix}\right)$, $\det A = 5$, $A^{-1} = \frac15\left(\begin{smallmatrix} 2 & -1 \\ -1 & 3 \end{smallmatrix}\right)$. Check: $AA^{-1} = \frac15\left(\begin{smallmatrix} 6 - 1 & -3 + 3 \\ 2 - 2 & -1 + 6 \end{smallmatrix}\right) = I$.

The inverse matrix solves systems of equations. The system $3x + y = 7$, $x + 2y = 4$ is a single vector equation $A\mathbf x = \mathbf b$ with $A = \left(\begin{smallmatrix} 3 & 1 \\ 1 & 2 \end{smallmatrix}\right)$ and $\mathbf b = (7, 4)$. We’re looking for the vector that $A$ sends to $\mathbf b$. Multiply both sides on the left by $A^{-1}$: $\mathbf x = A^{-1}\mathbf b = \frac15\,(2 \cdot 7 - 4, -7 + 3 \cdot 4) = (2, 1)$. Check: $3 \cdot 2 + 1 = 7$, $2 + 2 \cdot 1 = 4$. If $A^{-1}\mathbf b$ is written out in letters, we get exactly Cramer’s rule from chapter 11, and the condition $\Delta \ne 0$ from there now reads: the transformation doesn’t squash the plane.

Solve the system $2x + y = 5$, $3x + 2y = 8$ using the inverse matrix. Write the answer as a point $(x, y)$.

$A = \left(\begin{smallmatrix} 2 & 1 \\ 3 & 2 \end{smallmatrix}\right)$, $\det A = 4 - 3 = 1$, so $A^{-1} = \left(\begin{smallmatrix} 2 & -1 \\ -3 & 2 \end{smallmatrix}\right)$, a matrix of integers. Then $\mathbf x = A^{-1}\,(5, 8) = (10 - 8, -15 + 16) = (2, 1)$. Check: $4 + 1 = 5$, $6 + 2 = 8$.

How do you undo a chain? In the morning you put on socks, then shoes. In the evening you take them off in the reverse order: first shoes, then socks. It’s the same with matrices.

The trick is to walk the conveyor backwards: the operation done last is undone first.

The boat goes through the conveyor: first $\p2{B}$ (here a twofold stretch along the $x$-axis), then $\p1{A}$ (rotation through $90^\circ$). At the output is the boat transformed by the matrix $\p1{A}\p2{B}$. To go back, first take off the last thing done: apply $\p1{A^{-1}}$. It undoes $A$ and returns the boat to the intermediate station: $A^{-1}(AB\mathbf v) = (A^{-1}A)(B\mathbf v) = I(B\mathbf v) = B\mathbf v$ by the associative law. Then apply $\p2{B^{-1}}$ and undo the stretch: $B^{-1}(B\mathbf v) = \mathbf v$. The boat is home. So the chain “first $A^{-1}$, then $B^{-1}$”, that is, the matrix $B^{-1}A^{-1}$, undoes $AB$: $(B^{-1}A^{-1})(AB) = I$. In the other order it’s the same: $(AB)(B^{-1}A^{-1}) = A(BB^{-1})A^{-1} = AA^{-1} = I$. Both equations of the definition hold, so $(AB)^{-1} = B^{-1}A^{-1}$. But trying to undo in the original order, $A^{-1}B^{-1}$, usually leads astray: in the picture the boat comes back twice as long and half as high, and in the example from the conveyor $A^{-1}B^{-1} = \left(\begin{smallmatrix} 0 & 1 \\ -1 & 1 \end{smallmatrix}\right)$, while $(AB)^{-1} = \left(\begin{smallmatrix} 1 & 1 \\ -1 & 0 \end{smallmatrix}\right)$.

What was done last is undone first. In the expression $B^{-1}A^{-1}$ the matrix $A^{-1}$ stands further right, closer to the vector, so it acts first. The first operation of the chain is undone last. Example: $A = \left(\begin{smallmatrix} 0 & -1 \\ 1 & 0 \end{smallmatrix}\right)$, $B = \left(\begin{smallmatrix} 1 & 1 \\ 0 & 1 \end{smallmatrix}\right)$, $AB = \left(\begin{smallmatrix} 0 & -1 \\ 1 & 1 \end{smallmatrix}\right)$. By the formula for the inverse $(AB)^{-1} = \left(\begin{smallmatrix} 1 & 1 \\ -1 & 0 \end{smallmatrix}\right)$, and $B^{-1}A^{-1} = \left(\begin{smallmatrix} 1 & -1 \\ 0 & 1 \end{smallmatrix}\right)\left(\begin{smallmatrix} 0 & 1 \\ -1 & 0 \end{smallmatrix}\right) = \left(\begin{smallmatrix} 1 & 1 \\ -1 & 0 \end{smallmatrix}\right)$ gives the same.

You can get your hand in with matrix calculations, multiplying by a vector, products, determinants and inverses, in the trainer. And the solver will take any matrix of your own apart step by step.

The third dimension

In space everything works the same way, only there are three levers. A linear transformation of space is determined by the images of the vectors $\mathbf e_1 = (1, 0, 0)$, $\mathbf e_2 = (0, 1, 0)$, $\mathbf e_3 = (0, 0, 1)$; the proof of the two-levers theorem goes through word for word with three. The matrix is a $3 \times 3$ table of three columns $A\mathbf e_1$, $A\mathbf e_2$, $A\mathbf e_3$, multiplication by a vector is the same combination of columns, the product the same composition. The cube with sides $\mathbf e_1$, $\mathbf e_2$, $\mathbf e_3$ turns into a parallelepiped on the three columns.

Rotation through the angle $\varphi$ about the $z$-axis leaves $\mathbf e_3$ in place and rotates the $xy$-plane, so its matrix is the matrix of a plane rotation framed by zeros and a one: $\left(\begin{smallmatrix} \cos\varphi & -\sin\varphi & 0 \\ \sin\varphi & \cos\varphi & 0 \\ 0 & 0 & 1 \end{smallmatrix}\right)$. Projection onto the $xy$-plane sends $\mathbf e_3$ to zero: $\left(\begin{smallmatrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 0 \end{smallmatrix}\right)$.

The role of area is played by volume. The parallelepiped on the vectors $\mathbf a$, $\mathbf b$, $\mathbf c$ has volume “base area times height”, and the cross product from chapter 22 lets us compute it from coordinates.

The trick is that the cross product $\mathbf a \times \mathbf b$ gives both the area of the base (by its length) and the direction of the height.

The base is the parallelogram on $\p1{\mathbf a}$ and $\p1{\mathbf b}$. The vector $\p2{\mathbf n} = \mathbf a \times \mathbf b$ is perpendicular to the base, and its length equals the area of the base (chapter 22): $S = |\mathbf n|$. The height is the distance from the tip of $\p3{\mathbf c}$ to the plane of the base, that is, the length of the shadow of the vector $\mathbf c$ on the line of $\mathbf n$: $h = \frac{|\mathbf c \cdot \mathbf n|}{|\mathbf n|}$ (the projection from chapter 22). The volume is $V = S \cdot h = |\mathbf n| \cdot \frac{|\mathbf c \cdot \mathbf n|}{|\mathbf n|} = |\mathbf c \cdot (\mathbf a \times \mathbf b)|$. The number $\mathbf c \cdot (\mathbf a \times \mathbf b)$ is positive when $\mathbf c$ points to the same side of the base as $\mathbf a \times \mathbf b$, and that’s exactly what it means for the triple to be right-handed. Substitute the coordinates: $\mathbf a \times \mathbf b = (a_2b_3 - a_3b_2, a_3b_1 - a_1b_3, a_1b_2 - a_2b_1)$, and $\mathbf c \cdot (\mathbf a \times \mathbf b) = c_1(a_2b_3 - a_3b_2) + c_2(a_3b_1 - a_1b_3) + c_3(a_1b_2 - a_2b_1)$. Multiplying out gives exactly the six terms of the rule of Sarrus.

Three products along the main diagonal and the “broken diagonals” parallel to it, with a plus sign. Three products along the other diagonal and those parallel to it, with a minus sign. Each of the six products contains exactly one entry from each row and each column. Example: the matrix with columns $(1, 0, 1)$, $(2, 1, 0)$, $(0, 1, 1)$ has determinant $1 \cdot 1 \cdot 1 + 2 \cdot 1 \cdot 1 + 0 - 0 - 1 \cdot 1 \cdot 0 - 2 \cdot 0 \cdot 1 = 3$: the unit cube becomes a parallelepiped of volume $3$, and orientation is preserved. The projection onto the $xy$-plane has determinant $0$: the cube is squashed into a square.

All the theorems about area carry over to volume word for word, only with little cubes instead of little squares: a transformation of space multiplies volumes by $|\det A|$, and $\det(AB) = \det A \cdot \det B$. And the equation $\det A^{\mathsf T} = \det A$ for $3 \times 3$ can be seen from the rule of Sarrus: transposing swaps the roles of rows and columns and takes each of the six products to a product from the same list with the same sign.

Where next

Solving the system $A\mathbf x = \mathbf b$ means finding the vector that the transformation $A$ sends to $\mathbf b$, that is, undoing the transformation. For two unknowns the formula for the inverse matrix was enough. But in January 1801 the astronomer Giuseppe Piazzi found a new heavenly body, Ceres, followed it for a few weeks and lost it in the glare of the Sun. To compute its orbit from a handful of observations, the twenty-four-year-old Gauss solved systems with many unknowns. Formulas with determinants are hopeless here: a $20 \times 20$ determinant, written out on the pattern of the rule of Sarrus, consists of $20! \approx 2.4 \cdot 10^{18}$ products, and a computer doing a billion multiplications a second would take more than seventy years to compute them. How to undo a transformation in a few thousand operations is the subject of chapter 36.