Multiplying matrices and linear transformations of space – Matrices and Determinants of Order 3 (Senior 5 Mathematics) | Ganzaa

Matrices and Determinants of Order 3

Lessons

Senior 5 · Mathematics

Matrices and Determinants of Order 3

5 lessons

Lesson 2 of 5

Multiplying matrices and linear transformations of space

40 min

From bags to francs

Back at the cooperative of lesson 1, suppose a bag of beans sells for 40,000 FRW, a bag of maize for 25,000 FRW and a bag of potatoes for 15,000 FRW (illustrative prices). The income of the Musanze market in week 1 is $12(40\,000) + 20(25\,000) + 35(15\,000)$ : a row of the sales matrix "times" the column of prices. Doing this for every row at once is matrix multiplication. The same row-by-column rule moves points in space, which is why every 3-D drawing program, from an architect's model to a phone game, multiplies by $3 \times 3$ matrices millions of times a second.

Definition

Product of matrices. If $A$ is an $m \times n$ matrix and $B$ is an $n \times p$ matrix, their product $A B$ is the $m \times p$ matrix with entries

(AB)_{ij} = a_{i1}b_{1j} + a_{i2}b_{2j} + \dots + a_{in}b_{nj} = \sum_{k=1}^{n} a_{ik}b_{kj} ,

that is, row $i$ of $A$ multiplied, entry by entry, with column $j$ of $B$ , and the products added. The product exists only when the number of columns of $A$ equals the number of rows of $B$ . Two $3 \times 3$ matrices can always be multiplied, and the result is $3 \times 3$ .

Worked example 1: income per market

With the week-1 sales $A$ of lesson 1 and the price column $p$ , find the income of each market.

Ap = \begin{pmatrix} 12 & 20 & 35 \\ 8 & 15 & 40 \\ 10 & 18 & 25 \end{pmatrix} \begin{pmatrix} 40\,000 \\ 25\,000 \\ 15\,000 \end{pmatrix} = \begin{pmatrix} 480\,000 + 500\,000 + 525\,000 \\ 320\,000 + 375\,000 + 600\,000 \\ 400\,000 + 450\,000 + 375\,000 \end{pmatrix} = \begin{pmatrix} 1\,505\,000 \\ 1\,295\,000 \\ 1\,225\,000 \end{pmatrix} .

Musanze earned 1,505,000 FRW, Huye 1,295,000 FRW and Rubavu 1,225,000 FRW. A $3 \times 3$ matrix times a $3 \times 1$ column gives a $3 \times 1$ column.

Worked example 2: the order of the factors matters

Find $P Q$ and $Q P$ for $P = \begin{pmatrix} 1 & 2 & 0 \\ 0 & 1 & -1 \\ 3 & 0 & 2 \end{pmatrix}$ and $Q = \begin{pmatrix} 2 & -1 & 1 \\ 1 & 0 & 3 \\ 0 & 4 & -2 \end{pmatrix}$ .

Solution. The first entry of $P Q$ is row 1 of $P$ times column 1 of $Q$ : $1 (2) + 2 (1) + 0 (0) = 4$ . Continuing in the same way,

PQ = \begin{pmatrix} 4 & -1 & 7 \\ 1 & -4 & 5 \\ 6 & 5 & -1 \end{pmatrix}, \qquad QP = \begin{pmatrix} 5 & 3 & 3 \\ 10 & 2 & 6 \\ -6 & 4 & -8 \end{pmatrix} .

$PQ \ne QP$ : matrix multiplication is not commutative.

Important

Properties of the product (for matrices whose products exist, and a real number $k$ ):

associative: $(A B) C = A (B C)$ , so $A^{2} = A A$ , $A^{3} = A A A$ are well defined;
distributive: $A (B + C) = A B + A C$ and $(A + B) C = A C + B C$ ;
identity: $A I_{3} = I_{3} A = A$ for every $A \in M_3(\mathbb{R})$ ;
scalars move freely: $(k A) B = A (k B) = k (A B)$ ;
transpose of a product: $(A B)^{T} = B^{T} A^{T}$ , with the order reversed;
not commutative: in general $AB \ne BA$ , and $A B = O$ can happen with $A \ne O$ and $B \ne O$ .

Reason for the transpose rule. The $(i, j)$ entry of $(A B)^{T}$ is the $(j, i)$ entry of $A B$ , namely $\sum_k a_{jk}b_{ki}$ . The $(i, j)$ entry of $B^{T} A^{T}$ is $\sum_k (B^T)_{ik}(A^T)_{kj} = \sum_k b_{ki}a_{jk}$ , the same number.

Worked example 3: a non-zero matrix whose cube is zero

Let $N = \begin{pmatrix} 0 & 1 & 2 \\ 0 & 0 & 3 \\ 0 & 0 & 0 \end{pmatrix}$ . Find $N^{2}$ and $N^{3}$ .

Solution.

N^2 = \begin{pmatrix} 0 & 0 & 3 \\ 0 & 0 & 0 \\ 0 & 0 & 0 \end{pmatrix}, \qquad N^3 = N^2N = \begin{pmatrix} 0 & 0 & 0 \\ 0 & 0 & 0 \\ 0 & 0 & 0 \end{pmatrix} .

So $N^{3} = O$ although $N \ne O$ . With numbers, $x^{3} = 0$ forces $x = 0$ ; with matrices it does not. A matrix with some power equal to $O$ is called nilpotent.

Definition

A linear transformation of space is a function $f : \mathbb{R}^3 \to \mathbb{R}^3$ such that, for all vectors $u, v$ and real $k$ ,

f(u + v) = f(u) + f(v) \quad \text{and} \quad f(ku) = k f(u) .

Its matrix (in the standard basis $i = (1, 0, 0)$ , $j = (0, 1, 0)$ , $k = (0, 0, 1)$ ) is the $3 \times 3$ matrix whose columns are $f (i)$ , $f (j)$ and $f (k)$ . Then the image of any point is a product:

f\begin{pmatrix} x \\ y \\ z \end{pmatrix} = M \begin{pmatrix} x \\ y \\ z \end{pmatrix} .

Reason. $(x, y, z) = x i + y j + z k$ , so by linearity $f (x, y, z) = x f (i) + y f (j) + z f (k)$ , which is exactly the product of $M$ with the column $(x, y, z)$ .

Seeing a transformation: where the unit cube goes

The figure shows the shear $S (x, y, z) = (x, y + z, z)$ . Its columns are $S (i) = (1, 0, 0)$ , $S (j) = (0, 1, 0)$ , $S (k) = (0, 1, 1)$ , so its matrix is $\begin{pmatrix} 1 & 0 & 0 \\ 0 & 1 & 1 \\ 0 & 0 & 1 \end{pmatrix}$ . Every horizontal layer slides in the $y$ direction by its height $z$ , and the cube becomes a slanted box (a parallelepiped).

Transformation	Image of $(x, y, z)$	Matrix
Enlargement, centre $O$ , scale factor $k$	$(k x, k y, k z)$	$k I_{3}$
Reflection in the $x y$ -plane	$(x, y, - z)$	$\begin{pmatrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & -1 \end{pmatrix}$
Reflection in the $x z$ -plane	$(x, - y, z)$	$\begin{pmatrix} 1 & 0 & 0 \\ 0 & -1 & 0 \\ 0 & 0 & 1 \end{pmatrix}$
Orthogonal projection onto the $x y$ -plane	$(x, y, 0)$	$\begin{pmatrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 0 \end{pmatrix}$
Rotation through $\theta$ about the $z$ -axis	$(x\cos\theta - y\sin\theta,\ x\sin\theta + y\cos\theta,\ z)$	$\begin{pmatrix} \cos\theta & -\sin\theta & 0 \\ \sin\theta & \cos\theta & 0 \\ 0 & 0 & 1 \end{pmatrix}$

The rotation keeps $z$ and turns the $x y$ -plane exactly as the $2 \times 2$ rotation matrix of S4 does; the angle is anticlockwise when you look down the $z$ -axis towards the origin.

Important

Composition is multiplication. If $f$ has matrix $M_{1}$ and $g$ has matrix $M_{2}$ , then "first $f$ , then $g$ ", that is $g \circ f$ , has matrix $M_{2} M_{1}$ : the transformation applied first is written on the right, next to the column it acts on.

Worked example 4: rotate, reflect, and the order of the two

$R$ is the rotation through $90^\circ$ about the $z$ -axis and $G$ is the reflection in the $x z$ -plane. (a) Find the matrix of $R$ and the image of $A (2, - 1, 3)$ . (b) Find the matrices of " $G$ then $R$ " and of " $R$ then $G$ ", and the image of $A$ under each.

Solution. (a) With $\cos 90^\circ = 0$ and $\sin 90^\circ = 1$ :

R = \begin{pmatrix} 0 & -1 & 0 \\ 1 & 0 & 0 \\ 0 & 0 & 1 \end{pmatrix}, \qquad R\begin{pmatrix} 2 \\ -1 \\ 3 \end{pmatrix} = \begin{pmatrix} 1 \\ 2 \\ 3 \end{pmatrix} .

(b) $G = \begin{pmatrix} 1 & 0 & 0 \\ 0 & -1 & 0 \\ 0 & 0 & 1 \end{pmatrix}$ . " $G$ then $R$ " has matrix

RG = \begin{pmatrix} 0 & 1 & 0 \\ 1 & 0 & 0 \\ 0 & 0 & 1 \end{pmatrix}, \quad RG\begin{pmatrix} 2 \\ -1 \\ 3 \end{pmatrix} = \begin{pmatrix} -1 \\ 2 \\ 3 \end{pmatrix}; \qquad GR = \begin{pmatrix} 0 & -1 & 0 \\ -1 & 0 & 0 \\ 0 & 0 & 1 \end{pmatrix}, \quad GR\begin{pmatrix} 2 \\ -1 \\ 3 \end{pmatrix} = \begin{pmatrix} 1 \\ -2 \\ 3 \end{pmatrix} .

The two orders send $A$ to different points, as $RG \ne GR$ predicts.

Warning

$A^{2}$ is not "square every entry". $A^{2} = A A$ uses the row-by-column rule.
$(A B)^{T} = B^{T} A^{T}$ , not $A^{T} B^{T}$ . The same reversal appears for inverses in lesson 4.
$(A + B)^{2} = A^{2} + A B + B A + B^{2}$ . It equals $A^{2} + 2 A B + B^{2}$ only when $A B = B A$ .
The columns, not the rows, of a transformation's matrix are the images of $i, j, k$ .

Tip

Before multiplying, write the orders side by side: $(3 \times \underline{3})(\underline{3} \times 1)$ . If the two inner numbers match, the product exists, and the two outer numbers give its order.

3Blue1Brown: Three-dimensional linear transformations, Chapter 5 of Essence of Linear Algebra (about 5 min). Watch the columns of a 3 × 3 matrix appear as the landing places of i, j and k, exactly as in this lesson's figure.

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Practice

Questions 1 to 3 are routine; 4 to 6 need more care; 7 asks for a reason.

Find $A B$ and $B A$ for $A = \begin{pmatrix} 2 & 1 & 0 \\ -1 & 3 & 2 \\ 1 & 0 & 4 \end{pmatrix}$ and $B = \begin{pmatrix} 1 & 0 & -2 \\ 3 & 1 & 0 \\ 0 & 2 & 1 \end{pmatrix}$ .
Find $A^{2}$ for the same $A$ .
$A$ is $3 \times 3$ , $C$ is $2 \times 3$ and $D$ is $3 \times 1$ . Which of $A C$ , $C A$ , $A D$ , $D A$ , $C D$ exist? Give the order of each that does.
Check that $(A B)^{T} = B^{T} A^{T}$ for the matrices of question 1.
Write the matrix of the rotation through $90^\circ$ about the $x$ -axis, which sends $j$ to $k$ , and find the image of $(1, 2, 3)$ .
Let $J = \begin{pmatrix} 1 & 1 & 1 \\ 1 & 1 & 1 \\ 1 & 1 & 1 \end{pmatrix}$ . Show that $J^{2} = 3 J$ , and deduce $J^{3}$ and $J^{n}$ without multiplying again.
Find the matrix of "project onto the $x y$ -plane, then enlarge with centre $O$ and scale factor 3". Explain why doing the two in the other order gives the same matrix.

Show answersHide answers

$AB = \begin{pmatrix} 5 & 1 & -4 \\ 8 & 7 & 4 \\ 1 & 8 & 2 \end{pmatrix}$ , $BA = \begin{pmatrix} 0 & 1 & -8 \\ 5 & 6 & 2 \\ -1 & 6 & 8 \end{pmatrix}$ .
$A^2 = \begin{pmatrix} 3 & 5 & 2 \\ -3 & 8 & 14 \\ 6 & 1 & 16 \end{pmatrix}$
$C A$ is $2 \times 3$ ; $A D$ is $3 \times 1$ ; $C D$ is $2 \times 1$ . $A C$ and $D A$ do not exist.
Both equal $\begin{pmatrix} 5 & 8 & 1 \\ 1 & 7 & 8 \\ -4 & 4 & 2 \end{pmatrix}$ .
$\begin{pmatrix} 1 & 0 & 0 \\ 0 & 0 & -1 \\ 0 & 1 & 0 \end{pmatrix}$ ; image $(1, - 3, 2)$ .
Every entry of $J^{2}$ is $1 + 1 + 1 = 3$ . Then $J^{3} = J^{2} J = 3 J^{2} = 9 J$ , and in general $J^n = 3^{n-1}J$ .
$3I_3 \begin{pmatrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 0 \end{pmatrix} = \begin{pmatrix} 3 & 0 & 0 \\ 0 & 3 & 0 \\ 0 & 0 & 0 \end{pmatrix}$ . The enlargement's matrix is $3 I_{3}$ , and $(3 I_{3}) M = 3 M = M (3 I_{3})$ for every $M$ , so the order does not matter.

Sources and credits

Written by Ganzaa: the explanations, the diagram (drawn from computed coordinates), the worked examples, the practice items and the quiz questions. Every product was checked by computer. The prices and sales are illustrative.
Video: Three-dimensional linear transformations | Chapter 5, Essence of linear algebra, 3Blue1Brown (Grant Sanderson), embedded from YouTube.

1From bags to francs

2Seeing a transformation: where the unit cube goes

From bags to francs

Seeing a transformation: where the unit cube goes