Multivariable-Calculus · Unit 3 · Video 4 · Interactive Practice

Right to Left: What AB Does, What I Does, and the Matrix That Rotates the Plane

IKey Formulas

Formula	Name	What it says
$(AB)X = A(BX)$	Associativity	$AB$ means: apply $B$ first, then $A$
$AB \neq BA$ , and $BA$ need not exist	Order matters	$AB$ is defined when the width of $A$ equals the height of $B$
$IX = X$ for every $X$	The identity matrix	Ones down the diagonal, zeros elsewhere — one $I$ for each size
$R = \begin{pmatrix} 0 & -1 \\ 1 & 0 \end{pmatrix}, \quad R\begin{pmatrix} x \\ y \end{pmatrix} = \begin{pmatrix} -y \\ x \end{pmatrix}$	Rotation by $90\degree$	Its columns are $R\hat{\imath} = \hat{\jmath}$ and $R\hat{\jmath} = -\hat{\imath}$

Key Insight: Reading runs left to right, but acting runs right to left — in $AB$ the factor nearest the input goes first. And the columns of any matrix are exactly where the basis vectors land, because multiplying by $\hat{\imath}$ picks out column 1.

IIVisualization 1 — AB Means B First

In $AB$ the factor nearest the input acts first: $B$ transforms $X$ , then $A$ transforms the result.

A B Order Progress 100%

💡 Both orders exist here only because every factor is $2 \times 2$ . If $A$ is $3 \times 4$ and $B$ is $4 \times 2$ , then $AB$ is defined and is $3 \times 2$ , while $BA$ is not defined at all.

IIIVisualization 2 — The Columns Are the Images of the Basis Vectors

The two columns of $M$ are nothing but the places where $\hat{\imath}$ and $\hat{\jmath}$ land.

Mî: x 0 Mî: y 1 Mĵ: x −1 Mĵ: y 0

💡 The same reading works in every size: column $k$ of $M$ is $M$ applied to the $k$ -th basis vector, which is why the $3 \times 3$ identity — columns $\hat{\imath}, \hat{\jmath}, \hat{k}$ — returns every column of three entries unchanged.

IVVisualization 3 — Multiplying R by Itself

Composing the quarter turn with itself is ordinary matrix multiplication: $R^2$ , $R^3$ , $R^4$ .

Quarter turns 1

💡 Because $R^4 = I$ , three more quarter turns undo one: $R^3 R = I$ , so $R^3$ is the matrix that undoes $R$ — the first example of an inverse matrix.

VQuiz Questions

Problem 1 · Applying the Rotation Matrix

Given: $R = \begin{pmatrix} 0 & -1 \\ 1 & 0 \end{pmatrix}$ — compute $R\begin{pmatrix} 3 \\ 2 \end{pmatrix}$ .

✅ Correct!

R

sends

(x, y)

(-y, x)

, so

(3, 2)

lands on

(-2, 3)

— a quarter turn counterclockwise.

❌ Close, but that is the turn the other way.

(y, -x)

is a clockwise quarter turn; it comes from dotting the input with the columns of

R

instead of the rows.

❌ That negates $x$ and leaves $y$ alone. That is a flip across the

y

-axis. Row one of

R

(0, -1)

, so the first entry of the answer is

-y

, never

-x

❌ That is two quarter turns.

(-3, -2) = -(3, 2)

R^2 = -I

applied to the input, not

R

❌ Not quite. Entry one is row one dotted with the column:

0 \cdot 3 + (-1) \cdot 2

. Entry two is row two:

1 \cdot 3 + 0 \cdot 2

Show solution

Each entry of the answer is a row of $R$ dotted with the input column:

R\begin{pmatrix} 3 \\ 2 \end{pmatrix} = \begin{pmatrix} 0 \cdot 3 + (-1) \cdot 2 \\ 1 \cdot 3 + 0 \cdot 2 \end{pmatrix} = \begin{pmatrix} -2 \\ 3 \end{pmatrix}

This matches the general rule $(x, y) \mapsto (-y, x)$ with $x = 3$ and $y = 2$ .

Check the lengths: $|(3, 2)| = \sqrt{13}$ and $|(-2, 3)| = \sqrt{13}$ — a rotation cannot change a length.

Problem 2 · Which Product Even Exists

Given: $A$ is $3 \times 4$ and $B$ is $4 \times 2$ . Which statement is true?

✅ Correct!

(3 \times 4)(4 \times 2)

: the inner

4

s match, and the answer takes the outer sizes,

3 \times 2

. Reversed, the inner sizes are

2

and

3

— no match.

❌ Check $BA$ . It reads

(4 \times 2)(3 \times 4)

: the width of

B

2

and the height of

A

3

, so the rows of

B

and the columns of

A

have different lengths and no dot product can be formed.

❌ You kept the inner sizes. The inner pair must match; the product inherits the outer pair, so

(3 \times 4)(4 \times 2)

3 \times 2

❌ Check $AB$ again. The width of

A

4

and the height of

B

4

: the inner sizes agree, so

AB

certainly exists. It is

BA

that fails.

❌ Not quite. Compare the width of the left factor with the height of the right factor in each order.

Show solution

$AB$ : $A$ is $3 \times 4$ , $B$ is $4 \times 2$ . The width of $A$ is $4$ and the height of $B$ is $4$ , so each row of $A$ and each column of $B$ have four entries and their dot product makes sense. The product has one entry for each row of $A$ and each column of $B$ :

(3 \times 4)(4 \times 2) = 3 \times 2

$BA$ : now the width of $B$ is $2$ and the height of $A$ is $3$ .

(4 \times 2)(3 \times 4) \quad \text{undefined}

So $AB$ and $BA$ are not merely different answers — only one of them is even a question.

Problem 3 · Both Orders of the Same Pair

Given: the quarter turn $A = \begin{pmatrix} 0 & -1 \\ 1 & 0 \end{pmatrix}$ and the shear $B = \begin{pmatrix} 1 & 1 \\ 0 & 1 \end{pmatrix}$ — compute both products.

What is $AB$ ?

What is $BA$ ?

✅ Correct! Two genuinely different matrices:

AB

shears and then turns,

BA

turns and then shears.

❌ Check $AB$ . Entry

(i, j)

is row

i

A

dotted with column

j

B

. Row one of

A

(0, -1)

; column two of

B

(1, 1)

❌ Check $BA$ . Now the rows come from

B

and the columns from

A

. Row one of

B

(1, 1)

; column one of

A

(0, 1)

Show solution

$AB$ — rows of $A$ against columns of $B$ :

AB = \begin{pmatrix} 0 \cdot 1 + (-1) \cdot 0 & 0 \cdot 1 + (-1) \cdot 1 \\ 1 \cdot 1 + 0 \cdot 0 & 1 \cdot 1 + 0 \cdot 1 \end{pmatrix} = \begin{pmatrix} 0 & -1 \\ 1 & 1 \end{pmatrix}

$BA$ — rows of $B$ against columns of $A$ :

BA = \begin{pmatrix} 1 \cdot 0 + 1 \cdot 1 & 1 \cdot (-1) + 1 \cdot 0 \\ 0 \cdot 0 + 1 \cdot 1 & 0 \cdot (-1) + 1 \cdot 0 \end{pmatrix} = \begin{pmatrix} 1 & -1 \\ 1 & 0 \end{pmatrix}

Read them as transformations. Column one of $AB$ is $(0, 1)$ , so $AB$ sends $\hat{\imath}$ to $\hat{\jmath}$ : the shear leaves $\hat{\imath}$ alone and the turn lifts it. Column one of $BA$ is $(1, 1)$ , so $BA$ sends $\hat{\imath}$ to $(1, 1)$ : the turn lifts it to $\hat{\jmath}$ first, and the shear then slides it sideways.

Problem 4 · From a Formula to a Matrix, and to a Power of R

Given: a transformation of the plane sends $(x, y)$ to $(y, -x)$ — write it as a matrix, then name it as a power of $R = \begin{pmatrix} 0 & -1 \\ 1 & 0 \end{pmatrix}$ .

Which matrix is it?

Which power of $R$ is it?

✅ Correct! Three counterclockwise quarter turns is one clockwise quarter turn, and

R^3 = R^2 R = (-I)R = -R

❌ Check the coefficients. Write both outputs with every coefficient showing:

u = 0 \cdot x + 1 \cdot y

and

v = (-1) \cdot x + 0 \cdot y

, then read the rows off.

❌ Count the quarter turns.

R^2 = -I

sends

(x, y)

(-x, -y)

and

R^4 = I

changes nothing, so neither of those is

(y, -x)

Show solution

Step 1 — write the outputs with every coefficient in place. Calling the outputs $u$ and $v$ :

u = y = 0 \cdot x + 1 \cdot y, \qquad v = -x = (-1) \cdot x + 0 \cdot y

Step 2 — read the coefficients off, row by row:

M = \begin{pmatrix} 0 & 1 \\ -1 & 0 \end{pmatrix}

Step 3 — identify it as a power of $R$ . Since $R^2 = -I$ ,

R^3 = R^2 R = (-I)R = -R = \begin{pmatrix} 0 & 1 \\ -1 & 0 \end{pmatrix} = M

Check with the basis vectors. $M\hat{\imath}$ is column one, $(0, -1) = -\hat{\jmath}$ ; three counterclockwise quarter turns do carry $\hat{\imath}$ to $-\hat{\jmath}$ . And $M R = R^3 R = R^4 = I$ , so this is exactly the transformation that undoes $R$ .

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