Chapter 3 · Matrices

Exercise 3.3 · Q5: Verify $(AB)'=B'A'$ for a column times a row

Exercise 3.3Transpose of a product · Section 3.5

The question

For the matrices $A$ and $B$ ⁠, verify that $(AB)' = B'A'$ ⁠, where

(i)

A = \begin{bmatrix} 1 \\ -4 \\ 3 \end{bmatrix}, \quad B = \begin{bmatrix} -1 & 2 & 1 \end{bmatrix}

(ii)

A = \begin{bmatrix} 0 \\ 1 \\ 2 \end{bmatrix}, \quad B = \begin{bmatrix} 1 & 5 & 7 \end{bmatrix}

Exam answer

(i)

\begin{aligned} AB &= \begin{bmatrix} 1 \\ -4 \\ 3 \end{bmatrix}\begin{bmatrix} -1 & 2 & 1 \end{bmatrix} \\ &= \begin{bmatrix} 1(-1) & 1(2) & 1(1) \\ -4(-1) & -4(2) & -4(1) \\ 3(-1) & 3(2) & 3(1) \end{bmatrix} \\ &= \begin{bmatrix} -1 & 2 & 1 \\ 4 & -8 & -4 \\ -3 & 6 & 3 \end{bmatrix} \end{aligned}

\text{LHS} = (AB)' = \begin{bmatrix} -1 & 4 & -3 \\ 2 & -8 & 6 \\ 1 & -4 & 3 \end{bmatrix}

\begin{aligned} &\text{RHS} = B'A' \\ &= \begin{bmatrix} -1 \\ 2 \\ 1 \end{bmatrix}\begin{bmatrix} 1 & -4 & 3 \end{bmatrix} \\ &= \begin{bmatrix} -1(1) & -1(-4) & -1(3) \\ 2(1) & 2(-4) & 2(3) \\ 1(1) & 1(-4) & 1(3) \end{bmatrix} \\ &= \begin{bmatrix} -1 & 4 & -3 \\ 2 & -8 & 6 \\ 1 & -4 & 3 \end{bmatrix} \end{aligned}

$\therefore$ LHS = RHS, so $(AB)' = B'A'$ ⁠.

(ii)

\begin{aligned} AB &= \begin{bmatrix} 0 \\ 1 \\ 2 \end{bmatrix}\begin{bmatrix} 1 & 5 & 7 \end{bmatrix} \\ &= \begin{bmatrix} 0(1) & 0(5) & 0(7) \\ 1(1) & 1(5) & 1(7) \\ 2(1) & 2(5) & 2(7) \end{bmatrix} \\ &= \begin{bmatrix} 0 & 0 & 0 \\ 1 & 5 & 7 \\ 2 & 10 & 14 \end{bmatrix} \end{aligned}

\text{LHS} = (AB)' = \begin{bmatrix} 0 & 1 & 2 \\ 0 & 5 & 10 \\ 0 & 7 & 14 \end{bmatrix}

\begin{aligned} &\text{RHS} = B'A' \\ &= \begin{bmatrix} 1 \\ 5 \\ 7 \end{bmatrix}\begin{bmatrix} 0 & 1 & 2 \end{bmatrix} \\ &= \begin{bmatrix} 1(0) & 1(1) & 1(2) \\ 5(0) & 5(1) & 5(2) \\ 7(0) & 7(1) & 7(2) \end{bmatrix} \\ &= \begin{bmatrix} 0 & 1 & 2 \\ 0 & 5 & 10 \\ 0 & 7 & 14 \end{bmatrix} \end{aligned}

$\therefore$ LHS = RHS, so $(AB)' = B'A'$ ⁠. Hence verified.

Want the reason behind every step, the basics and the checks?

The one idea this problem is built on

The transpose of a product is the product of the transposes in reverse order: $(AB)' = B'A'$ ⁠. The reversal is forced by the orders. Here $A$ is $3\times1$ and $B$ is $1\times3$ ⁠, so $AB$ is $3\times3$ ⁠. Then $B'$ is $3\times1$ and $A'$ is $1\times3$ ⁠, so $B'A'$ is again a column times a row, $3\times3$ ⁠. The other order, $A'B'$ ⁠, would be a row times a column: a single number, which cannot equal a $3\times3$ matrix.

Given, to find, and the plan

Given: Two pairs: a $3\times1$ column $A$ and a $1\times3$ row $B$ ⁠.
To show: $(AB)' = B'A'$ for each pair.
Plan: For each pair: (1) check orders; (2) compute every entry of $AB$ ⁠, then transpose; (3) write $B'$ and $A'$ and compute every entry of $B'A'$ ⁠; (4) compare.

Basics you need

1When a product is defined, and its order

$AB$ is defined when the number of columns of $A$ equals the number of rows of $B$ ⁠. If $A$ is $m\times n$ and $B$ is $n\times p$ ⁠, then $AB$ is $m\times p$ ⁠. Basics: pair up, multiply, add

2The entries of a product

The $(i,k)$ entry of $AB$ is row $i$ of $A$ times column $k$ of $B$ ⁠: multiply matching entries and add, $c_{ik} = a_{i1}b_{1k} + a_{i2}b_{2k}$ $+ \dots + a_{in}b_{nk}$ ⁠. When $A$ has only one column ( $n = 1$ ⁠), there is only one term: $c_{ik} = a_{i1}b_{1k}$ ⁠. Lesson: matrix multiplication

3The transpose

$(A')_{ij} = a_{ji}$ ⁠; a column turns into a row and a row into a column. Basics: reading a table sideways

Part (i): $A$ = column $1, -4, 3$ ⁠; $B$ = row $-1, 2, 1$

(i) Check the orders

$A$ is $3\times1$ and $B$ is $1\times3$ ⁠. Columns of $A$ = 1 = rows of $B$ ⁠, so $AB$ is defined and has order $3\times3$ ⁠. Its transpose is $3\times3$ ⁠.

$B'$ is $3\times1$ and $A'$ is $1\times3$ ⁠. Columns of $B'$ = 1 = rows of $A'$ ⁠, so $B'A'$ is defined and is $3\times3$ ⁠. Both sides have the same order.

Product $m\times n$ times $n\times p$ is $m\times p$ (Basic 1). Basics: pair up, multiply, add

(i) Compute $AB$ ⁠, entry by entry

Each row of $A$ holds one number and each column of $B$ holds one number, so each entry is a single product $c_{ik} = a_{i1}b_{1k}$ ⁠. In the table, row $i$ and column $k$ hold $c_{ik}$ ⁠:

column 1	column 2	column 3
$1\cdot(-1) = -1$	$1\cdot 2 = 2$	$1\cdot1 = 1$
$(-4)(-1) = 4$	$(-4)\cdot2 = -8$	$(-4)\cdot1 = -4$
$3\cdot(-1) = -3$	$3\cdot 2 = 6$	$3\cdot 1 = 3$

Assemble:

AB = \begin{bmatrix} -1 & 2 & 1 \\ 4 & -8 & -4 \\ -3 & 6 & 3 \end{bmatrix}.

Each row of

AB

is the row

B

scaled by one entry of

A

⁠: row 1 is

1\times B

⁠, row 2 is

-4\times B

⁠, row 3 is

3\times B

⁠. That gives a quick check on the nine products.

c_{ik}

= row

i

A

times column

k

B

⁠, Section 3.4.5 (Basic 2). Lesson: matrix multiplication

(i) Transpose $AB$

The rows $-1, 2, 1$ ⁠; $4, -8, -4$ ⁠; $-3, 6, 3$ become columns:
$(AB)' = \begin{bmatrix} -1 & 4 & -3 \\ 2 & -8 & 6 \\ 1 & -4 & 3 \end{bmatrix}.$

(i) Compute $B'A'$ ⁠, entry by entry

Transposing turns the row $B$ into a column and the column $A$ into a row:

B' = \begin{bmatrix} -1 \\ 2 \\ 1 \end{bmatrix}_{3\times1}, \quad A' = \begin{bmatrix} 1 & -4 & 3 \end{bmatrix}_{1\times3}.

Writing $d_{ik}$ for the entries of $B'A'$ ⁠, each is again a single product, (entry $i$ of $B'$ ⁠) × (entry $k$ of $A'$ ⁠); row $i$ ⁠, column $k$ of the table holds $d_{ik}$ ⁠:

column 1	column 2	column 3
$(-1)\cdot 1 = -1$	$(-1)(-4) = 4$	$(-1)\cdot 3 = -3$
$2\cdot1 = 2$	$2(-4) = -8$	$2\cdot3 = 6$
$1\cdot 1 = 1$	$1(-4) = -4$	$1\cdot 3 = 3$

B'A' = \begin{bmatrix} -1 & 4 & -3 \\ 2 & -8 & 6 \\ 1 & -4 & 3 \end{bmatrix}.

We compute

B'A'

from scratch rather than assuming the answer. The purpose of "verify" is that the two sides are found independently and then compared.

(i) Compare

Part (i), Steps 3 and 4 give the same $3\times3$ matrix, entry for entry. So $(AB)' = B'A'$ for the pair in (i).

Equality of matrices, Definition 2.

Part (ii): $A$ = column $0, 1, 2$ ⁠; $B$ = row $1, 5, 7$

(ii) Orders

Exactly as in Part (i): $A$ is $3\times1$ ⁠, $B$ is $1\times3$ ⁠, so $AB$ is $3\times3$ ⁠; $B'$ is $3\times1$ ⁠, $A'$ is $1\times3$ ⁠, so $B'A'$ is $3\times3$ ⁠.

(ii) Compute $AB$ and transpose it

Each entry is $c_{ik} = a_{i1}b_{1k}$ ⁠:

column 1	column 2	column 3
$0\cdot1 = 0$	$0\cdot5 = 0$	$0\cdot7 = 0$
$1\cdot1 = 1$	$1\cdot5 = 5$	$1\cdot7 = 7$
$2\cdot1 = 2$	$2\cdot5 = 10$	$2\cdot7 = 14$

AB = \begin{bmatrix} 0 & 0 & 0 \\ 1 & 5 & 7 \\ 2 & 10 & 14 \end{bmatrix}.

Rows become columns:

(AB)' = \begin{bmatrix} 0 & 1 & 2 \\ 0 & 5 & 10 \\ 0 & 7 & 14 \end{bmatrix}.

The first row of

AB

is all zeros because the first entry of

A

is 0: a zero row times anything is zero. After transposing, that zero row becomes a zero column.

(ii) Compute $B'A'$

B' = \begin{bmatrix} 1 \\ 5 \\ 7 \end{bmatrix}, \quad A' = \begin{bmatrix} 0 & 1 & 2 \end{bmatrix}.

Each entry is (entry $i$ of $B'$ ⁠) × (entry $k$ of $A'$ ⁠); row $i$ ⁠, column $k$ of the table holds $d_{ik}$ ⁠:

column 1	column 2	column 3
$1\cdot0 = 0$	$1\cdot1 = 1$	$1\cdot2 = 2$
$5\cdot0 = 0$	$5\cdot1 = 5$	$5\cdot2 = 10$
$7\cdot0 = 0$	$7\cdot1 = 7$	$7\cdot2 = 14$

B'A' = \begin{bmatrix} 0 & 1 & 2 \\ 0 & 5 & 10 \\ 0 & 7 & 14 \end{bmatrix}.

(ii) Compare

Both sides are the same $3\times3$ matrix, so $(AB)' = B'A'$ for the pair in (ii).
Why the order must reverse

Look at one entry in general. The $(i,k)$ entry of $(AB)'$ is the $(k,i)$ entry of $AB$ ⁠, which is $a_{k1}b_{1i}$ ⁠. The $(i,k)$ entry of $B'A'$ is $(B')_{i1}(A')_{1k} = b_{1i}a_{k1}$ ⁠. These are equal because numbers commute, $a_{k1}b_{1i} = b_{1i}a_{k1}$ ⁠. In the wrong order, $A'B'$ is ( $1\times3$ ⁠)( $3\times1$ ⁠), a $1\times1$ matrix: in (i) it is $[1(-1) + (-4)(2) + 3(1)] = [-6]$ ⁠, which is not even the same size as $(AB)'$ ⁠.

Matrices do not commute in general, so the order in a product matters. The reversal in $(AB)' = B'A'$ is like taking off socks and shoes: what goes on last comes off first.

$(AB)' = B'A'$ ⁠, Section 3.5.1, property (iv); Example 21 is the same kind of check. Lesson: transpose

Result

(i) Both sides equal

\begin{bmatrix} -1 & 4 & -3 \\ 2 & -8 & 6 \\ 1 & -4 & 3 \end{bmatrix}.

(ii) Both sides equal

\begin{bmatrix} 0 & 1 & 2 \\ 0 & 5 & 10 \\ 0 & 7 & 14 \end{bmatrix}.

So $(AB)' = B'A'$ is verified in both cases.

Check it with numbers

In (i), the (2,3) entry of $(AB)'$ should be the (3,2) entry of $AB$ ⁠, which is $a_{31}b_{12} = 3\cdot 2 = 6$ ⁠. Both sides show 6 there. ✓ In (ii), the (3,3) entry is $a_{31}b_{13} = 2\cdot 7 = 14$ ⁠. ✓

Common mistakes

Writing $(AB)' = A'B'$ ⁠. The order must reverse. Here $A'B'$ is $1\times1$ and cannot equal a $3\times3$ matrix.
Confusing $AB$ with $BA$ ⁠. $BA$ is a row times a column, $1\times1$ ⁠; in (i) it is $[-6]$ ⁠.
Sign slips in products. $(-4)(-1) = 4$ and $(-1)(-4) = 4$ ⁠.

What to take away

$(AB)' = B'A'$ ⁠: the transpose of a product reverses the order.
Column ( $3\times1$ ⁠) times row ( $1\times3$ ⁠) is a $3\times3$ matrix with entries $a_i b_k$ ⁠; row times column is a single number.
Always check orders before multiplying; they often tell you which order is the only possible one.

Every fact used, and where it comes from

Fact	Used in	Source
$AB$ defined when columns of $A$ = rows of $B$ ⁠; order $m\times p$	Part (i), Step 1; Part (ii), Step 1	Section 3.4.5 (Lesson: multiplication)
Entry of a product: row times column	Parts (i), (ii), Steps 2–4	Section 3.4.5, definition of the product
Transpose $(A')_{ij} = a_{ji}$	Throughout	Definition 3, Section 3.5
$(AB)' = B'A'$	Part (ii), Step 5	Section 3.5.1, property (iv); Example 21
Matrix multiplication is not commutative	Part (ii), Step 5	Section 3.4.5, Examples 13 and 14

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Exercise 3.3 · Q5: Verify (AB)′=B′A′(AB)'=B'A'(AB)′=B′A′ for a column times a row

Basics you need

Part (i): AAA = column 1,−4,31, -4, 31,−4,3⁠; BBB = row −1,2,1-1, 2, 1−1,2,1

(i) Check the orders

(i) Compute ABABAB⁠, entry by entry

(i) Transpose ABABAB

(i) Compute B′A′B'A'B′A′⁠, entry by entry

(i) Compare

Part (ii): AAA = column 0,1,20, 1, 20,1,2⁠; BBB = row 1,5,71, 5, 71,5,7

(ii) Orders

(ii) Compute ABABAB and transpose it

(ii) Compute B′A′B'A'B′A′

(ii) Compare

Why the order must reverse