Example 21: Verify $(AB)'=B'A'$

Basics you need

Solution

1. Check that $AB$ is defined and find its order

   $A$ is $3\times1$ and $B$ is $1\times3$⁠. The number of columns of $A$ (which is 1) equals the number of rows of $B$ (also 1), so $AB$ is defined, and its order is (rows of $A$⁠) $\times$ (columns of $B$⁠) $= 3\times3$⁠.

   A product can only be formed when each row of the first matrix has exactly as many entries as each column of the second, so that they can be paired off. Checking the orders first also tells us the shape of the answer before we compute anything.
   Section 3.4.5: if $A$ is $m\times n$ and $B$ is $n\times p$⁠, then $AB$ is $m\times p$⁠. Lesson: matrix multiplication

2. Compute all nine entries of $AB$

   Row $i$ of $A$ is the single number $a_{i1}$ and column $k$ of $B$ is the single number $b_{1k}$⁠, so $c_{ik} = a_{i1}b_{1k}$⁠. With $a_{11} = -2$⁠, $a_{21} = 4$⁠, $a_{31} = 5$ and $b_{11} = 1$⁠, $b_{12} = 3$⁠, $b_{13} = -6$⁠:

   $$\begin{aligned} c_{11} &= (-2)(1) = -2,\\ c_{12} &= (-2)(3) = -6,\\ c_{13} &= (-2)(-6) = 12,\\ c_{21} &= (4)(1) = 4,\\ c_{22} &= (4)(3) = 12,\\ c_{23} &= (4)(-6) = -24,\\ c_{31} &= (5)(1) = 5,\\ c_{32} &= (5)(3) = 15,\\ c_{33} &= (5)(-6) = -30. \end{aligned}$$ $$AB = \begin{bmatrix} -2 & -6 & 12 \\ 4 & 12 & -24 \\ 5 & 15 & -30 \end{bmatrix}.$$
   With an inner size of 1, "row times column" has only one pair to multiply, so there is nothing to add. Row $i$ of $AB$ is therefore $a_{i1}$ times the row $B$⁠: row 1 is $-2$ times $1, 3, -6$⁠, and so on. Watch the signs: $(-2)(-6) = +12$⁠.
   The $(i,k)$ entry of $AB$ is $\sum_j a_{ij}b_{jk}$⁠; here $j$ takes only the value 1. Basics: pair up, multiply, add

3. Transpose $AB$⁠: the left side

   $AB$ is $3\times3$⁠, so $(AB)'$ is $3\times3$⁠. Row 1 of $AB$ ($-2, -6, 12$⁠) becomes column 1, row 2 ($4, 12, -24$⁠) becomes column 2, and row 3 ($5, 15, -30$⁠) becomes column 3:

   $$(AB)' = \begin{bmatrix} -2 & 4 & 5 \\ -6 & 12 & 15 \\ 12 & -24 & -30 \end{bmatrix}.$$
   For a square matrix the order does not change under transpose, but the entries off the main diagonal swap across it: the $(1,2)$ entry $-6$ of $AB$ becomes the $(2,1)$ entry of $(AB)'$⁠. The diagonal entries $-2, 12, -30$ stay in place.
   Definition 3, Section 3.5: $(i,j)$ entry of $X'$ is the $(j,i)$ entry of $X$⁠. Lesson: transpose

4. Find $A'$ and $B'$⁠, and check $B'A'$ is defined

   The column $A$ becomes a row, and the row $B$ becomes a column:

   $$A' = \left[\, -2 \quad 4 \quad 5 \,\right] \ (1\times3), \qquad B' = \begin{bmatrix} 1 \\ 3 \\ -6 \end{bmatrix} \ (3\times1).$$

   $B'$ is $3\times1$ and $A'$ is $1\times3$⁠: the columns of $B'$ (1) match the rows of $A'$ (1), so $B'A'$ is defined and has order $3\times3$⁠, the same as $(AB)'$⁠.

   Checking the order now is a first test of the rule: if $B'A'$ were not $3\times3$ it could not possibly equal $(AB)'$⁠. It also shows why the reverse order is needed (see the Common mistakes box).

5. Compute all nine entries of $B'A'$⁠: the right side

   Write $d_{ik}$ for the $(i,k)$ entry. Each is (entry $i$ of the column $B'$⁠) times (entry $k$ of the row $A'$⁠):

   $$\begin{aligned} d_{11} &= (1)(-2) = -2,\\ d_{12} &= (1)(4) = 4,\\ d_{13} &= (1)(5) = 5,\\ d_{21} &= (3)(-2) = -6,\\ d_{22} &= (3)(4) = 12,\\ d_{23} &= (3)(5) = 15,\\ d_{31} &= (-6)(-2) = 12,\\ d_{32} &= (-6)(4) = -24,\\ d_{33} &= (-6)(5) = -30. \end{aligned}$$ $$B'A' = \begin{bmatrix} -2 & 4 & 5 \\ -6 & 12 & 15 \\ 12 & -24 & -30 \end{bmatrix}.$$
   This is computed from scratch, without looking at $(AB)'$⁠, so the comparison in the next step is a real check rather than a copy.

6. Compare

   Both matrices are $3\times3$⁠, and all nine entries agree: row 1 is $-2, 4, 5$⁠, row 2 is $-6, 12, 15$⁠, row 3 is $12, -24, -30$ in each. Hence $(AB)' = B'A'$.

   Why it works in general: the $(i,k)$ entry of $(AB)'$ is the $(k,i)$ entry of $AB$⁠, that is $\sum_j a_{kj}b_{ji}$⁠. The $(i,k)$ entry of $B'A'$ is $\sum_j b'_{ij}a'_{jk} = \sum_j b_{ji}a_{kj}$⁠. These are the same products of numbers, only written in the other order, and multiplication of numbers is commutative.
   Property (iv) of Section 3.5.1: $(AB)' = B'A'$⁠.

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