Example 16: Find $A(BC)$ and $(AB)C$⁠; show they are equal

Basics you need

Solution

1. Check that every product is defined and track the orders

   Product	Orders	Inner sizes	Result
   $AB$	$(3\times3)(3\times2)$	$3=3$	$3\times2$
   $(AB)C$	$(3\times2)(2\times4)$	$2=2$	$3\times4$
   $BC$	$(3\times2)(2\times4)$	$2=2$	$3\times4$
   $A(BC)$	$(3\times3)(3\times4)$	$3=3$	$3\times4$

   Both sides are defined and both are $3\times4$⁠, so comparing them makes sense.

   The associative law is stated "whenever both sides of the equality are defined", so checking this first is part of applying it. It also tells us how many entries to expect: 6 in $AB$⁠, 12 in each of the others.
   Order of a product, Section 3.4.5. Lesson: matrix multiplication

2. Compute $AB$

   Rows of $A$⁠: $[1\ \ 1\ \ {-1}]$⁠, $[2\ \ 0\ \ 3]$⁠, $[3\ \ {-1}\ \ 2]$⁠. Columns of $B$⁠: $(1,0,-1)$ and $(3,2,4)$⁠. Write $AB=[p_{ik}]$⁠:

   $$\begin{aligned} p_{11} &= 1\cdot 1 + 1\cdot 0 + (-1)\cdot(-1)\\ &= 1 + 0 + 1 = 2\\ p_{12} &= 1\cdot 3 + 1\cdot 2 + (-1)\cdot 4\\ &= 3 + 2 - 4 = 1\\ p_{21} &= 2\cdot 1 + 0\cdot 0 + 3\cdot(-1)\\ &= 2 + 0 - 3 = -1\\ p_{22} &= 2\cdot 3 + 0\cdot 2 + 3\cdot 4\\ &= 6 + 0 + 12 = 18\\ p_{31} &= 3\cdot 1 + (-1)\cdot 0 + 2\cdot(-1)\\ &= 3 + 0 - 2 = 1\\ p_{32} &= 3\cdot 3 + (-1)\cdot 2 + 2\cdot 4\\ &= 9 - 2 + 8 = 15 \end{aligned}$$ $$AB = \begin{bmatrix} 2 & 1 \\ -1 & 18 \\ 1 & 15 \end{bmatrix}$$
   Each entry pairs a row of $A$ (3 entries) with a column of $B$ (3 entries), so each is a sum of three products. Writing each product with its sign before adding keeps track of the many negative entries.
   $p_{ik}=a_{i1}b_{1k}+a_{i2}b_{2k}+a_{i3}b_{3k}$⁠. Basics: pair up, multiply, add

3. Compute $(AB)C$

   Rows of $AB$⁠: $[2\ \ 1]$⁠, $[-1\ \ 18]$⁠, $[1\ \ 15]$⁠. Columns of $C$⁠: $(1,2)$⁠, $(2,0)$⁠, $(3,-2)$⁠, $(-4,1)$⁠. Write $(AB)C=[q_{ik}]$⁠. Row 1:

   $$\begin{aligned} q_{11} &= 2\cdot 1 + 1\cdot 2 = 2 + 2 = 4\\ q_{12} &= 2\cdot 2 + 1\cdot 0 = 4 + 0 = 4\\ q_{13} &= 2\cdot 3 + 1\cdot(-2) = 6 - 2 = 4\\ q_{14} &= 2\cdot(-4) + 1\cdot 1 = -8 + 1 = -7 \end{aligned}$$

   Row 2:

   $$\begin{aligned} q_{21} &= (-1)\cdot 1 + 18\cdot 2\\ &= -1 + 36 = 35\\ q_{22} &= (-1)\cdot 2 + 18\cdot 0\\ &= -2 + 0 = -2\\ q_{23} &= (-1)\cdot 3 + 18\cdot(-2)\\ &= -3 - 36 = -39\\ q_{24} &= (-1)\cdot(-4) + 18\cdot 1\\ &= 4 + 18 = 22 \end{aligned}$$

   Row 3:

   $$\begin{aligned} q_{31} &= 1\cdot 1 + 15\cdot 2\\ &= 1 + 30 = 31\\ q_{32} &= 1\cdot 2 + 15\cdot 0\\ &= 2 + 0 = 2\\ q_{33} &= 1\cdot 3 + 15\cdot(-2)\\ &= 3 - 30 = -27\\ q_{34} &= 1\cdot(-4) + 15\cdot 1\\ &= -4 + 15 = 11 \end{aligned}$$ $$(AB)C = \begin{bmatrix} 4 & 4 & 4 & -7 \\ 35 & -2 & -39 & 22 \\ 31 & 2 & -27 & 11 \end{bmatrix}$$
   Now the first factor is the matrix $AB$ we just found. Its rows have 2 entries and the columns of $C$ have 2 entries, so each entry is a sum of two products. The result is $3\times4$⁠, as predicted in Step 1.

4. Compute $BC$

   Now the other bracketing. Rows of $B$⁠: $[1\ \ 3]$⁠, $[0\ \ 2]$⁠, $[-1\ \ 4]$⁠. Columns of $C$ as before. Write $BC=[r_{ik}]$⁠. Row 1:

   $$\begin{aligned} r_{11} &= 1\cdot 1 + 3\cdot 2 = 1 + 6 = 7\\ r_{12} &= 1\cdot 2 + 3\cdot 0 = 2 + 0 = 2\\ r_{13} &= 1\cdot 3 + 3\cdot(-2) = 3 - 6 = -3\\ r_{14} &= 1\cdot(-4) + 3\cdot 1 = -4 + 3 = -1 \end{aligned}$$

   Row 2:

   $$\begin{aligned} r_{21} &= 0\cdot 1 + 2\cdot 2 = 0 + 4 = 4\\ r_{22} &= 0\cdot 2 + 2\cdot 0 = 0 + 0 = 0\\ r_{23} &= 0\cdot 3 + 2\cdot(-2) = 0 - 4 = -4\\ r_{24} &= 0\cdot(-4) + 2\cdot 1 = 0 + 2 = 2 \end{aligned}$$

   Row 3:

   $$\begin{aligned} r_{31} &= (-1)\cdot 1 + 4\cdot 2 = -1 + 8 = 7\\ r_{32} &= (-1)\cdot 2 + 4\cdot 0 = -2 + 0 = -2\\ r_{33} &= (-1)\cdot 3 + 4\cdot(-2)\\ &= -3 - 8 = -11\\ r_{34} &= (-1)\cdot(-4) + 4\cdot 1 = 4 + 4 = 8 \end{aligned}$$ $$BC = \begin{bmatrix} 7 & 2 & -3 & -1 \\ 4 & 0 & -4 & 2 \\ 7 & -2 & -11 & 8 \end{bmatrix}$$
   $BC$ is $3\times4$⁠, a completely different matrix from $AB$ (which is $3\times2$⁠). The associative law says that, even so, multiplying $A$ onto it will land on the same final matrix.

5. Compute $A(BC)$

   Rows of $A$ as in Step 2. Columns of $BC$⁠: $(7,4,7)$⁠, $(2,0,-2)$⁠, $(-3,-4,-11)$⁠, $(-1,2,8)$⁠. Write $A(BC)=[s_{ik}]$⁠. Row 1, from $[1\ \ 1\ \ {-1}]$⁠:

   $$\begin{aligned} s_{11} &= 1{\cdot} 7 + 1{\cdot} 4 + (-1){\cdot} 7\\ &= 7 + 4 - 7 = 4\\ s_{12} &= 1{\cdot} 2 + 1{\cdot} 0 + (-1){\cdot}(-2)\\ &= 2 + 0 + 2 = 4\\ s_{13} &= 1{\cdot}(-3) + 1{\cdot}(-4) + (-1){\cdot}(-11)\\ &= -3 - 4 + 11 = 4\\ s_{14} &= 1{\cdot}(-1) + 1{\cdot} 2 + (-1){\cdot} 8\\ &= -1 + 2 - 8 = -7 \end{aligned}$$

   Row 2, from $[2\ \ 0\ \ 3]$⁠:

   $$\begin{aligned} s_{21} &= 2{\cdot} 7 + 0{\cdot} 4 + 3{\cdot} 7\\ &= 14 + 0 + 21 = 35\\ s_{22} &= 2{\cdot} 2 + 0{\cdot} 0 + 3{\cdot}(-2)\\ &= 4 + 0 - 6 = -2\\ s_{23} &= 2{\cdot}(-3) + 0{\cdot}(-4) + 3{\cdot}(-11)\\ &= -6 + 0 - 33 = -39\\ s_{24} &= 2{\cdot}(-1) + 0{\cdot} 2 + 3{\cdot} 8\\ &= -2 + 0 + 24 = 22 \end{aligned}$$

   Row 3, from $[3\ \ {-1}\ \ 2]$⁠:

   $$\begin{aligned} s_{31} &= 3{\cdot} 7 + (-1){\cdot} 4 + 2{\cdot} 7\\ &= 21 - 4 + 14 = 31\\ s_{32} &= 3{\cdot} 2 + (-1){\cdot} 0 + 2{\cdot}(-2)\\ &= 6 + 0 - 4 = 2\\ s_{33} &= 3{\cdot}(-3) + (-1){\cdot}(-4) + 2{\cdot}(-11)\\ &= -9 + 4 - 22 = -27\\ s_{34} &= 3{\cdot}(-1) + (-1){\cdot} 2 + 2{\cdot} 8\\ &= -3 - 2 + 16 = 11 \end{aligned}$$ $$A(BC) = \begin{bmatrix} 4 & 4 & 4 & -7 \\ 35 & -2 & -39 & 22 \\ 31 & 2 & -27 & 11 \end{bmatrix}$$
   Each entry is now a sum of three products, because rows of $A$ and columns of $BC$ have 3 entries. The arithmetic is entirely different from Step 3 (for example $q_{23}=-3-36$ but $s_{23}=-6+0-33$⁠), yet the totals agree.

6. Compare the two results

   Both matrices are $3\times4$⁠, and comparing the twelve entries of Step 3 with those of Step 5:

   $$\begin{aligned} &\text{row 1: } 4,\ 4,\ 4,\ -7 \ \text{ in both}\\ &\text{row 2: } 35,\ -2,\ -39,\ 22 \ \text{ in both}\\ &\text{row 3: } 31,\ 2,\ -27,\ 11 \ \text{ in both} \end{aligned}$$

   Same order and all corresponding entries equal, so $(AB)C=A(BC)$.

   This is what the definition of equal matrices requires: equal orders and $q_{ik}=s_{ik}$ for every $i,k$⁠. Checking all twelve, not just a few, is what makes the conclusion valid for these matrices.
   Associative law, Section 3.4.6, property 1. Lesson: properties of the operations

7. Why the two groupings always agree

   The textbook states the associative law without proof, but the reason is short. Entry $(i,l)$ of $(AB)C$ is

   $$\sum_{k}\Big(\sum_{j} a_{ij}b_{jk}\Big)c_{kl} = \sum_{j}\sum_{k} a_{ij}b_{jk}c_{kl},$$

   and entry $(i,l)$ of $A(BC)$ is

   $$\sum_{j} a_{ij}\Big(\sum_{k} b_{jk}c_{kl}\Big) = \sum_{j}\sum_{k} a_{ij}b_{jk}c_{kl}.$$

   Both are the sum of $a_{ij}b_{jk}c_{kl}$ over every pair $(j,k)$⁠, just added up in a different order.

   Real-number multiplication is associative and distributes over addition, and a finite sum can be added in any order. So the two double sums are equal for any matrices of compatible orders. Here $j$ runs over 3 values and $k$ over 2, so each entry is a sum of 6 triple products. For entry $(2,3)$⁠, the four terms with $j=1$ or $j=3$ are $$\begin{aligned} &2\cdot 1\cdot 3 + 2\cdot 3\cdot(-2)\\ &+ 3\cdot(-1)\cdot 3 + 3\cdot 4\cdot(-2)\\ &= 6 - 12 - 9 - 24 = -39, \end{aligned}$$ and the two terms with $j=2$ are 0 because $a_{22}=0$⁠.

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