Example 17: Verify $(A+B)C=AC+BC$

Basics you need

Solution

1. Check the orders

   $A$ and $B$ are both $3\times3$⁠, so $A+B$ is defined and is $3\times3$⁠. Then $(A+B)C$⁠, $AC$ and $BC$ are each $(3\times3)(3\times1)$⁠: inner sizes $3=3$⁠, so each is defined and is a $3\times1$ column. Finally $AC+BC$ adds two $3\times1$ columns, which is defined and $3\times1$⁠.

   The law is stated "whenever both sides of equality are defined". Here both sides are $3\times1$ columns, so they can be compared.
   Sum needs equal orders; product needs columns of the first = rows of the second. Basics: rows, columns and the address

2. Compute $A+B$

   Add matching entries, $s_{ij}=a_{ij}+b_{ij}$⁠:

   $$\begin{aligned} s_{11} &= 0 + 0 = 0 & s_{12} &= 6 + 1 = 7\\ s_{13} &= 7 + 1 = 8 & s_{21} &= -6 + 1 = -5\\ s_{22} &= 0 + 0 = 0 & s_{23} &= 8 + 2 = 10\\ s_{31} &= 7 + 1 = 8 & s_{32} &= -8 + 2 = -6\\ s_{33} &= 0 + 0 = 0 \end{aligned}$$ $$A+B = \begin{bmatrix} 0 & 7 & 8 \\ -5 & 0 & 10 \\ 8 & -6 & 0 \end{bmatrix}$$
   To use the left side $(A+B)C$ we need the matrix $A+B$ itself. Each entry uses only the two entries in the same position.
   Section 3.4.1, addition of matrices. Lesson: addition and scalar multiplication

3. Compute $(A+B)C$

   The column $C$ is $(2,-2,3)$⁠. Write $(A+B)C=[u_{i1}]$⁠:

   $$\begin{aligned} u_{11} &= 0\cdot 2 + 7\cdot(-2) + 8\cdot 3\\ &= 0 - 14 + 24 = 10\\ u_{21} &= (-5)\cdot 2 + 0\cdot(-2) + 10\cdot 3\\ &= -10 + 0 + 30 = 20\\ u_{31} &= 8\cdot 2 + (-6)\cdot(-2) + 0\cdot 3\\ &= 16 + 12 + 0 = 28 \end{aligned}$$ $$(A+B)C = \begin{bmatrix} 10 \\ 20 \\ 28 \end{bmatrix}$$
   The product has a single column because $C$ has a single column; each of its three entries is one row of $A+B$ times $C$⁠.
   $u_{i1}=s_{i1}c_{11}+s_{i2}c_{21}+s_{i3}c_{31}$⁠. Basics: pair up, multiply, add

4. Compute $AC$

   Write $AC=[v_{i1}]$⁠:

   $$\begin{aligned} v_{11} &= 0\cdot 2 + 6\cdot(-2) + 7\cdot 3\\ &= 0 - 12 + 21 = 9\\ v_{21} &= (-6)\cdot 2 + 0\cdot(-2) + 8\cdot 3\\ &= -12 + 0 + 24 = 12\\ v_{31} &= 7\cdot 2 + (-8)\cdot(-2) + 0\cdot 3\\ &= 14 + 16 + 0 = 30 \end{aligned}$$ $$AC = \begin{bmatrix} 9 \\ 12 \\ 30 \end{bmatrix}$$
   Watch the sign in $v_{31}$⁠: $(-8)\cdot(-2)=+16$⁠, since a product of two negatives is positive.

5. Compute $BC$

   Write $BC=[w_{i1}]$⁠:

   $$\begin{aligned} w_{11} &= 0\cdot 2 + 1\cdot(-2) + 1\cdot 3\\ &= 0 - 2 + 3 = 1\\ w_{21} &= 1\cdot 2 + 0\cdot(-2) + 2\cdot 3\\ &= 2 + 0 + 6 = 8\\ w_{31} &= 1\cdot 2 + 2\cdot(-2) + 0\cdot 3\\ &= 2 - 4 + 0 = -2 \end{aligned}$$ $$BC = \begin{bmatrix} 1 \\ 8 \\ -2 \end{bmatrix}$$
   Same column $C$⁠, now with the rows of $B$⁠.

6. Add $AC$ and $BC$⁠, and compare

   Both are $3\times1$⁠. Adding matching entries, $t_{i1}=v_{i1}+w_{i1}$⁠:

   $$\begin{aligned} t_{11} &= 9 + 1 = 10\\ t_{21} &= 12 + 8 = 20\\ t_{31} &= 30 + (-2) = 28 \end{aligned}$$ $$AC+BC = \begin{bmatrix} 10 \\ 20 \\ 28 \end{bmatrix}$$

   This is the same order as $(A+B)C$ from Step 3, and the entries agree: $10=10$⁠, $20=20$⁠, $28=28$⁠. Hence $(A+B)C=AC+BC$.

   Equal order and all three corresponding entries equal is exactly the definition of equal matrices, so the law is verified for these matrices.
   Distributive law, Section 3.4.6, property 2 (ii). Lesson: properties of the operations

7. Why it always works

   Take any $A=[a_{ij}]$ and $B=[b_{ij}]$ of order $m\times n$ and any $C=[c_{jk}]$ of order $n\times p$⁠. Entry $(i,k)$ of $(A+B)C$ is

   $$\begin{aligned} &\textstyle\sum_{j}(a_{ij}+b_{ij})\,c_{jk}\\ &= \textstyle\sum_{j} a_{ij}c_{jk} + \sum_{j} b_{ij}c_{jk}, \end{aligned}$$

   which is entry $(i,k)$ of $AC$ plus entry $(i,k)$ of $BC$⁠.

   The only fact used is the distributive law for numbers, $(a+b)c=ac+bc$⁠, applied to each term, followed by regrouping a finite sum. Since the matrices were arbitrary (of compatible orders), this proves $(A+B)C=AC+BC$ in general. In row 1 here, the middle term $(6+1)\cdot(-2)$ splits as $6\cdot(-2)+1\cdot(-2)$⁠, and likewise for the other terms, so the entry 10 splits as 9 (from $A$⁠) plus 1 (from $B$⁠).

Every fact used, and where it comes from