Exercise 3.2 · Q14: Show that $AB \ne BA$ for two pairs of matrices

Basics you need

Part (i): the $2\times2$ pair

Here

1. Both products are defined

   $A$ and $B$ are both $2\times2$⁠. For $AB$⁠: columns of $A$ (2) = rows of $B$ (2), so $AB$ is defined and is $2\times2$⁠. For $BA$⁠: columns of $B$ (2) = rows of $A$ (2), so $BA$ is defined and is $2\times2$⁠.

   If one product were undefined, or the two had different orders, they could not be equal and there would be nothing to compute. Here both are $2\times2$⁠, so we must compare entries.
   Basics: when a product is defined

2. Compute $AB$⁠, entry by entry

   Rows of $A$⁠: $(5, -1)$ and $(6, 7)$⁠. Columns of $B$⁠: $(2, 3)$ and $(1, 4)$⁠, read top to bottom.

   $$\begin{aligned} c_{11} &= 5\cdot 2 + (-1)\cdot 3 = 10 - 3 = 7\\ c_{12} &= 5\cdot 1 + (-1)\cdot 4 = 5 - 4 = 1\\ c_{21} &= 6\cdot 2 + 7\cdot 3 = 12 + 21 = 33\\ c_{22} &= 6\cdot 1 + 7\cdot 4 = 6 + 28 = 34 \end{aligned}$$ $$AB = \begin{bmatrix} 7 & 1 \\ 33 & 34 \end{bmatrix}$$
   Entry $c_{ik}$ uses row $i$ of the first factor and column $k$ of the second. The figure below shows $c_{21}$⁠.
   $c_{ik} = a_{i1}b_{1k} + a_{i2}b_{2k}$⁠. Basics: pair up, multiply, add

3. Compute $BA$⁠, entry by entry

   Now the rows come from $B$⁠: $(2, 1)$ and $(3, 4)$⁠; the columns come from $A$⁠: $(5, 6)$ and $(-1, 7)$⁠.

   $$\begin{aligned} d_{11} &= 2\cdot 5 + 1\cdot 6 = 10 + 6 = 16\\ d_{12} &= 2\cdot(-1) + 1\cdot 7 = -2 + 7 = 5\\ d_{21} &= 3\cdot 5 + 4\cdot 6 = 15 + 24 = 39\\ d_{22} &= 3\cdot(-1) + 4\cdot 7 = -3 + 28 = 25 \end{aligned}$$ $$BA = \begin{bmatrix} 16 & 5 \\ 39 & 25 \end{bmatrix}$$
   Swapping the factors swaps which matrix supplies the rows, so every entry is a different sum.

4. Compare

   Both products are $2\times2$⁠, but in position $(1,1)$⁠, $AB$ has 7 and $BA$ has 16. Since $7 \ne 16$⁠, $AB \ne BA$. (In fact all four positions differ.)

   Equality would need all four entries to agree; one disagreement is enough to rule it out.

Part (ii): the $3\times3$ pair

Here

1. Both products are defined

   Both matrices are $3\times3$⁠. Columns of $A$ (3) = rows of $B$ (3), so $AB$ is defined and $3\times3$⁠; columns of $B$ (3) = rows of $A$ (3), so $BA$ is defined and $3\times3$⁠.

   As in Part (i), Step 1: same order, so the comparison must be made entry by entry.

2. Compute $AB$⁠, row by row

   Columns of $B$⁠, top to bottom: $(-1, 0, 2)$⁠, $(1, -1, 3)$⁠, $(0, 1, 4)$⁠.

   Row 1 of $A$ is $(1, 2, 3)$⁠:

   $$\begin{aligned} c_{11} &= 1(-1) + 2(0) + 3(2) = 5\\ c_{12} &= 1(1) + 2(-1) + 3(3) = 8\\ c_{13} &= 1(0) + 2(1) + 3(4) = 14 \end{aligned}$$

   Row 2 of $A$ is $(0, 1, 0)$⁠:

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

   Row 3 of $A$ is $(1, 1, 0)$⁠:

   $$\begin{aligned} c_{31} &= 1(-1) + 1(0) + 0(2) = -1\\ c_{32} &= 1(1) + 1(-1) + 0(3) = 0\\ c_{33} &= 1(0) + 1(1) + 0(4) = 1 \end{aligned}$$ $$AB = \begin{bmatrix} 5 & 8 & 14 \\ 0 & -1 & 1 \\ -1 & 0 & 1 \end{bmatrix}$$
   For example $c_{11} = -1 + 0 + 6 = 5$ and $c_{12} = 1 - 2 + 9 = 8$⁠. Row 2 of $A$ is $(0, 1, 0)$⁠, so it simply picks out row 2 of $B$⁠: $(0, -1, 1)$⁠.
   $c_{ik} = a_{i1}b_{1k} + a_{i2}b_{2k} + a_{i3}b_{3k}$⁠. Basics: pair up, multiply, add

3. Compute $BA$⁠, row by row

   Now the columns come from $A$⁠, top to bottom: $(1, 0, 1)$⁠, $(2, 1, 1)$⁠, $(3, 0, 0)$⁠.

   Row 1 of $B$ is $(-1, 1, 0)$⁠:

   $$\begin{aligned} d_{11} &= -1(1) + 1(0) + 0(1) = -1\\ d_{12} &= -1(2) + 1(1) + 0(1) = -1\\ d_{13} &= -1(3) + 1(0) + 0(0) = -3 \end{aligned}$$

   Row 2 of $B$ is $(0, -1, 1)$⁠:

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

   Row 3 of $B$ is $(2, 3, 4)$⁠:

   $$\begin{aligned} d_{31} &= 2(1) + 3(0) + 4(1) = 6\\ d_{32} &= 2(2) + 3(1) + 4(1) = 11\\ d_{33} &= 2(3) + 3(0) + 4(0) = 6 \end{aligned}$$ $$BA = \begin{bmatrix} -1 & -1 & -3 \\ 1 & 0 & 0 \\ 6 & 11 & 6 \end{bmatrix}$$
   For example $d_{32} = 4 + 3 + 4 = 11$⁠. Every entry is a different sum from the one in the same position of $AB$⁠, because rows now come from $B$⁠.

4. Compare

   Both are $3\times3$⁠, but in position $(1,1)$⁠, $AB$ has 5 and $BA$ has $-1$⁠. Since $5 \ne -1$⁠, $AB \ne BA$.

   A single differing position settles inequality. (Comparing position by position, in fact all nine entries differ: for instance $(2,3)$ has $1$ in $AB$ and $0$ in $BA$⁠.)

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