Example 15: $AB=O$ with $A\ne O$⁠, $B\ne O$

Basics you need

Solution

1. Check the product is defined, and that neither factor is zero

   $A$ and $B$ are both $2\times2$⁠: inner sizes $2=2$⁠, so $AB$ is defined and is $2\times2$⁠.

   $A\ne O$ because its $(1,2)$ entry is $-1\ne0$⁠. $B\ne O$ because its $(1,1)$ entry is $3\ne0$⁠.

   The point of the example is that a zero product can come from non-zero factors, so we first make sure the factors really are non-zero. One non-zero entry is enough for that.
   Zero matrix, Section 3.3 (vii): all entries are zero. Basics: rows, columns and the address

2. Compute the four entries of $AB$

   Rows of $A$⁠: $[0\ \ {-1}]$ and $[0\ \ 2]$⁠. Columns of $B$⁠: $(3,0)$ and $(5,0)$⁠.

   $$\begin{aligned} c_{11} &= 0\cdot 3 + (-1)\cdot 0 = 0 + 0 = 0\\ c_{12} &= 0\cdot 5 + (-1)\cdot 0 = 0 + 0 = 0\\ c_{21} &= 0\cdot 3 + 2\cdot 0 = 0 + 0 = 0\\ c_{22} &= 0\cdot 5 + 2\cdot 0 = 0 + 0 = 0 \end{aligned}$$ $$AB = \begin{bmatrix} 0 & 0 \\ 0 & 0 \end{bmatrix} = O$$
   Each sum has two products, and in every one of them a factor is 0. Notice the pattern: the first product always uses $a_{i1}$⁠, which is 0, and the second always uses $b_{2k}$⁠, which is 0.
   $c_{ik}=a_{i1}b_{1k}+a_{i2}b_{2k}$⁠. Basics: pair up, multiply, add

3. Why every entry is zero

   For every $i$ and $k$⁠,

   $$c_{ik} = \underbrace{a_{i1}}_{=\,0}\,b_{1k} + a_{i2}\,\underbrace{b_{2k}}_{=\,0} = 0.$$

   Column 1 of $A$ is all zeros, so the first term vanishes. Row 2 of $B$ is all zeros, so the second term vanishes. Hence $AB=O$ although $A\ne O$ and $B\ne O$⁠.

   This explains the result instead of just observing it. The second entry of each row of $A$ is the only non-zero part, and in the product it always meets the second entry of a column of $B$⁠, which is always zero. Any matrices with this pattern of zeros, whatever their other entries, would also multiply to $O$⁠.

4. Order still matters: $BA$ is not zero

   Rows of $B$⁠: $[3\ \ 5]$ and $[0\ \ 0]$⁠. Columns of $A$⁠: $(0,0)$ and $(-1,2)$⁠. Write $BA=[d_{ik}]$⁠:

   $$\begin{aligned} d_{11} &= 3\cdot 0 + 5\cdot 0 = 0\\ d_{12} &= 3\cdot(-1) + 5\cdot 2 = -3 + 10 = 7\\ d_{21} &= 0\cdot 0 + 0\cdot 0 = 0\\ d_{22} &= 0\cdot(-1) + 0\cdot 2 = 0 \end{aligned}$$ $$BA = \begin{bmatrix} 0 & 7 \\ 0 & 0 \end{bmatrix} \ne O$$
   This is not asked, but it is worth seeing: $AB=O$ tells us nothing about $BA$⁠. In the reverse order the non-zero row of $B$ meets the non-zero column of $A$⁠. It is one more instance of $AB\ne BA$ (compare Example 14).

5. What this means: no cancelling in matrix equations

   Since $AB=O$ and also $AO=O$⁠, we have $AB=AO$ with $A\ne O$⁠, yet $B\ne O$⁠. So from $AB=AC$ we may not conclude $B=C$ just because $A\ne O$⁠.

   With numbers, $ab=ac$ and $a\ne0$ give $b=c$ by dividing by $a$⁠. For matrices, "dividing by $A$" needs an inverse matrix, which a non-zero matrix need not have. This $A$ has none. Suppose some $2\times2$ matrix $X$ had $XA=I$⁠. Then $$\begin{aligned} B &= IB = (XA)B\\ &= X(AB) = XO = O, \end{aligned}$$ which contradicts $B\ne O$⁠. So no such $X$ exists, and there is nothing to "divide by".
   The identity matrix is defined in Section 3.3 (vi); inverse matrices come in Section 3.7. Basics: identity and undo (inverses)

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