Why A Times B Is Not Always B Times A for Matrices

Adding matrices works entry by entry, but multiplying them combines rows of one matrix with columns of the other. That one change makes the order matter.

This part covers matrix multiplication, why it is not commutative, the transpose, and symmetric and skew symmetric matrices with the uniqueness of the inverse.

**A product $AB$ exists only when the number of columns of $A$ equals the number of rows of $B$, and each entry of $AB$ is a row of $A$ multiplied entry by entry with a column of $B$ and added; matrix multiplication is associative and distributive over addition.

The rule.** If $A$ is $m\times n$ and $B$ is $n\times p$, then $AB$ is $m\times p$, with

$$c_{ik}=a_{i1}{b}_{1k}+a_{i2}{b}_{2k}+\cdots +a_{in}{b}_{nk}$$

Worked example. Let $A=\left[\begin{array}{cc}1& 2\\ 3& 4\end{array}\right]$ and $B=\left[\begin{array}{cc}2& 0\\ 1& 5\end{array}\right]$.

$$AB=\left[\begin{array}{cc}4& 10\\ 10& 20\end{array}\right]$$

For instance, row 1 with column 2 gives $1\cdot 0+2\cdot 5=10$.

Properties, whenever the products are defined:

- Associative — $(AB)C=A(BC)$
- Distributive — $A(B+C)=AB+AC$ and $(A+B)C=AC+BC$

An everyday example. A canteen's order matrix of idli and dosa plates for each class, multiplied by a price column, gives every class's bill in one step.

The substance. **A $2\times 3$ matrix can multiply a $3\times 4$ matrix, but not the other way round** — whether a product exists depends on the order.

**In general $AB\ne BA$, even when both products exist, and two non-zero matrices can have a zero product — so neither the commutative law nor the rule that a zero product needs a zero factor holds for matrices.

Worked example — order matters.** With the same $A$ and $B$:

$$BA=\left[\begin{array}{cc}2& 0\\ 1& 5\end{array}\right]\left[\begin{array}{cc}1& 2\\ 3& 4\end{array}\right]=\left[\begin{array}{cc}2& 4\\ 16& 22\end{array}\right]\ne AB$$

Worked example — a zero product.

$$\left[\begin{array}{cc}0& 1\\ 0& 0\end{array}\right]\left[\begin{array}{cc}1& 0\\ 0& 0\end{array}\right]=\left[\begin{array}{cc}0& 0\\ 0& 0\end{array}\right]$$

Consequences:

- $(A+B{)}^2=A^2+AB+BA+B^2$, which equals $A^2+2AB+B^2$ only if $AB=BA$
- From $AB=AC$ you cannot cancel $A$ to conclude $B=C$ in general

An everyday example. Putting on socks and then shoes is not the same as shoes and then socks — as with matrices, the order of the steps changes the result.

The substance. Some pairs do commute — every square matrix commutes with $I$ and with itself.

**The transpose $A^{\prime }$ turns the rows of $A$ into columns, so an $m\times n$ matrix becomes $n\times m$, and transposes obey $(A^{\prime }{)}^{\prime }=A$, $(A+B{)}^{\prime }=A^{\prime }+B^{\prime }$, $(kA{)}^{\prime }=k{A}^{\prime }$ and the reversal rule $(AB{)}^{\prime }=B^{\prime }{A}^{\prime }$.

Definition.** If $A=[a_{ij}{]}_{m\times n}$, then $A^{\prime }=[a_{ji}{]}_{n\times m}$.

Worked example — verify the reversal rule. From before, $AB=\left[\begin{array}{cc}4& 10\\ 10& 20\end{array}\right]$, so $(AB{)}^{\prime }=\left[\begin{array}{cc}4& 10\\ 10& 20\end{array}\right]$. Now

$$B^{\prime }{A}^{\prime }=\left[\begin{array}{cc}2& 1\\ 0& 5\end{array}\right]\left[\begin{array}{cc}1& 3\\ 2& 4\end{array}\right]=\left[\begin{array}{cc}4& 10\\ 10& 20\end{array}\right]=(AB{)}^{\prime }$$

In the other order, $A^{\prime }{B}^{\prime }=\left[\begin{array}{cc}2& 16\\ 4& 22\end{array}\right]$, which is why the order must reverse.

An everyday example. Turning a marks register sideways, so subjects become rows and students become columns, is exactly taking a transpose — the data stay the same, only the layout flips.

The substance. The reversal rule extends to longer products: $(ABC{)}^{\prime }=C^{\prime }{B}^{\prime }{A}^{\prime }$.

**A square matrix is symmetric if $A^{\prime }=A$ and skew symmetric if $A^{\prime }=-A$; every square matrix equals $\frac{1}{2}(A+A^{\prime })+\frac{1}{2}(A-A^{\prime })$, a symmetric plus a skew symmetric matrix, and an invertible matrix has exactly one inverse.

Identifying them:

- Symmetric** — $a_{ij}=a_{ji}$; entries mirrored across the main diagonal are equal
- Skew symmetric — $a_{ij}=-a_{ji}$; in particular $a_{ii}=-a_{ii}$, so **every diagonal entry is $0$

Worked example — the split.** Let $A=\left[\begin{array}{cc}3& 5\\ 1& -2\end{array}\right]$, so $A^{\prime }=\left[\begin{array}{cc}3& 1\\ 5& -2\end{array}\right]$.

$$P=\frac{1}{2}(A+A^{\prime })=\left[\begin{array}{cc}3& 3\\ 3& -2\end{array}\right]Q=\frac{1}{2}(A-A^{\prime })=\left[\begin{array}{cc}0& 2\\ -2& 0\end{array}\right]$$

$P$ is symmetric, $Q$ is skew symmetric, and $P+Q=A$.

Uniqueness of the inverse. If $B$ and $C$ are both inverses of $A$, then $AB=BA=I$ and $AC=CA=I$, so

$$B=BI=B(AC)=(BA)C=IC=C$$

An everyday example. A road distance chart between cities, where Pune to Nagpur equals Nagpur to Pune, is a symmetric matrix with zeros on its diagonal.

The substance. The zero matrix is the only matrix that is both symmetric and skew symmetric, since $A^{\prime }=A$ and $A^{\prime }=-A$ together force $A=-A$.