Why AB and BA Are Usually Different Matrices

Numbers can be multiplied in any order, and a product is zero only when a factor is zero. Matrices break both rules. Understanding why — and when a matrix can be undone by an inverse — is the key to solving equations with matrices.

This part covers matrix multiplication and non-commutativity, zero products of non-zero matrices, and singular matrices with the uniqueness of the inverse.

**The product AB exists only when the number of columns of A equals the number of rows of B; an $m\times n$ matrix times an $n\times p$ matrix is an $m\times p$ matrix whose entries are row-by-column sums of products, and in general $AB\ne BA$.

Rule.** $(AB{)}_{ij}=a_{i1}{b}_{1j}+a_{i2}{b}_{2j}+\cdots +a_{in}{b}_{nj}$.

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

$$AB=\left[\begin{array}{cc}2& 1\\ 4& 3\end{array}\right],BA=\left[\begin{array}{cc}3& 4\\ 1& 2\end{array}\right]$$

So $AB\ne BA$: multiplying by B on the right swaps the columns of A, while multiplying on the left swaps its rows.

Orders can block a product. If A is $2\times 3$ and B is $3\times 4$, then AB is $2\times 4$, but BA does not exist.

Properties that do hold: $(AB)C=A(BC)$, $A(B+C)=AB+AC$ and $AI=IA=A$.

An everyday example. **Two customers buying rice and dal in amounts $\left[\begin{array}{cc}2& 1\\ 1& 3\end{array}\right]$ kg, at prices $\left[\begin{array}{c}60\\ 120\end{array}\right]$ rupees per kg,** get the bills $\left[\begin{array}{c}240\\ 420\end{array}\right]$ — each row of quantities multiplied by the column of prices.

The substance. Non-commutativity changes familiar algebra — $(A+B{)}^2=A^2+AB+BA+B^2$, which equals $A^2+2AB+B^2$ only when $AB=BA$.

**Yes — unlike numbers, two non-zero matrices can have the zero matrix as their product, so $AB=O$ does not force $A=O$ or $B=O$.

Worked example.** Let $A=\left[\begin{array}{cc}1& 1\\ 1& 1\end{array}\right]$ and $B=\left[\begin{array}{cc}1& -1\\ -1& 1\end{array}\right]$.

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

Neither A nor B is the zero matrix.

Worked example 2. $C=\left[\begin{array}{cc}0& 1\\ 0& 0\end{array}\right]$ satisfies $C^2=O$, even though $C\ne O$.

Consequences:

- The cancellation law fails: $AB=AC$ does not imply $B=C$ unless A has an inverse
- An equation such as $X^2=O$ can have non-zero solutions

An everyday example. A stock-sold matrix and a matrix of price changes that exactly cancel across items can multiply to the zero matrix — a total revenue change of zero — even though neither matrix is zero.

The substance. This happens only with singular matrices — as the next section shows, a matrix with an inverse can always be cancelled.

**A square matrix A is singular when $|A|=0$ and non-singular when $|A|\ne 0$; only a non-singular matrix has an inverse, a matrix B with $AB=BA=I$, and that inverse is unique.

Singular or not.** For $A=\left[\begin{array}{cc}a& b\\ c& d\end{array}\right]$, $|A|=ad-bc$.

- $\left[\begin{array}{cc}2& 3\\ 4& 6\end{array}\right]$ has determinant $12-12=0$, so it is singular
- $\left[\begin{array}{cc}2& 1\\ 5& 3\end{array}\right]$ has determinant $6-5=1$, so it is non-singular

Worked example. For $A=\left[\begin{array}{cc}2& 1\\ 5& 3\end{array}\right]$, the inverse is $A^{-1}=\left[\begin{array}{cc}3& -1\\ -5& 2\end{array}\right]$. Check:

$$A{A}^{-1}=\left[\begin{array}{cc}6-5& -2+2\\ 15-15& -5+6\end{array}\right]=\left[\begin{array}{cc}1& 0\\ 0& 1\end{array}\right]$$

Proof of uniqueness. Suppose B and C are both inverses of A. Then

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

so the two inverses are the same matrix.

An everyday example. A matrix used to scramble a message must be non-singular, or the receiver has no way to decode it.

The substance. The rows of a singular matrix are dependent — in $\left[\begin{array}{cc}2& 3\\ 4& 6\end{array}\right]$ the second row is twice the first, which is why no inverse exists.