Matrices: Common Mistakes and Fixes (1)

Question

If $A = \begin{pmatrix} 2 & 3 \\ 1 & 4 \end{pmatrix}$ and $B = \begin{pmatrix} 1 & 0 \\ 2 & 1 \end{pmatrix}$, find $AB$, $BA$, and check whether $AB = BA$.

Solution — Step by Step

Final answer: $AB = \begin{pmatrix} 8 & 3 \\ 9 & 4 \end{pmatrix}$, $BA = \begin{pmatrix} 2 & 3 \\ 5 & 10 \end{pmatrix}$, and $AB \ne BA$.

Why This Works

Matrix multiplication is row-of-first dotted with column-of-second. Unlike scalar multiplication, the order changes the answer. Special cases where $AB = BA$ exist (e.g., diagonal matrices, $A$ and $A^{-1}$, $A$ and $I$), but in general expect non-commutativity.

This is why “matrix algebra” is genuinely different from ordinary algebra — every identity from school like $(A+B)^2 = A^2 + 2AB + B^2$ is wrong unless $AB = BA$.

Alternative Method

Use the column-as-linear-combination view: column $j$ of $AB$ equals $A$ acting on column $j$ of $B$. Sometimes faster when one matrix has many zero columns.