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

Question

Accepted Answer

Both matrices are $2	imes2$, so $AB$ exists and is $2	imes2$. Using the row-by-column rule: $$AB=\begin{pmatrix} 1(2)+2(1) & 1(0)+2(3) \ 3(2)+4(1) & 3(0)+4(3) \end{pmatrix}=\begin{pmatrix} 2+2 & 0+6 \ 6+4 & 0+12 \end{pmatrix}=\begin{pmatrix} 4 & 6 \ 10 & 12 \end{pmatrix}.$$

**Verification of the $(2,1)$ entry.** Row $2$ of $A$ is $(3,4)$ and column $1$ of $B$ is $(2,1)$: $3(2)+4(1)=6+4=10$, matching the value above.

> [!ANSWER]
> $AB=\begin{pmatrix} 4 & 6 \ 10 & 12 \end{pmatrix}$.

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

Concept understanding — Algebra of Matrices

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

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