Determine the product AB of the matrices A with rows (1, 3, 2), (2, -1, -3), (1, 1, 0) and B with rows (-3, 2, 4), (1, 0, 2), (0, 2, 3).

Step-by-step solution

1. 1

   Check the shapes and fix the rule. $A$ is $3\times 3$ and $B$ is $3\times 3$, so the inner dimensions match and $AB$ is $3\times 3$. The entry in row $i$, column $j$ is the dot product of row $i$ of $A$ with column $j$ of $B$:

   $(AB)_{ij}=\sum_{k=1}^{3}A_{ik}B_{kj}.$

   Multiplying entry by entry in the same position is the classic error — that is not matrix multiplication.

2. 2

   Compute row 1 of the product. Row $1$ of $A$ is $(1,3,2)$:

   $(1)(-3)+(3)(1)+(2)(0)=0,$
   $(1)(2)+(3)(0)+(2)(2)=6,$
   $(1)(4)+(3)(2)+(2)(3)=16.$

3. 3

   Compute row 2. Row $2$ of $A$ is $(2,-1,-3)$:

   $(2)(-3)+(-1)(1)+(-3)(0)=-7,$
   $(2)(2)+(-1)(0)+(-3)(2)=-2,$
   $(2)(4)+(-1)(2)+(-3)(3)=-3.$

4. 4

   Compute row 3. Row $3$ of $A$ is $(1,1,0)$, so the third column of $B$ never contributes through its bottom entry:

   $(1)(-3)+(1)(1)+(0)(0)=-2,$
   $(1)(2)+(1)(0)+(0)(2)=2,$
   $(1)(4)+(1)(2)+(0)(3)=6.$

5. 5

   Assemble the product.

   $AB=\begin{pmatrix}0&6&16\\ -7&-2&-3\\ -2&2&6\end{pmatrix}.$

6. 6

   Spot-check with a column vector. Multiplying $B$ by $(1,0,0)^{T}$ picks out its first column $(-3,1,0)^{T}$; then $A(-3,1,0)^{T}=(-3+3+0,\ -6-1+0,\ -3+1+0)^{T}=(0,-7,-2)^{T}$, which is exactly the first column of $AB$ ✓. Remember also that matrix multiplication is not commutative: $BA$ is a different matrix.