## The $$(i,j)$$-entry of the product of two matrices

Given a matrix $$A$$, the $$(i,j)$$-entry of $$A$$ is the entry in row $$i$$ and column $$j$$ of $$A$$ and is normally denoted by $$A_{i,j}$$.

Let $$A$$ be an $$m\times p$$ matrix and let $$B$$ be a $$p \times n$$ matrix. Let $$Q$$ denote the product $$AB$$.

Row $$i$$ of $$Q$$ is given by $$a_i B$$ where $$a_i$$ denotes the $$i$$th row of $$A$$.

For example, if $$A = \begin{bmatrix} 2 & 1 \\ 0 & 3 \\ 4 & 0 \end{bmatrix}$$ and $$B = \begin{bmatrix} -1 & 1 \\ 0 & 3 \end{bmatrix}$$, then the second row of $$AB$$ is given by $$\begin{bmatrix} 0 & 3 \end{bmatrix} \begin{bmatrix} -1 & 1 \\ 0 & 3\end{bmatrix} = \begin{bmatrix} 0 & 9 \end{bmatrix}$$.

Recall from the definition of matrix product that column $$j$$ of $$Q$$ is given by $$A B_j$$ where $$B_j$$ denotes the $$j$$th column of $$B$$.

Therefore, $$Q_{i,j}$$, which is given by column $$j$$ of $$a_iB$$, is $$a_i B_j = A_{i,1} B_{1,j} + A_{i,2} B_{2,j} + \cdots + A_{i,p}B_{p,j}$$.

For the example above, the $$(3,2)$$-entry of the product $$AB$$ is given by $$\begin{bmatrix} 4 & 0 \end{bmatrix} \begin{bmatrix} 1 \\ 3\end{bmatrix} = 4$$.

## Matrix multiplication is associative

Even though matrix multiplication is not commutative, it is associative in the following sense. If $$A$$ is an $$m\times p$$ matrix, $$B$$ is a $$p \times q$$ matrix, and $$C$$ is a $$q \times n$$ matrix, then $A(BC) = (AB)C.$ This important property makes simplification of many matrix expressions possible. In particular, we can simply write $$ABC$$ without having to worry about the order in which multiplication is performed.

To see this, first let $$a_i$$ denote the $$i$$th row of $$A$$. Let $$P$$ denote the product $$BC$$.

The $$(i,j)$$-entry of $$A(BC)$$ is given by $$a_iP_j = A_{i,1} P_{1,j} + A_{i,2} P_{2,j} + \cdots + A_{i,p} P_{p,j}.$$

But $$P_j = BC_j$$. Thus $$P_{s,j} = B_{s,1} C_{1,j} + B_{s,2} C_{2,j} + \cdots + B_{s,q} C_{q,j}$$, giving \begin{eqnarray} a_i P_j & = & A_{i,1} (B_{1,1} C_{1,j} + B_{1,2} C_{2,j} + \cdots + B_{1,q} C_{q,j}) \\ & & + A_{i,2} (B_{2,1} C_{1,j} + B_{2,2} C_{2,j} + \cdots + B_{2,q} C_{q,j}) \\ & & \vdots \\ & & + A_{i,p} (B_{p,1} C_{1,j} + B_{p,2} C_{2,j} + \cdots + B_{p,q} C_{q,j}) \\ & = & (A_{i,1} B_{1,1} + A_{i,2} B_{2,1} + \cdots + A_{i,p} B_{p,1}) C_{1,j} \\ & & + (A_{i,1} B_{1,2} + A_{i,2} B_{2,2} + \cdots + A_{i,p} B_{p,2}) C_{2,j} \\ & & \vdots \\ & & + (A_{i,1} B_{1,q} + A_{i,2} B_{2,q} + \cdots + A_{i,p} B_{p,q}) C_{q,j} \\ & = & (a_i B_1) C_{1,j} + (a_i B_2) C_{2,j} + \cdots + (a_i B_q) C_{q,j}. \end{eqnarray}

Now, let $$Q$$ denote the product $$AB$$. Then $$Q_{i,r} = a_i B_r$$. Hence, the $$(i,j)$$-entry of $$(AB)C$$ is given by $Q_{i,1} C_{1,j} + Q_{i,2} C_{2,j} + \cdots + Q_{i,q} C_{q,j} =(a_iB_1) C_{1,j} + (a_iB_2) C_{2,j} + \cdots + (a_iB_q) C_{q,j} = a_i P_j.$

Hence, the $$(i,j)$$-entry of $$A(BC)$$ is the same as the $$(i,j)$$-entry of $$(AB)C$$. It follows that $$A(BC) = (AB)C$$.

## Exercises

Give the $$(2,2)$$-entry of each of the following.

1. $$\begin{bmatrix} 2 & 1 \\ 0 & 3 \end{bmatrix} \begin{bmatrix} 2 & -1 \\ -1 & 2 \end{bmatrix}$$

2. $$\begin{bmatrix} 1 \\ 2 \\ 3 \\ 4 \end{bmatrix} \begin{bmatrix} 0 & 1 & 2 & 3 \end{bmatrix}$$