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This is from my textbook:

If $A=(a_{ij})\in M_{mn}(\Bbb F), B=(b_{ij})\in M_{np}(\Bbb F)$ then $C=A\times B=(c_{ij})\in M_{mp}(\Bbb F)$. $c_{ij}=\sum_{k=1}^{n} a_{ik}b_{kj}$ where $i=1,...m, j=1,...p$

I know how to multiply matrices but I don't understand this notation : $c_{ij}=\sum_{k=1}^{n} a_{ik}b_{kj}$

Can someone explain what that represents by giving me an example? And how did we get that formula?

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  • $\begingroup$ Example: To find $c_{11}$, you multiply the first row of matrix $A$ with the first column of matrix $B$. $\endgroup$ – Prahlad Vaidyanathan Dec 18 '16 at 10:18
  • $\begingroup$ $c_{ij}$ is the element on $i$-th row and $j$-th column of $C$. Similarly, $a_{ik}$, $b_{kj}$ are elements of matrices $B$ and $C$. $\endgroup$ – Peter Franek Dec 18 '16 at 10:19
  • $\begingroup$ If you know coding, this is how we multiply two matrices. $\endgroup$ – Rohan Dec 18 '16 at 10:20
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Visualisation might help. I'll use your notations and dimensions of the given matrices:

If $A=(a_{ij})\in M_{mn}(\Bbb F), B=(b_{ij})\in M_{np}(\Bbb F)$ then $C=A\times B=(c_{ij})\in M_{mp}(\Bbb F)$. $c_{ij}=\sum_{k=1}^{n} a_{ik}b_{kj}$ where $i=1,...m, j=1,...p$

You say you know how to multiply matrices, so take a look at one specific element in the product $C=AB$, namely the element on position $(i,j)$, i.e. in the $i$th row and $j$th column.

To obtain this element, you:

  • first multiply all elements of the $i$th row of the matrix $A$ pairwise with all the elements of the $j$th column of the matrix $B$;
  • and then you add these $n$ products.

You have to repeat this procedure for every element of $C$, but let's zoom in on that one specific (but arbitrary) element on position $(i,j)$ for now:

$$\begin{pmatrix} a_{11} &\ldots &a_{1n}\\ \vdots& \ddots &\vdots\\ \color{blue}{\mathbf{a_{i1}}} &\color{blue}{\rightarrow} &\color{blue}{\mathbf{a_{in}}}\\ \vdots& \ddots &\vdots\\ a_{m1} &\ldots &a_{mn} \end{pmatrix} \cdot \begin{pmatrix} b_{11}&\ldots &\color{red}{\mathbf{b_{1j}}} &\ldots &b_{1p}\\ \vdots& \ddots &\color{red}{\downarrow} & \ddots &\vdots\\ b_{n1}&\ldots &\color{red}{\mathbf{b_{nj}}}&\ldots &b_{np} \end{pmatrix} = \begin{pmatrix} c_{11}&\ldots& c_{1j} &\ldots &c_{1p}\\ \vdots& \ddots & & &\vdots\\ c_{i1}& & \color{purple}{\mathbf{c_{ij}}} & &c_{ip}\\ \vdots& & & \ddots &\vdots\\ c_{m1} &\ldots& c_{mj} &\ldots &c_{mp} \end{pmatrix}$$ with element $\color{purple}{\mathbf{c_{ij}}}$ equal to: $$\mathbf{\color{purple}{c_{ij}} = \color{blue}{a_{i1}} \color{red}{b_{1j}} + \color{blue}{a_{i2}} \color{red}{b_{2j}} + \cdots + \color{blue}{a_{in}} \color{red}{b_{nj}}}$$ Now notice that in the sum above, the left outer index is always $i$ ($i$th row of $A$) and the right outer index is always $j$ ($j$th column of $B$). The inner indices run from $1$ to $n$ so you can introduce a summation index $k$ and write this sum compactly using summation notation: $$\color{purple}{\mathbf{c_{ij}}}=\sum_{k=1}^{n} \color{blue}{\mathbf{a_{ik}}}\color{red}{\mathbf{b_{kj}}}$$ The formule above thus gives you the element on position $(i,j)$ in the product matrix $C=AB$ and therefore completely defines $C$ by letting $i=1,...,m$ and $j=1,...,p$.


Can someone explain what that represents by giving me an example? And how did we get that formula?

The illustration above should give you an idea of the general formula, but here's a concrete example where I took $3 \times 3$ matrices for $A$ and $B$ and focus on the element on position $(2,3)$:

$$\begin{pmatrix} a_{11} & a_{12} &a_{13}\\ \color{blue}{1} &\color{blue}{2} &\color{blue}{3}\\ a_{31} & a_{32} &a_{33} \end{pmatrix} \cdot \begin{pmatrix} b_{11}&b_{12} &\color{red}{6}\\ b_{21}&b_{22} &\color{red}{5}\\ b_{31}&b_{32} &\color{red}{4} \end{pmatrix} = \begin{pmatrix} c_{11}& c_{12} &c_{13}\\ c_{21}& c_{22} &\color{purple}{\mathbf{c_{23}}}\\ c_{31}& c_{32} &c_{33} \end{pmatrix}$$ with element $\color{purple}{\mathbf{c_{23}}}$ equal to: $$\begin{array}{rccccccc} \color{purple}{c_{23}} & = & \color{blue}{a_{21}} \color{red}{b_{13}} &+& \color{blue}{a_{22}} \color{red}{b_{23}} &+& \color{blue}{a_{23}} \color{red}{b_{33}} & = & \displaystyle \sum_{k=1}^{3} \color{blue}{a_{2k}}\color{red}{b_{k3}} \\ & = & \color{blue}{1} \cdot \color{red}{6} &+& \color{blue}{2} \cdot \color{red}{5} &+& \color{blue}{3} \cdot \color{red}{4} \\[8pt] & = & 6&+&10&+&12 & =& 28 \end{array}$$

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As you can see the sum run over $k$ from $1$ to $n$, but after you fix $i,j$. You said that you know how to multiply matrices, so you know that if $C=AB$, then $(i,j)-th$ element $c_{ij}$ is obtained fixing the $i-th$ row of $A$, the $j-th$ column of $B$, and then you multiply the elements $a_{i1}$ with $b_{1j}$, $a_{i2}$ with $b_{2j}$, $\dots$, $a_{in}$ with $b_{nj}$, and finally you sum all products. Thus

$$c_{ij}=\sum_{k=1}^n a_{ik}b_{kj}$$

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