# Matrices (Pt. 2): MM

## Matrix Multiplication

This is probably where 💩 gets slightly difficult, but it is also where the crux of all data transformation happens 😌. I will firstly walk-through the **mechanical operation**, and then subsequently **a few ways to think** about it in the next page.

### Matrix-Vector Multiplication Operation

*Each row* of matrix evaluates with its corresponding *column vector*

$$
\left\[\begin{array}{cc}
1 & 2\\
3 & 4
\end{array}\right]
\times
\left\[\begin{array}{cc}
5 \\
6
\end{array}\right]
==================

\left\[\begin{array}{cc}
(1\times5) + (2\times 6)\\
(3\times 5) + (4\times 6)
\end{array}\right]  =
\left\[\begin{array}{cc}
17\\
39
\end{array}\right]
$$

We can break-up the above operation to help assist in understanding.

#### Step 1:

![Row 1 of Matrix A, interacts with Col 1 of Matrix B](/files/-Lx1ngneR8Lxlu0C5cuF)

{% hint style="info" %}

* A(1, 1) multiplies with B(1, 1).  
* A(1, 2) multiplies with B(2, 1).
* The result of C(1, 1) is the summation of elements involved in the interaction from row 1 of Matrix A and col 1 of Matrix B
  {% endhint %}

#### Step 2:

![Row 2 of Matrix A, interacts with Col 1 of Matrix B](/files/-Lx1nsdXPvJ8wfkEMhsN)

{% hint style="info" %}

* A(2, 1) multiplies with B(1, 1).  
* A(2, 2) multiplies with B(2, 1).
* The result of C(2, 1) is the summation of elements involved in the interaction from row 2 of Matrix A and col 1 of Matrix B
  {% endhint %}

#### Test Yourself!

Now, try to perform the following problem before revealing its answer! 👍

{% tabs %}
{% tab title="Question" %}
$$
\left\[\begin{array}{cc}
3 & 2  & 1\\
4 & 5 & 6 \\
7 & 8 & 9
\end{array}\right]
\times
\left\[\begin{array}{cc}
2 \\
3 \\
4
\end{array}\right]\
\=  ?
$$
{% endtab %}

{% tab title="Answer" %}
$$
\left\[\begin{array}{cc}
3 & 2  & 1\\
4 & 5 & 6 \\
7 & 8 & 9
\end{array}\right]
\times
\left\[\begin{array}{cc}
2 \\
3 \\
4
\end{array}\right]\
\\=
\left\[\begin{array}{cc}
(3\times 2) + (2\times 3) + (1\times 4)\\
(4\times 2) + (5\times 3) + (6\times 4)\\
(7\times 2) + (8 \times 3) + (9\times 4)
\end{array}\right]
==================

\left\[\begin{array}{cc}
16\\
47\\
74
\end{array}\right]
$$
{% endtab %}
{% endtabs %}

### Matrix-Matrix Multiplication Operation

*Each row* of matrix evaluates with each of its *respective* *column vector*.

* Number of columns (n) of left matrix must be equal to number of rows (m) of right matrix

$$
\left\[\begin{array}{cc}
1 & 4  & 7\\
2 & 5 & 8 \\
3 & 6 & 9
\end{array}\right]
\times
\left\[\begin{array}{cc}
2 & 6 \\
1 & 3  \\
6 & 9
\end{array}\right]
==================

\left\[\begin{array}{cc}
48 & 81 \\
77 &  99 \\
66 & 117
\end{array}\right]
$$

Let us break up this problem like we've had previously. 💔

1. We begin with the **first row** of the *left-matrix*, and 
2. Multiply it with the **first column** of the *right-matrix*, with 
3. The output element thus occupying the space (1, 1) 

![Output element (1, 1)](/files/-Lx6G8HCC4mT7y3cxYV4)

![Output element (1, 2)](/files/-Lx6GYaoUpj8Bmq9cr48)

![Output element (1, 3)](/files/-Lx6HGrIaOzLwHsLQvg7)

![Output element (2, 1)](/files/-Lx6Hr2PpRaOpgVoBKR2)

![Output element (2, 2)](/files/-Lx6IV0J7fgE0WgU2TmZ)

![Output element (2, 3)](/files/-Lx6I_zhEHXvHEegBdpR)

## Useful Tricks

### Size of output

With the previous example laid out, you might question **"What would happen if there are not enough elements in my right-matrix to interact with my left?".**

With that being a valid question, math simply answers back with a "**you can't**" 🥁. Hence, the size of both matrices and the non-commutative property (discussed at next page) are important to ensure valid operations. 

To recap, we would require 

* The same amount of *columns* in the *left-matrix*, to  
* the *rows* of the *right-matrix*

For example, a left-matrix of size **m x n,** has to interact with a right-matrix of size **n x o**, with m and o being any integers greater than or equals to 1. 

Finally, the following might be a trivial, but yet a useful trick to find the size of the output vector.

![Size of output](/files/-Lx4pQ-xlqJirXwH8dQx)

As it is a requirement for **"n"** (value 2) to be the same,

* With the values of **"m"** (value 2), and
* **"o"** (value 1),
* The size of output variable of **m x o** (2 x 1) can be seen

{% hint style="success" %}
The output matrix size is the rows of the left-matrix, by the columns of the right-matrix **(m x o)**. 
{% endhint %}

### Standard Method (rows times columns)

To recap from the previous page, an element has the notation of **a(i,j)**, 

* where **i** is the *row number*,
* and **j** is the *column number*

![Notation of Element](/files/-Lx4vy5cs4SlcPN4ETTd)

Supposed the fact that the above matrix is the *output matrix X* in the operation (a . b = x)*,* 

{% hint style="info" %}
What then were the **factors of A and B** used for the computation of a particular element in X?
{% endhint %}

You may then recall that we previously used the count of *row numbers* of the left-matrix, and the count of *column numbers* of the right-matrix to determine the output matrix size. 

Similarly, for each output element x(i, j), its factors are the resultant interaction between the 

* **"i-th" row** of the left matrix, and 
* **the "j-th" column** of the right-matrix.

This would otherwise be also the **standard way** to think about matrix multiplication, with:

$$
x\_{\mathrm{i,j}} =
\sum\_{\mathrm{k=1}}^k
a\_{\mathrm{i,n}}b\_{\mathrm{n,j}}
$$

The following question helps illustrates the discussion. 

Let's say we arbitrarily wanted to check the value and the method we obtained x(2, 3). 

![Element of Output Matrix x(2, 3)](/files/-Lx4wSJfsQlCBxOJQVwT)

How do we go about doing so? Easy!

We can do so simply via the **2nd row** of the left-matrix, and the **3rd column** of the right-matrix.

![Factors involved in computing x(2, 3)](/files/-Lx4wnRzoYr7ellNH--s)

### Columns Method

The *output matrix X* can be alternatively thought of as 

* A combination of 3 separate columns, from
* Its **corresponding column** in *right-matrix B,*
* Multiplied by *matrix A*

![(A\*Cols of B) represents Cols of X](/files/-Lx6U_x2ARCuVQDQxNhm)

### Row Method

Alternatively, the *output matrix X* can be alternatively thought of as 

* A combination of 3 separate rows, from
* Its **corresponding row** in *left-matrix* A*,*
* Multiplied by *matrix B*

![(Rows of A\*B) represents Rows of X](/files/-Lx6XLE0yZrpE6jw_TOG)

#### **More on the next page ⏭**


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