# Tutorial Video

Lets first look at the addition of scalars i.e. 1 by 1 matrices. Let's now for the sake of conceptualisation prescribe the values we have to objects. If we have 3 pens at home and order 5 more pens then when our order arrives we will have:

3 <span style="color: #3366ff;">pens</span> + 5 <span style="color: #3366ff;">pens</span> = 8 <span style="color: #3366ff;">pens</span>

$\displaystyle \left[ {\text{3 pens}} \right]+\left[ {\text{5 pens}} \right]=\left[ {8\text{ pens}} \right]$

$\displaystyle \left[ 3 \right]+\left[ 5 \right]=\left[ 8 \right]$

$\displaystyle 3+5=8$

To write this in MATLAB we would use

3+5

Now let us conceptualise a more complicated example. Assume we have 3 pens and 2 pads at home, we then make a order for 5 more pens and 3 more pads.

Since pens and pads are separate objects we classify them as such. Let's essentially treat each item as a scalar:

<span style="background-color: #f2f4f5; color: #222222;">3 </span><span style="color: #3366ff;">pens</span><span style="background-color: #f2f4f5; color: #222222;"> + 5 </span><span style="color: #3366ff;">pens</span><span style="background-color: #f2f4f5; color: #222222;"> = 8 </span><span style="color: #3366ff;">pens</span>

$\displaystyle \left[ {\text{3 pens}} \right]+\left[ {\text{5 pens}} \right]=\left[ {8\text{ pens}} \right]$

$\displaystyle \left[ 3 \right]+\left[ 5 \right]=\left[ 8 \right]$

$\displaystyle 3+5=8$

<span style="background-color: #f2f4f5; color: #222222;">2 </span><span style="color: #ff0000;">pads</span><span style="background-color: #f2f4f5; color: #222222;"> + 3 </span><span style="color: #ff0000;">pads</span><span style="background-color: #f2f4f5; color: #222222;"> = 5 </span><span style="color: #ff0000;">pads</span>

$\displaystyle \left[ {\text{2 pads}} \right]+\left[ {\text{3 pads}} \right]=\left[ {5\text{ pads}} \right]$

$\displaystyle \left[ 2 \right]+\left[ 3 \right]=\left[ 5 \right]$

$\displaystyle 2+3=5$

If these are instead written as a column vector we will have:

$\displaystyle \left[ {\begin{array}{*{20}{c}} {3\text{ pens}} \\ {2\text{ pads}} \end{array}} \right]+\left[ {\begin{array}{*{20}{c}} {\text{5 pens}} \\ {3\text{ pads}} \end{array}} \right]=\left[ {\begin{array}{*{20}{c}} {8\text{ pens}} \\ {5\text{ pads}} \end{array}} \right]$

$\displaystyle \left[ {\begin{array}{*{20}{c}} 3 \\ 2 \end{array}} \right]+\left[ {\begin{array}{*{20}{c}} 5 \\ 3 \end{array}} \right]=\left[ {\begin{array}{*{20}{c}} 8 \\ 5 \end{array}} \right]$

To write this in MATLAB we would use

[3;2]+[5;3]

If I instead had 2 pens on my desk and I ordered 3 pads then in column vector notation I would have:

$\displaystyle \left[ {\begin{array}{*{20}{c}} {\text{2 pens}} \\ {0\text{ pads}} \end{array}} \right]+\left[ {\begin{array}{*{20}{c}} {\text{0 pens}} \\ {3\text{ pads}} \end{array}} \right]=\left[ {\begin{array}{*{20}{c}} {2\text{ pens}} \\ {3\text{ pads}} \end{array}} \right]$

With this notation bare in mind that pens and pads are separate objects meaning a pen cannot transform into a pad and a pad cannot transform into a pen!

$\displaystyle \left[ {\begin{array}{*{20}{c}} 2 \\ 0 \end{array}} \right]+\left[ {\begin{array}{*{20}{c}} 0 \\ 3 \end{array}} \right]=\left[ {\begin{array}{*{20}{c}} 2 \\ 3 \end{array}} \right]$

To write this in MATLAB we would use

[2;0]+[0;3]

Note how we have to use zeros here to denote that we don't have a pad or pen in the first and second column vectors respectively. Element by Element operations will only work if the Arrays have the same amount of Elements.Â

Okay so far, so good. Let's make things slightly more complicated now. Assume there are two neighbours, neighbour 1 lives in the red house and he has 3 pens and 2 pads, he makes an order for 5 pens and 3 pads and neighbour 2 lives in the green house and she has 2 pens and 2 pads, she makes an order for 7 pens and 5 pads. In this case after the order the man in the red house would have 8 pens and 5 pads whilst the woman would have 9 pens and 7 pads.

$\displaystyle \left[ {\begin{array}{*{20}{c}} {\text{3 pens}} & {2\text{ pens}} \\ {\text{2 pads}} & {\text{2 pads}} \end{array}} \right]+\left[ {\begin{array}{*{20}{c}} {\text{5 pens}} & {\text{7 pens}} \\ {\text{3 pads}} & {5\text{ pads}} \end{array}} \right]=\left[ {\begin{array}{*{20}{c}} {\text{8 pens}} & {\text{9 pens}} \\ {5\text{ pads}} & {7\text{ pads}} \end{array}} \right]$

$\displaystyle \left[ {\begin{array}{*{20}{c}} 3 & 2 \\ 2 & 2 \end{array}} \right]+\left[ {\begin{array}{*{20}{c}} 5 & 7 \\ 3 & 5 \end{array}} \right]=\left[ {\begin{array}{*{20}{c}} 8 & 9 \\ 5 & 7 \end{array}} \right]$

To write this in MATLAB we would use

[3,2;2,2]+[5,7;3,5]

Now obviously we can conceive of more complicated scenarios where there are say 5 neighbours and they each have and order 7 different item types of stationary which would give a 7 by 5 matrix opposed to the 2 by 2 case above.

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