Well, we're starting up again. Not sure whether to count this as take 2, 3, or 4, so I took the average. Starting with the same post as last time.

So the last thread, I accidentally turned what was supposed to be a "school" on linear algebra into an exploration of linear algebra through the lens of category theory. Which was useful to me, but probably less useful to the people who were watching the thread. This thread is for those people.

So! Linear algebra.

What is linear algebra? Linear algebra is the study of what you can do when you are restricted to two basic operations - adding things together, and multiplying everything by a number.

Seems restrictive, right? Well, that's a lot of how math works - by restricting what you think you can do, you can make more situations similar - and so you can study things much more deeply.

The thing is, we actually haven't restricted ourselves all too much. Or rather, we've restricted ourselves to a pair of very useful things we can do. Don't believe me?

Every day, Alice buys two apples and one banana for breakfast. Every day, Bob buys one apple and two bananas for breakfast. Last month, Alice paid $5, while Bob paid $4. This month, Alice paid $5, while Bob paid $7.

How did the prices change?

Let the price of an apple be a dollars, and the price of a banana be b dollars. We then get the equations for last month:

2a + b = 5

a + 2b = 4

So let's solve. Double the second equation - multiply all the numbers in it by 2. That still gives us a valid equation. The cost of 2 apples and 4 bananas would be 8 dollars.

2a + b = 5

2a + 4b = 8

Now, if it cost 5 dollars to get 2 apples and a banana, and 8 dollars to get 2 apples and 4 bananas, it stands to reason (assuming no discounts) that the difference in cost comes from the difference in goods. So we can subtract the first equation from the second, to get:

2a + b = 5

3b = 3

Again assuming no discounts, if 3 bananas cost $3, then one banana would cost $1.

2a + b = 5

b = 1

Finally, we can remove bananas from the equation and divide by 2 to find that an apple cost $2 last month.

2a = 4

b = 1

a = 2

b = 1

Notice that the only things we did were listed above - we multiplied (or divided) by numbers, and then we added (or subtracted) two things.

OK. Now let's do the calculation for this month.

2a + b = 5

a + 2b = 7

2a + b = 5

2a + 4b = 14

2a + b = 5

3b = 9

2a + b = 5

b = 3

2a = 2

b = 3

a = 1

b = 3

We have found that the new prices are $1 for an apple, and $3 for a banana. But one thing to notice - we in fact did the exact same calculations as we did before, just with different numbers on the right. This will be important later.

Hopefully I have already convinced you that we already use linear equations commonly. Next up, the process of solving generally, and some notation.

So the last thread, I accidentally turned what was supposed to be a "school" on linear algebra into an exploration of linear algebra through the lens of category theory. Which was useful to me, but probably less useful to the people who were watching the thread. This thread is for those people.

So! Linear algebra.

What is linear algebra? Linear algebra is the study of what you can do when you are restricted to two basic operations - adding things together, and multiplying everything by a number.

Seems restrictive, right? Well, that's a lot of how math works - by restricting what you think you can do, you can make more situations similar - and so you can study things much more deeply.

The thing is, we actually haven't restricted ourselves all too much. Or rather, we've restricted ourselves to a pair of very useful things we can do. Don't believe me?

Every day, Alice buys two apples and one banana for breakfast. Every day, Bob buys one apple and two bananas for breakfast. Last month, Alice paid $5, while Bob paid $4. This month, Alice paid $5, while Bob paid $7.

How did the prices change?

Let the price of an apple be a dollars, and the price of a banana be b dollars. We then get the equations for last month:

2a + b = 5

a + 2b = 4

So let's solve. Double the second equation - multiply all the numbers in it by 2. That still gives us a valid equation. The cost of 2 apples and 4 bananas would be 8 dollars.

2a + b = 5

2a + 4b = 8

Now, if it cost 5 dollars to get 2 apples and a banana, and 8 dollars to get 2 apples and 4 bananas, it stands to reason (assuming no discounts) that the difference in cost comes from the difference in goods. So we can subtract the first equation from the second, to get:

2a + b = 5

3b = 3

Again assuming no discounts, if 3 bananas cost $3, then one banana would cost $1.

2a + b = 5

b = 1

Finally, we can remove bananas from the equation and divide by 2 to find that an apple cost $2 last month.

2a = 4

b = 1

a = 2

b = 1

Notice that the only things we did were listed above - we multiplied (or divided) by numbers, and then we added (or subtracted) two things.

OK. Now let's do the calculation for this month.

2a + b = 5

a + 2b = 7

2a + b = 5

2a + 4b = 14

2a + b = 5

3b = 9

2a + b = 5

b = 3

2a = 2

b = 3

a = 1

b = 3

We have found that the new prices are $1 for an apple, and $3 for a banana. But one thing to notice - we in fact did the exact same calculations as we did before, just with different numbers on the right. This will be important later.

Hopefully I have already convinced you that we already use linear equations commonly. Next up, the process of solving generally, and some notation.