#ret
Suppose A = \(\begin{pmatrix} 1 & 3\\ 2 & -1 \end{pmatrix}\) and B = \(\begin{pmatrix} 0 & -1\\ 2 & 1 \end{pmatrix}\). Multiply \(AB\) and \(BA\). What do you notice???
Nothing. Matrix multiplication is not commutative.
Use matrices to solve the system: \(\begin{aligned}2x-y=3\\x+3y=5\end{aligned}\)
\[
\begin{aligned} &&\left[\begin{matrix}2&-1\\1&3\end{matrix}\right] \left[\begin{matrix}x\\y\end{matrix}\right]&= &\left[\begin{matrix}3\\5\end{matrix}\right] \\ &\left[\begin{matrix}3&0\\0&1\end{matrix}\right] &\left[\begin{matrix}2&-1\\1&3\end{matrix}\right] \left[\begin{matrix}x\\y\end{matrix}\right]&= &\left[\begin{matrix}3&0\\0&1\end{matrix}\right] \left[\begin{matrix}3\\5\end{matrix}\right] \\ \left[\begin{matrix}1&1\\0&1\end{matrix}\right] &\left[\begin{matrix}3&0\\0&1\end{matrix}\right] &\left[\begin{matrix}2&-1\\1&3\end{matrix}\right] \left[\begin{matrix}x\\y\end{matrix}\right]&= \left[\begin{matrix}1&1\\0&1\end{matrix}\right] &\left[\begin{matrix}3&0\\0&1\end{matrix}\right] \left[\begin{matrix}3\\5\end{matrix}\right] \\ \left[\begin{matrix}1&1\\0&1\end{matrix}\right] &&\left[\begin{matrix}6&-3\\1&3\end{matrix}\right] \left[\begin{matrix}x\\y\end{matrix}\right]&= \left[\begin{matrix}1&1\\0&1\end{matrix}\right] &\left[\begin{matrix}9\\5\end{matrix}\right] \\ &&\left[\begin{matrix}7&0\\1&3\end{matrix}\right] \left[\begin{matrix}x\\y\end{matrix}\right]&= &\left[\begin{matrix}14\\5\end{matrix}\right] \\ &&\left[\begin{matrix}7x\\x+3y\end{matrix}\right]&= &\left[\begin{matrix}14\\5\end{matrix}\right] \\ &\left[\begin{matrix}\frac{1}{7}&0\\0&1\end{matrix}\right] &\left[\begin{matrix}7x\\x+3y\end{matrix}\right]&= &\left[\begin{matrix}\frac{1}{7}&0\\0&1\end{matrix}\right] \left[\begin{matrix}14\\5\end{matrix}\right] \\ &&\left[\begin{matrix}x\\x+3y\end{matrix}\right]&= &\left[\begin{matrix}2\\5\end{matrix}\right] \end{aligned}\] \[
\begin{aligned} x = 2 \\ x + 3 y = 5 \end{aligned}\] I'm not sure how to solve the rest of it with matrices, so I'll just do it normally: \[
\begin{aligned} x &= 2\\ x + 3y &= 5\\ 2 + 3y &= 5\\ 3y &= 3\\ y &= 1\\ \end{aligned}\] 3. > Do 2x2 matrices form a group under > a. addition? > b. multiplication?
See KBe2020math530refGroups I'll assume that our matrices have real numbers in them.
| Operation Property | Closed | Identity | Inverse | Associative? | Final |
|---|---|---|---|---|---|
| Addition | Yes | \(e=\left[\begin{matrix}0&0\\0&0\end{matrix}\right]\) | \(\left[\begin{matrix}a&b\\c&d\end{matrix}\right] + \left[\begin{matrix}-a&-b\\-c&-d\end{matrix}\right]=e\) | "Inherits from addition" | Yes |
| Multiplication | Yes | \(e=\left[\begin{matrix}1&0\\0&1\end{matrix}\right]\) | Maybe? | Yes, see below | Undecided |
Associativity of 2x2 matrices under multiplication: \[
\begin{aligned} \left(\left[\begin{matrix}a&b\\c&d\end{matrix}\right] \left[\begin{matrix}e&f\\g&h\end{matrix}\right]\right) \left[\begin{matrix}i&j\\k&l\end{matrix}\right] = \left[\begin{matrix}ae+bg&af+bh \\ ce+dg&cf+dh \end{matrix}\right] \left[\begin{matrix}i&j\\k&l\end{matrix}\right] \\= \left[\begin{matrix}aei+bgi+afk+bhk&aej+bgj+afl+bhl\\cei+dgi+cfk+dhk&cej+dgj+cfl+dhl\end{matrix}\right] \\= \left[\begin{matrix}a&b\\c&d\end{matrix}\right] \left[\begin{matrix}ei+fk&ej+fl\\gi+hk&gj+hl\end{matrix}\right] = \left[\begin{matrix}a&b\\c&d\end{matrix}\right] \left(\left[\begin{matrix}e&f\\g&h\end{matrix}\right] \left[\begin{matrix}i&j\\k&l\end{matrix}\right]\right) \end{aligned}\]
I can't figure out if 2x2 matrices have multiplicative inverses… I tried to work it out using algebra but kept proving identities. Not sure what the right way to go about this is…
I spent far too long on this assignment (1.6h), so I probably won't spend as much time LaTeXing my homework in the future.