Is there any proof? Because I don't like to blindly follow any mathematical formula without any proof.
#Solving a system of three linear equations using matrices.
tardy eagle
rich nest
I will try to find a proof when I get home, but I am thinking of finding the intersection between the three planes you get from the system.
I am unsure on whether this is the most efficient way but it is intuitive for me to view things geometrically
gilded valley
Is there any proof? Because I don't like to blindly follow any mathematical form...
Okay, so. You accept that if you add, subtract, multiply, or divide the same thing on both sides of a true equation, the result is a true equation, right?
That is, if a = b, then a + c = b + c, etc. for the other operations?
wintry wasp
continue im invested rn
rich nest
I will try to find a proof when I get home, but I am thinking of finding the int...
Ok so plotting 3D planes on a graph turns out to be quite tedious and I have no skills with the required tooling, I give up for now
By the way, I think you are talking about Cramer's method of solving a of three linear equation, but for the sake of elminating ambiguity, could you clarify?
tardy eagle
Let's say you have three equations,
$\a_1x+b_1y+c_1z=l \
a_2x+b_2y+c_2z=m\
a_3x+b_3y+c_3z=o\$
desert lanternBOT
ExplorerAC™
tardy eagle
In matrix for it can be written as, $ \
\begin{bmatrix}
a_1 & b_1 & c_1 \
a_2 & b_2 & c_2 \
a_3 & b_3 & c_3 \
\end{bmatrix}
\begin{bmatrix}
x \
y \
z \
\end{bmatrix}
\begin{bmatrix}
a_1 x + b_1 y + c_1 z \
a_2 x + b_2 y + c_2 z \
a_3 x + b_3 y + c_3 z \
\end{bmatrix}$
desert lanternBOT
ExplorerAC™
tardy eagle
Which is $AX=D$
desert lanternBOT
ExplorerAC™
tardy eagle
Now multiply $A^{-1}$ both sides, you will get ur answer
desert lanternBOT
ExplorerAC™
tardy eagle
Okay, so. You accept that if you add, subtract, multiply, or divide the same thi...
but sir one thing I didn't understand is why does A^-1 = adj(A)/|A|
gilded valley
but sir one thing I didn't understand is why does ```A^-1 = adj(A)/|A|```
...what?
tardy eagle
if I want to find an inverse of a matrix
tardy eagle
**ExplorerAC™**
@gilded valley if I want to solve this
to find inverse of matrix
sorry if its a dumb question
gilded valley
<@352737152154468352> if I want to solve this
You're not even OP.
rich nest
Which is $AX=D$
if you're asking how to solve this
here is the proof
sry im still figuring this latex thing out
stray sedge
but sir one thing I didn't understand is why does ```A^-1 = adj(A)/|A|```
(For the cofactor notation)
tardy eagle
Lmaoo
tardy eagle
Tysm
civic narwhal
Does every result need to have a proof behind it or sometimes it could be just pattern recognition?
Like in this case adj(A)A = Det(A)I
We could prove it by taking LHS and RHS
But is there any derivation for this too?
stray sedge
We could prove it by taking LHS and RHS
wdym
Like the proof that adj(A)A=det(A)I is literally computing adj(A)A, then observing the diagonal entries are det(A), and the rest are 0, hence it's det(A)I
civic narwhal
Like the proof that adj(A)A=det(A)I is literally computing adj(A)A, then observi...
Sorry idk how to correctly say this but yeah is there any other proof other than to solve and observe?
stray sedge
Maybe
civic narwhal
I'm not good with mathematical terminologies but this is a formula right?
civic narwhal
So there should be a derivation too right?
stray sedge
yes, the proof
proof and derivation in this context are synonymous
(unless you mean something else which isn't being articulated)
civic narwhal
(unless you mean something else which isn't being articulated)
Yeah sorry my articulation is bad. Basically you might prove that each matrix follows this property by putting one into it and observing that it indeed is correct. But how'd one proof it any other way? Idk if I'm correctly saying this or not. Like how could someone come to this conclusion if they are not given a Matrix with numbers as elements inside but less say elements like a ij
Maybe I made it more confusing
stray sedge
If you computed adj(A)A for a single matrix, and saw that it was det(A)I, that's not a proof
civic narwhal
That's called observation right?
stray sedge
yes
you've noticed that that particular matrix A
has that property
you've not said anything about it holding for all A
now if you did for like, 10/20/30 matrices and kept getting the same result, you'd likely suspect it'll hold for all A (or in your testing, you just never hit a matrix that it didn't hold for)
civic narwhal
Yeah so you're recognising patterns there
And to prove that it is right you can plug any matrix into it the answer would be correct
But is there any other way to derive this formula/come to conclusion that this works other than by observation?
Actually nevermind even if there is one maybe I won't understand it cuz it must be pretty advanced and that would be just a waste of time