Skip to main content

[Community Question] Linear-algebra: 3x3 matrix operations intuition help

One of our user asked:

So I understand the intuition of taking the determinant of a 2x2 matrix, but what is the intuition for taking the determinant of 3x3, matrix? It makes zero intuitive sense just looking at it. Also, when finding the inverse of a matrix, why do we need to find the cofactor matrix and the adjugate matrix and transpose the cofactor matrix, and what is the point of the checkerboard matrix with + and - signs?


Comments

Popular posts from this blog

[Community Question] Calculus: Manifold with boundary - finding the boundary

One of our user asked: I have the manifold with boundary $M:= \lbrace (x_1,x_2,x_3) \in \mathbb R^3 : x_1\geq 0, x_1^2+x_2^2+x_3^2=1\rbrace \cup\lbrace (x_1,x_2,x_3) \in \mathbb R^3 : x_1= 0, x_1^2+x_2^2+x_3^2\leq1\rbrace$ and I need to find the boundary of this manifold. I think it is $\lbrace (x_1,x_2,x_3) \in \mathbb R^n : x_1= 0, x_2^2+x_3^2=1\rbrace$ , the other option is that the boundary is the empty set? I think the first is right? Am I wrong?

[Community Question] Linear-algebra: Are linear transformations between infinite dimensional vector spaces always differentiable?

One of our user asked: In class we saw that every linear transformation is differentiable (since there's always a linear approximation for them) and we also saw that a differentiable function must be continuous, so it must be true that all linear operators are continuous, however, I just read that between infinite dimensional vector spaces this is not necessarily true. I would like to know where's the flaw in my reasoning (I suspect that linear transformations between infinite dimensional vector spaces are not always differentiable).

Order of elements of the Prüfer groups $\mathbb{Z}(p^{\infty})$

Let $\mathbb{Z}(p^{\infty})$ be defined by $\mathbb{Z}(p^{\infty}) = \{ \overline{a/b} \in \mathbb{Q}/ \mathbb{Z} / a,b \in \mathbb{Z}, b=p^i$ $ with$ $ i \in \mathbb{N} \}$ , I wish show that any element in $\mathbb{Z}(p^{\infty})$ has order $p^n$ with $n \in \mathbb{N}$ . i try several ways but I have not been successful, some help ?? thank you