Recent content by Victor Feitosa

Homework Statement Exercise 4.15 on the image. The exercise asks for the value of the current I using superposition Homework Equations Superposition theorem, nodal and mesh analysis The Attempt at a Solution I think i am doing the mesh analysis wrong, but can't see where. Can someone...

Just got it! Finally. I'll upload my answer and mark topic as solve

Homework Statement The problemas statement is: "Find Vo and Io by the superposition principle" Homework Equations Nodal Analysis and The Superposition Theorem. The Attempt at a Solution So, I tried to shutdown first the independent 30V tension font. It led me to Vo = 8V. Then, turning off...

Thank yu both, guys! It took long to understand but thanks to you all I managed to grasp it. And I think it'll stick now!

But the answer says that the statement T(x,y) = (x, 3x+y) for all x,y∈ℝ? is True. Also, [T]b.(x,y)b return the answer in Basis B, not?

Thanks everyone who helped! I think i figured how to do it and will try soon and post my result!

I did what you said, but I get that T(x,y) = (-x+y, 3x-2y) How I convert it to standard basis again? I tried to multiply T(x,y).B , but my result is (4x -3y, 3x-2y) I don't know what I am doing wrong..

Sorry, I just want to know if the following statement is true: True or False: T(x,y) = (x, 3x+y) for all x,y∈ℝ?

Homework Statement Being T: ℝ2 → ℝ2 the linear operator which matrix in relation to basis B = {(-1, 1), (0, 1)} IS: [T]b = \begin{bmatrix} 1 & 0\\ -3 & 1 \end{bmatrix} True or False: T(x,y) = (x, 3x+y) for all x,y∈ℝ? Homework EquationsThe Attempt at a Solution 3 [/B] So first I convert (x,y)...

Ok, so this is my matrix equation in original post: # \begin{pmatrix} -1 & D+a \\ D+2 & -1 \end{pmatrix} \quad \begin{pmatrix} v1(t)\\ v2(t) \end{pmatrix} \quad = \begin{pmatrix} es(t)\\ 0 \end{pmatrix}

Homework Statement The Matricial Circuit of a certain circuit in time-domain is: [ -1 D+a] [v1(t)] = [es(s)] [D+2 -1] [v2(t)] [0 ] where a is constant >0. Find G1(s) = V1(s)/Es(s). Put your result in a relation of polinomials N(s)/D(s), with the D(s) monic polinomial...

Oooooh, so logic and true! Thank you. So, a matrix is singular if it's determinant is zero hence it have no inverse, is it right?

I see why this is false! My matrix A is a echeloned matrix with trace 0. So it's det is 0 and it's not invertible. Sorry for not posting how i found this. I will try to edit and post my anwser! Thanks PF.

Hello, my friend! Thanks for your feedback. I thought that A is non-singular because if it was, A.x = x' could not be true, because it would have infinite solutions. But thinking right now i don't see how this is true at all. No, it's not P5, it's P8. I thought it was a generic polynomial...

Homework Statement True or false? If T: ℙ8(ℝ) → ℙ8(ℝ) is defined by T(p) = p', so exists a basis of ℙ8(ℝ) such that the matrix of T in relation to this basis is inversible. Homework EquationsThe Attempt at a Solution So i think that my equations is of the form: A.x = x' hence A is...