Recent content by Michael_0039

Yes, my apologies.

I am very sorry, I was careless on this post. Correction: -x + 4y - 2z = 1 -2x + 10y + (2λ - 4)z = 6 3x - 11y + (λ2+6)z = 5λ - 1

Yes, my mistake I mean "∈" and also for the 2nd equation I forgot the z, you are right. The final augnebted matrix it came after calculations between rows, as shown bellow: 1) R2 → R2 - 2R1 2) R3 → R3 + 3R1 3) R3 → R3 - R2

Thread moved from technical math section, so there is no homework template. (∀λ∃ℝ) -x + y - z = 1 -2x + 10y + (2λ + 6) = 6 3x + 11y + (λ2+6)z = 5λ - 1 after gaussian elimination I have this: -1 4 -2 | 1 0 1 λ | 2 0 0 λ(λ-1) | 5λ So, for λ=0 ⇒ ∞ solutions, for λ=1...

So for the 2nd question: Is A reversible or/and diagonal ? The answer if A-1 exists is λ1*λ2≠0 and is non-diagonal because λ2 has reached one time but has 2 eigenvectors (?)

So the answer will be: λ1=6 with u1=[ 1 2 1 ]T λ2=3 with u2=[ 1 -1 1 ]T and u3=[ 2 -1 0 ]T (?)

Matrix A is not given, only those equations.

Assume a table A(3x3) with the following: A [ 1 2 1 ]^T = 6 [ 1 2 1 ]^T A [ 1 -1 1 ]^T = 3 [ 1 -1 1 ]^T A [ 2 -1 0]^T = 3 [ 1 -1 1]^T Find the Eigenvalues and eigenvectors: I have in mind to start with the Av=λv or det(A-λI)v=0.... Also, the first 2 equations seems to have the form Av=λv...

Hi, This is my try. Are you in agreement with this ? Thanks.

Hi, this is my try: I would appreciate any confirmation or correction . -- Thanks

You are right, I forgot (10+(10*I2))/12

Hi! I wonder if it is correct. Can someone confirm (maybe with a software) ? I(L) = I2 I(C) = I2 - I3 -- Thanks.

Hi ! This is my try: Is that correct ? Thanks

So integrate from 0 to π: V=(π^2)/8

Oops my mistake, it is: 0 ≤ x ≤ π I have to fix it. Thanks