Recent content by fuzzyorama

\left( \begin{array}{cc} 2 & 2 \\ 3 & 1 \end{array} \right)\left( \begin{array}{c} 2 \\ -3 \end{array} \right) = -\left( \begin{array}{c} 2 \\ -3 \end{array} \right) 2(2) + 2(-3) = -2 3(2) + 1(-3) = 3 I GOT IT RIGHT! YOU GUYS ARE THE AWESOMEST! THANKS :!) :!) :!)

I got it right? Understood! So it can also be \left( {\begin{array}{*{20}c} 4 \\ { - 6} \\\end{array}}\right) ?

Thanks for pointing that out:smile: Now I will redo everything. A = \left( \begin{array}{cc} 2 & 2 \\ 3 & 1 \end{array} \right) \left\vert \begin{array}{cc} 2 - \lambda & 2 \\ 3 & 1 - \lambda \end{array} \right\vert = 0 (2 - \lambda )(1 - \lambda) - (3)(2) = 0 2 - 2\lambda - \lambda +...

Oh wait! \left( \begin{array}{cc} 2 & 2 \\ 3 & 1 \end{array} \right)\left(\begin{array}{c}1 \\ 3\end{array}\right) = 4\left(\begin{array}{c}1 \\ 3\end{array}\right) \left(\begin{array}{c}8 \\ 6\end{array}\right) = \left(\begin{array}{c} 4 \\ 12 \end{array}\right) But it doesn't...

Ahh.. so I don't actually need to multiply the eigenvector with the eigenvalue to see if they are equal?

OMG I got a typo. a22 should be -3. Apologies. \left( \begin{array}{cc} -2 & 2 \\ 3 & -3 \end{array} \right)\left( \begin{array}{c} x \\ y \end{array} \right) = 4\left( \begin{array}{c} x \\ y \end{array} \right) From what I understand, by that you mean from the equation \left(...

I thought that just by replacing the y in \left( \begin{array}{cc} -2 & 2 \\ 3 & -2 \end{array} \right)\left( \begin{array}{c} x \\ y \end{array} \right) = 4\left( \begin{array}{c} x \\ y \end{array} \right) with 3x would give me the eigenvector. I then multiplied it with the eigenvalue...

Eigenvalues & Eigenvectors !SOLVED! Homework Statement Find the eigenvalues and eigenvectors of matrix A = \left( \begin{array}{cc} 2 & 2 \\ 3 & 1 \end{array} \right) Homework Equations Ax = \lambda x The Attempt at a Solution Solving \left\vert \begin{array}{cc} 2 - \lambda &...

Oh no I failed at the maths again. How about this? x = -\frac{3Mg}{7k} I can latex!

Heh no problem yo :cool: With spring A taken into account, I can write it as Mg + kAxA = -kBCx ... or can't I? Because kA = k, and kBC = 4k/3, Mg + kx = -4kx/3 (skipped) x = -\frac{3kMg}{7} I'm getting confused with the + - signs :confused:

Err, I'm sorry.. could you be more specific? If the effective spring constant for springs B and C is keff = (\frac{1}{k_{B}} + \frac{1}{k_{C}})^{-1} and kB = 2k, kC = 4k, then keff = (\frac{1}{2k} + \frac{1}{4k})^{-1} = (\frac{2k+4k}{2k*4k})^{-1} = \frac{8k^{2}}{6k} so eventually...

Thank you. Oh?! OK so kBC = (2k*4k)/(2k+4k) = 4k/3 mg = -kBCx mg = -(4k/3)*x and x = -(3mg/4k) Is this correct? I kinda have a pretty good feeling about this:) If it is, I'm guessing the force on the table's surface is F = -kBCx = -(4k/3)*(-3mg/4k) = mg ?

Homework Statement Three springs, A (spring constant k), B (spring constant 2k) and C (spring constant 4k) are serially joined and are vertically hanged from a ceiling with spring A at the top, spring B in the middle & spring C at the bottom. All three have the same relaxed length of L. The...