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I want to row reduce the following matrix into an echelon form: \begin{pmatrix} -(-\beta+\alpha)^{1/2} & 0 & 1 & 0\\ 0& -(-\beta+\alpha)^{1/2}& 0 & 1\\ -\beta & \alpha & -(-\beta+\alpha)^{1/2}&0 \\ \alpha & -\beta & 0 & -(-\beta+\alpha)^{1/2} \end{pmatrix}

where $\beta = k(1/M + 1/m)$, $\alpha = k/m$

I have no idea. I have been sitting here in the library for two hours doing calculations like $-\beta + \alpha = -k/M$, trying to add rows together, having the $-(-\beta+\alpha)^{1/2} = -i\sqrt{k/M}$, letting $-(-\beta+\alpha)^{1/2} = \gamma$ and I just don't know anymore

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    $\begingroup$ What do you mean by solving the matrix? Do you want to find the determinant of the matrix? $\endgroup$ – Paul Jun 4 '14 at 2:47
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    $\begingroup$ What does "solve" mean? There is no equation here...row-reduce? Find the determinant? Find eigenvalues? $\endgroup$ – afedder Jun 4 '14 at 2:50
  • $\begingroup$ Echo Paul and afedder. And it should be $-(-\beta + \alpha)^{1/2} = -i\sqrt{k/M}$. $\endgroup$ – M. Vinay Jun 4 '14 at 2:59
  • $\begingroup$ Sorry guys long night $\endgroup$ – Don Larynx Jun 4 '14 at 13:56
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Call the rows $(1)$ through $(4)$. I will explicitly tell you the steps to produce row-reduced echelon form, assuming $\alpha \neq \beta$ and $\alpha \neq 0.$ $$$$ Step 1: Add $-\beta(\alpha - \beta)^{-1/2} \times (1)$ to $(3)$.

Step 2: Add $\alpha(\alpha - \beta)^{-1/2} \times (1)$ to $(4)$.

Step 3: Subtract $(2)$ from $(3)$.

Step 4: Add $-\beta \times (2)$ to $(4)$.

Step 5: Divide $(4)$ by $\alpha$.

Step 6: Subtract $(3)$ from $(4)$.

Step 7: Subtract $(3)$ from $(1)$.

Step 8: Divide $(2)$ by $-(\alpha - \beta)^{1/2}$.

Step 9: Divide $(1)$ by $-(\alpha - \beta)^{1/2}$.

This will produce the following matrix: $$\begin{pmatrix} 1 & 0 & 0 & -(\alpha - \beta)^{-\frac{1}{2}} \\ 0 & 1 & 0 & -(\alpha - \beta)^{-\frac{1}{2}} \\ 0 & 0 & 1 & -1 \\ 0 & 0 & 0 & 0 \end{pmatrix}$$

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  • $\begingroup$ How did you see that this was the solution? $\endgroup$ – Don Larynx Jun 4 '14 at 16:31
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    $\begingroup$ @DonLarynx The steps are clear, right? You just eliminate one entry at a time... $\endgroup$ – rschwieb Jun 4 '14 at 16:35
  • $\begingroup$ Just worked it out...honestly first thing I did was try to get rid of $\alpha$ and $-\beta$ in $(3)$ and $(4)$. It wasn't too difficult once I did this. @DonLarynx $\endgroup$ – afedder Jun 4 '14 at 16:36
  • $\begingroup$ I tried that yesterday, maybe I just suck at math. I am doing the steps right now, will report back if I have a question. $\endgroup$ – Don Larynx Jun 4 '14 at 16:39
  • $\begingroup$ My suggestion is to look at what you can eliminate one row at a time until you have something manageable, where you can see the rest of the row-reduction easily. @DonLarynx $\endgroup$ – afedder Jun 4 '14 at 16:40
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Set $\gamma=-(-\beta+\alpha)^{1/2}$; then your matrix becomes $$ \begin{pmatrix} \gamma & 0 & 1 & 0\\ 0& \gamma & 0 & 1\\ -\beta & \alpha & \gamma &0 \\ \alpha & -\beta & 0 & \gamma \end{pmatrix} $$ Now, multiply the first row by $\gamma^{-1}$: $$ \begin{pmatrix} 1 & 0 & \gamma^{-1} & 0\\ 0& \gamma & 0 & 1\\ -\beta & \alpha & \gamma &0 \\ \alpha & -\beta & 0 & \gamma \end{pmatrix} $$ Add to the third row the first multiplied by $\beta$; then add to the fourth row the first multiplied by $-\alpha$: $$ \begin{pmatrix} 1 & 0 & \gamma^{-1} & 0\\ 0& \gamma & 0 & 1\\ 0 & \alpha & \gamma+\beta\gamma^{-1} &0 \\ 0 & -\beta & -\alpha\gamma^{-1} & \gamma \end{pmatrix} $$ Multiply the second row by $\gamma^{-1}$: $$ \begin{pmatrix} 1 & 0 & \gamma^{-1} & 0\\ 0 & 1 & 0 & \gamma^{-1}\\ 0 & \alpha & \gamma+\beta\gamma^{-1} &0 \\ 0 & -\beta & -\alpha\gamma^{-1} & \gamma \end{pmatrix} $$ Add to the third row the second row multiplied by $-\alpha$; add to the fourth row the second row multiplied by $\beta$: $$ \begin{pmatrix} 1 & 0 & \gamma^{-1} & 0\\ 0 & 1 & 0 & \gamma^{-1}\\ 0 & 0 & \gamma+\beta\gamma^{-1} &-\alpha\gamma^{-1} \\ 0 & 0 & -\alpha\gamma^{-1} & \gamma+\beta\gamma^{-1} \end{pmatrix} $$ Now observe that $$ \gamma+\beta\gamma^{-1}=\frac{\gamma^2+\beta}{\gamma}=\alpha\gamma^{-1} $$ Thus the matrix you got is $$ \begin{pmatrix} 1 & 0 & \gamma^{-1} & 0\\ 0 & 1 & 0 & \gamma^{-1}\\ 0 & 0 & \alpha\gamma^{-1} &-\alpha\gamma^{-1} \\ 0 & 0 & -\alpha\gamma^{-1} & \alpha\gamma^{-1} \end{pmatrix} $$ Multiply the third row by $\alpha^{-1}\gamma$: $$ \begin{pmatrix} 1 & 0 & \gamma^{-1} & 0\\ 0 & 1 & 0 & \gamma^{-1}\\ 0 & 0 & 1 & -1 \\ 0 & 0 & -\alpha\gamma^{-1} & \alpha\gamma^{-1} \end{pmatrix} $$ Finally, add to the fourth row the third row multiplied by $\alpha\gamma^{-1}$: $$ \begin{pmatrix} 1 & 0 & \gamma^{-1} & 0\\ 0 & 1 & 0 & \gamma^{-1}\\ 0 & 0 & 1 & -1 \\ 0 & 0 & 0 & 0 \end{pmatrix} $$

The reduced row echelon form is obtained now by adding to the first row the third row multiplied by $-\gamma^{-1}$: $$ \begin{pmatrix} 1 & 0 & 0 & \gamma^{-1}\\ 0 & 1 & 0 & \gamma^{-1}\\ 0 & 0 & 1 & -1 \\ 0 & 0 & 0 & 0 \end{pmatrix} $$

Of course, this assumes that $\gamma\ne0$ (that is $\alpha\ne\beta$) and $\alpha\ne0$. Depending on your assumptions, these cases should be examined separately.

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Maybe this will help you:

$$\left[\begin{array}{cccc} 1 & & & \\ 0 & 1 & & \\ 0 & \sqrt{\alpha-\beta} & -\sqrt{\alpha-\beta} & \\ 0 & 0 & 0 & -\frac{\beta}{\sqrt{\alpha-\beta}} \end{array}\right]\left[\begin{array}{cccc} 1\\ 0 & 1\\ 0 & 0 & \frac{\beta}{\alpha}\\ 0 & 0 & -\frac{\beta}{\alpha} & \frac{\alpha}{\beta} \end{array}\right]\left[\begin{array}{cccc} 1\\ 0 & 1\\ 1 & 0 & -\frac{\sqrt{\alpha-\beta}}{\beta}\\ 1 & 0 & 0 & \frac{\sqrt{\alpha-\beta}}{\alpha} \end{array}\right]\\\cdot\left[\begin{array}{cccc} -\sqrt{-\beta+\alpha} & 0 & 1 & 0\\ 0 & -\sqrt{-\beta+\alpha} & 0 & 1\\ -\beta & \alpha & -\sqrt{-\beta+\alpha} & 0\\ \alpha & -\beta & 0 & -\sqrt{-\beta+\alpha} \end{array}\right]=$$

$$=-\sqrt{\alpha-\beta}\left[\begin{array}{cccc} 1 & 0 & -\frac{1}{\sqrt{\alpha-\beta}} & 0\\ 0 & 1 & 0 & -\frac{1}{\sqrt{\alpha-\beta}}\\ 0 & 0 & 1 & -1\\ 0 & 0 & 1 & -1 \end{array}\right]$$

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  • $\begingroup$ I dont quite follow this $\endgroup$ – Don Larynx Jun 4 '14 at 17:55
  • $\begingroup$ The end result is a row reduced matrix. The pre-multiply matrices represent the steps needed to get there. So if you pre-multiply the right hand side (which is missing from the post) with the same matrices you will end up with a row reduced system of equation. $\endgroup$ – John Alexiou Jun 4 '14 at 18:46

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