Find the row echelon form of a 4x4 matrix 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
 A: 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}$$
A: Set $\gamma=-(-\beta+\alpha)^{1/2}$; then your matrix becomes
$$
\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}
$$
Of course, this assumes that $\gamma\ne0$ (that is $\alpha\ne\beta$) and $\alpha\ne0$. Depending on your assumptions, these cases should be examined separately.
A: Maybe this will help you:
$$\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]$$
