Q:

Row reducing this matrix

I have this matrix:
$$\begin{bmatrix}\sqrt3 -\frac{1}{2}(\sqrt3 + i)&0&-1\\0&-1-\frac{1}{2}(\sqrt3 + i)&0\\1&0&-\frac{1}{2}(\sqrt3 + i)\end{bmatrix}$$
I've tried the following row operation:
$$R_3\rightarrow R_3 - \frac{\sqrt3+i}{2}R_1$$
Leading to:
$$\begin{bmatrix}\sqrt3 -\frac{1}{2}(\sqrt3 + i)&0&-1\\0&-1-\frac{1}{2}(\sqrt3 + i)&0\\0&0&0\end{bmatrix}$$
What constant can be used to reduce every leading coefficient to 1?

A:

The $(1,1)$ entry is
$$
\frac{\sqrt{3}}{2}-\frac{1}{2}i
$$
whose inverse is
$$
\frac{\sqrt{3}}{2}+\frac{1}{2}i
$$
If you multiply the first row by $\frac{\sqrt{3}}{2}+\frac{1}{2}i$, you get
$$
\begin{bmatrix}
1 & 0 & -\frac{\sqrt{3}}{2}-\frac{1}{2}i \\
0 & -1-\frac{1}{2}(\sqrt3 + i) & 0 \\
1 & 0 &-\frac{1}{2}(\sqrt3 + i)
\end{bmatrix}
$$
Subtract the first row from the third and multiply the second row by the inverse of the $(2,2)$ entry to find the row reduced echelon form
$$
\begin{bmatrix}
1 & 0 & -\frac{\sqrt{3}}{2}-\frac{1}{2}i \\
0 & 1 & 0 \\
0 & 0 & 0
\end{bmatrix}
$$

