Row reducing this matrix

Question

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?

Answer 1

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 $$ \begin{bmatrix} 1 & 0 & -\frac{\sqrt{3}}{2}-\frac{1}{2}i \\ 0 & 1 & 0 \\ 0 & 0 & 0 \end{bmatrix} $$

Answer 2

It can't be reduced further by row operations since we have only one entry for each column.

We can multiply the rows by some constant but this wouldn't simplify more than that.