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$\Delta=\begin{vmatrix} \sqrt 6& 2i& 3+\sqrt 6 \\ \sqrt{12}&\sqrt 3 +\sqrt8i &3\sqrt 2 +\sqrt 6i \\ \sqrt{18} &\sqrt 2+ \sqrt {12}i &\sqrt {27}+2i \end{vmatrix}$

taking $\sqrt 6$ out from the first column and performing the following operations

$$R_2\rightarrow R_2 -\sqrt 2 R_1$$

$$R_3 \rightarrow R_3-\sqrt 3 R_1$$

$$\sqrt 6\begin {vmatrix} 1&2i&3+\sqrt 6 \\\ 0&\sqrt 3&\sqrt 6i-2\sqrt 3 \\\ 0&\sqrt 2&2i-3\sqrt 2 \end {vmatrix}$$

What should I do next?

Aditya
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3 Answers3

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From here

$$\sqrt 6\begin {vmatrix} 1&2i&3+\sqrt 6 \\\ 0&\sqrt 3&\sqrt 6i-2\sqrt 3 \\\ 0&\sqrt 2&2i-3\sqrt 2 \end {vmatrix}$$

Pull out a factor of $\sqrt{3}$ from the second row and a factor of $\sqrt{2}$ from the third to get

$$6\begin {vmatrix} 1&2i&3+\sqrt 6 \\\ 0& 1 &\sqrt 2i-2 \\\ 0& 1 & \sqrt{2}i-3 \end {vmatrix}$$

Now you can compute it easily by expanding along the first row or by subtracting the second row from the third. Lets subtract.

$$6\begin {vmatrix} 1&2i&3+\sqrt 6 \\\ 0& 1 &\sqrt 2i-2 \\\ 0& 1 & \sqrt{2}i-3 \end {vmatrix} = 6\begin {vmatrix} 1&2i&3+\sqrt 6 \\\ 0& 1 &\sqrt 2i-2 \\\ 0& 0 & -1 \end {vmatrix}$$

The determinant is just the product of the diagonal entries, so $6 \cdot 1 \cdot 1 \cdot -1 = -6$.

SescoMath
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  • What do you mean by ‘determinant is just the product of diagonal entries’ (Sorry, I am new to this) – Aditya May 18 '20 at 17:16
  • The determinant of a triangular matrix is just the product of the entries along the diagonal. See how we reduced the problem by pulling out factor and row reducing until we reached an upper triangular matrix, this makes finding the determinant very easy. – SescoMath May 18 '20 at 17:18
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The obvious way is to simply find the determinant right away. But the fact that you have asked this must mean that must not be allowed. So to prove this without finding the determinant at all:

Take a factor of $\sqrt3$ from $R_2$ and $\sqrt2$ from $R_3$.

$$\Delta=6\begin{vmatrix} 1 & 2i & 3+\sqrt6 \\ 0 & 1 & \sqrt2 i-2\\ 0 &1 & \sqrt2 i-3 \end{vmatrix}$$

Now perform $R_2 \to R_2-R_3$

$$=6\begin{vmatrix} 1 & 2i & 3+\sqrt6\\ 0 &0 & 1\\ 0&1& \sqrt2 i -3 \end{vmatrix}$$

Now $C_2 \to C_2- 2iC_1$, $C_3 \to C_3 -(3+\sqrt6)C_1$

$$=6\begin{vmatrix} 1 &0 & 0 \\ 0 & 0 & 1 \\ 0 & 1 & \sqrt2 i-3 \end{vmatrix}$$

Finally : $C_3 \to C_3 - (\sqrt2 i - 3)C_2 $

$$\Delta = 6\begin{vmatrix} 1 & 0 & 0 \\ 0 & 0& 1 \\ 0&1&0 \end{vmatrix}$$

Now this determinant will obviously be a integral value (as it will be a linear combination of product of integers)

NOTE: I am trying to get rid of the non-integral terms using the 1 0 0 columns.

Saket Gurjar
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Expand along the first column. Thanks to the two zeroes there, you will only have $\sqrt{6}\cdot1\cdot(2\times2\text{ determinant})$. And you can calculate that small determinate directly.

zipirovich
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