# Is there any easy way to calculate the value of this determinant?

$$\begin{vmatrix} 0 & 3 & 1 & 2 & 10! & e^{-7}\\ 1 & 2 & -1 & 2 & \sqrt{2} & 2 \\ -1 & -2 & 3 & -3 & 1 & -\frac{1}{5} \\ -2 & -1 & 3 & 2 & -2 & -9 \\ 0 & 0 & 0 & 0 & 4 & 2 \\ 0 & 0 & 0 & 0 & 1 & 1 \\ \end{vmatrix}$$

I still can't see any easy way to compute the determinant above I would appreciate any kind of help. Thanks.

• A tip to compute determinants is to pick the row with the most zeros. But the problem is still tedious. I would recommend giving it to Mathematica. Jun 30, 2020 at 13:26
• If you expand the last two rows it is $2$ times the determinant of the upper left $4 \times 4$. You can replace the upper four rows in the last two columns with $0$s to save typing. I don't see an easy way to simplify the upper left $4 \times 4$. Jun 30, 2020 at 13:40
• As a quick observation, it is clear that the absurd values like $10!$ and $e^{-7}$ and so on in the top-right corner have absolutely no relevance to the determinant whatsover. This can be seen from the definition and noting that the only patterns with a nonzero product will be those which utilize the $4,1$ in the bottom right or the $2,1$ in the bottom right. Beyond that, you can save a bit of time with what you know about block matrices and their determinants and treat this like a block matrix thanks to the large block of zeroes in the bottom left. Jun 30, 2020 at 13:40
• Also, you can add multiples of a column to another (same with rows) and the determinant won't change. You can use this in order to get a more sparse matrix. Jun 30, 2020 at 13:41
• Those entries $10!, e^{-7},\sqrt{2},-1/5$ are red herrings. As @maxbp noted, they do not effect the result. Jun 30, 2020 at 14:02

• $$R_5 \leftarrow R_5 - 4 R_6$$
• $$R_3 \leftarrow R_3 + R_2$$
• $$R_4 \leftarrow R_4 + 2 R_2$$
• $$R_4 \leftarrow R_4 - R_1$$
Then flip $$R_6$$ and $$R_5$$, $$R_1$$ and $$R_2$$ (each operation multiplies the determinant by $$-1$$, so it remains unchanged), and you get a triangular matrix with diagonal $$(1, 3, 2, 4, 1, -2)$$, so you have a determinant of $$- 48$$.